OPTIMIZATION PROBLEMS 4 APPLICATIONS OF DIFFERENTIATION 4.5 OPTIMIZATION PROBLEMS
Optimization Problems In solving certain problems (maximizing areas/volumes/profits and minimizing distances/times/costs, etc.) the hardest part may be extracting the information from the problem needed to construct a function to be maximized or minimized.
Optimization Problems GENERAL PROCEDURE FOR SOLVING OPTIMIZATION PROBLEMS
Optimization Problems GENERAL PROCEDURE FOR SOLVING OPTIMIZATION PROBLEMS Understand the Problem: Read problem carefully until clearly understood.
Optimization Problems GENERAL PROCEDURE FOR SOLVING OPTIMIZATION PROBLEMS Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given?
Optimization Problems GENERAL PROCEDURE FOR SOLVING OPTIMIZATION PROBLEMS Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram.
Optimization Problems GENERAL PROCEDURE FOR SOLVING OPTIMIZATION PROBLEMS Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram. 3. Use Variables: Use a variable for the quantity (let’s say Q) to be optimized. Also select letters or symbols for other unknowns and label the diagram using these.
Optimization Problems GENERAL PROCEDURE FOR SOLVING OPTIMIZATION PROBLEMS Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram. 3. Use Variables: Use a variable for the quantity (let’s say Q) to be optimized. Also select letters or symbols for other unknowns and label the diagram using these.
Optimization Problems GENERAL PROCEDURE FOR SOLVING OPTIMIZATION PROBLEMS Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram. 3. Use Variables: Use a variable for the quantity (let’s say Q) to be optimized. Also select letters or symbols for other unknowns and label the diagram using these. 4. Write an equation: Express quantity Q in terms of letters/symbols in Step 3.
Optimization Problems GENERAL PROCEDURE FOR SOLVING OPTIMIZATION PROBLEMS Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram. 3. Use Variables: Use a variable for the quantity (let’s say Q) to be optimized. Also select letters or symbols for other unknowns and label the diagram using these. 4. Write an equation: Express quantity Q in terms of letters/symbols in Step 3. 5. Simplify, if needed: If Q is a function of more than one letter/symbol, then use information given to find equations relating them. Using these equations, eliminate all but one letter/symbol in the expression for Q.
Optimization Problems GENERAL PROCEDURE FOR SOLVING OPTIMIZATION PROBLEMS Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram. 3. Use Variables: Use a variable for the quantity (let’s say Q) to be optimized. Also select letters or symbols for other unknowns and label the diagram using these. 4. Write an equation: Express quantity Q in terms of letters/symbols in Step 3. 5. Simplify, if needed: If Q is a function of more than one letter/symbol, then use information given to find equations relating them. Using these equations, eliminate all but one letter/symbol in the expression for Q.
Optimization Problems GENERAL PROCEDURE FOR SOLVING OPTIMIZATION PROBLEMS Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram. 3. Use Variables: Use a variable for the quantity (let’s say Q) to be optimized. Also select letters or symbols for other unknowns and label the diagram using these. 4. Write an equation: Express quantity Q in terms of letters/symbols in Step 3. 5. Simplify, if needed: If Q is a function of more than one letter/symbol, then use information given to find equations relating them. Using these equations, eliminate all but one letter/symbol in the expression for Q. 6. Find Domain: We now have a function of the form 𝑄=𝑓 𝑥 . Find its domain.
Optimization Problems GENERAL PROCEDURE FOR SOLVING OPTIMIZATION PROBLEMS Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram. 3. Use Variables: Use a variable for the quantity (let’s say Q) to be optimized. Also select letters or symbols for other unknowns and label the diagram using these. 4. Write an equation: Express quantity Q in terms of letters/symbols in Step 3. 5. Simplify, if needed: If Q is a function of more than one letter/symbol, then use information given to find equations relating them. Using these equations, eliminate all but one letter/symbol in the expression for Q. 6. Find Domain: We now have a function of the form 𝑄=𝑓 𝑥 . Find its domain. 7. Find Extrema: Find the absolute max and/or min needed to solve problem.
Optimization Problems Let’s illustrate the procedure.
Optimization Problems
Optimization Problems Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given?
Optimization Problems Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram.
Optimization Problems Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram. 3. Use Variables: Use a variable for the quantity (let’s say Q) to be optimized. Also select letters or symbols for other unknowns and label the diagram using these.
Optimization Problems Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram. 3. Use Variables: Use a variable for the quantity (let’s say Q) to be optimized. Also select letters or symbols for other unknowns and label the diagram using these.
Optimization Problems Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram. 3. Use Variables: Use a variable for the quantity (let’s say Q) to be optimized. Also select letters or symbols for other unknowns and label the diagram using these. 4. Write an equation: Express quantity Q in terms of letters/symbols in Step 3.
Optimization Problems Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram. 3. Use Variables: Use a variable for the quantity (let’s say Q) to be optimized. Also select letters or symbols for other unknowns and label the diagram using these. 4. Write an equation: Express quantity Q in terms of letters/symbols in Step 3.
Optimization Problems Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram. 3. Use Variables: Use a variable for the quantity (let’s say Q) to be optimized. Also select letters or symbols for other unknowns and label the diagram using these. 4. Write an equation: Express quantity Q in terms of letters/symbols in Step 3.
Optimization Problems Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram. 3. Use Variables: Use a variable for the quantity (let’s say Q) to be optimized. Also select letters or symbols for other unknowns and label the diagram using these. 4. Write an equation: Express quantity Q in terms of letters/symbols in Step 3.
Optimization Problems Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given? 2. Draw a Diagram: Often it is useful to draw a diagram and to identify the given and required quantities on that diagram. 3. Use Variables: Use a variable for the quantity (let’s say Q) to be optimized. Also select letters or symbols for other unknowns and label the diagram using these. 4. Write an equation: Express quantity Q in terms of letters/symbols in Step 3.
Optimization Problems 5. Simplify, if needed: If Q is a function of more than one letter/symbol, then use information given to find equations relating them. Using these equations, eliminate all but one letter/symbol in the expression for Q.
Optimization Problems 5. Simplify, if needed: If Q is a function of more than one letter/symbol, then use information given to find equations relating them. Using these equations, eliminate all but one letter/symbol in the expression for Q.
Optimization Problems 6. Find Domain: We now have a function of the form 𝑄=𝑓 𝑥 . Find its domain.
Optimization Problems 6. Find Domain: We now have a function of the form 𝑄=𝑓 𝑥 . Find its domain.
Optimization Problems 6. Find Domain: We now have a function of the form 𝑄=𝑓 𝑥 . Find its domain. 7. Find Extrema: Find the absolute max and/or min needed to solve problem.
Optimization Problems 6. Find Domain: We now have a function of the form 𝑄=𝑓 𝑥 . Find its domain. 7. Find Extrema: Find the absolute max and/or min needed to solve problem.
Optimization Problems 6. Find Domain: We now have a function of the form 𝑄=𝑓 𝑥 . Find its domain. 7. Find Extrema: Find the absolute max and/or min needed to solve problem.
Optimization Problems 6. Find Domain: We now have a function of the form 𝑄=𝑓 𝑥 . Find its domain. 7. Find Extrema: Find the absolute max and/or min needed to solve problem.
Optimization Problems 6. Find Domain: We now have a function of the form 𝑄=𝑓 𝑥 . Find its domain. 7. Find Extrema: Find the absolute max and/or min needed to solve problem.
Optimization Problems 6. Find Domain: We now have a function of the form 𝑄=𝑓 𝑥 . Find its domain. 7. Find Extrema: Find the absolute max and/or min needed to solve problem.
Optimization Problems
Optimization Problems
Optimization Problems
Optimization Problems
Optimization Problems
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Optimization Problems CONTINUE HERE NEXT TIME
Optimization Problems What do these two rectangular solids (boxes) have in common? 40 10 10 10 5 5
Optimization Problems What do these two rectangular solids (boxes) have in common? VOLUME = L x W x H 40 10 10 10 5 5
Optimization Problems What do these two rectangular solids (boxes) have in common? VOLUME = L x W x H 40 10 10 10 5 5 V = 5 x 5 x 40 = 1000 V = 10 x 10 x 10 = 1000
Optimization Problems What do these two rectangular solids (boxes) have in common? VOLUME = L x W x H 40 10 10 10 5 5 V = 5 x 5 x 40 = 1000 V = 10 x 10 x 10 = 1000 Does this mean that it would take the same amount of material to construct each one?
Optimization Problems What do these two rectangular solids (boxes) have in common? VOLUME = L x W x H 40 10 10 10 5 5 V = 5 x 5 x 40 = 1000 V = 10 x 10 x 10 = 1000 Does this mean that it would take the same amount of material to construct each one? NO!
Optimization Problems What do these two rectangular solids (boxes) have in common? VOLUME = L x W x H 40 10 10 10 5 5 V = 5 x 5 x 40 = 1000 V = 10 x 10 x 10 = 1000 Does this mean that it would take the same amount of material to construct each one? NO! It would take four 40 x 5 pieces and two 5 x 5 pieces for a total of 800 + 50 = 850 square feet of material to construct the one on the left.
Optimization Problems What do these two rectangular solids (boxes) have in common? VOLUME = L x W x H 40 10 10 10 5 5 V = 5 x 5 x 40 = 1000 V = 10 x 10 x 10 = 1000 Does this mean that it would take the same amount of material to construct each one? NO! It would take four 40 x 5 pieces and two 5 x 5 pieces for a total of 800 + 50 = 850 square feet of material to construct the one on the left. It would take six 10 x 10 pieces for a total of 600 square feet of material to construct the one on the right.
Optimization Problems What do these two rectangular solids (boxes) have in common? VOLUME = L x W x H 40 Keeping what we have learned here in mind, let’s now move on to another problem. 10 10 10 5 5 V = 5 x 5 x 40 = 1000 V = 10 x 10 x 10 = 1000 Does this mean that it would take the same amount of material to construct each one? NO! It would take four 40 x 5 pieces and two 5 x 5 pieces for a total of 800 + 50 = 850 square feet of material to construct the one on the left. It would take six 10 x 10 pieces for a total of 600 square feet of material to construct the one on the right.
Optimization Problems Let’s illustrate the general problem-solving procedure with another example.
Optimization Problems Let’s illustrate the general problem-solving procedure with another example. A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 2.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 2.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 3. 2.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. Let 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 3. 2.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. Let 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. 2.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. Let 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. 2.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. Let 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. 2.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 2. 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 𝑉=1000=𝜋 𝑟 2 ℎ
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟 1000 𝜋 𝑟 2 =2𝜋 𝑟 2 + 2000 𝑟
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟 1000 𝜋 𝑟 2 =2𝜋 𝑟 2 + 2000 𝑟
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 𝐴=2𝜋 𝑟 2 + 2000 𝑟
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 𝐴=2𝜋 𝑟 2 + 2000 𝑟
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 𝐴′=4𝜋𝑟−2000 𝑟 −2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 𝐴′=4𝜋𝑟−2000 𝑟 −2 4𝜋𝑟− 2000 𝑟 2 =0 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 𝐴′=4𝜋𝑟−2000 𝑟 −2 4𝜋𝑟= 2000 𝑟 2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 𝐴′=4𝜋𝑟−2000 𝑟 −2 4𝜋𝑟= 2000 𝑟 2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 4𝜋 𝑟 3 =2000 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 𝐴′=4𝜋𝑟−2000 𝑟 −2 4𝜋𝑟= 2000 𝑟 2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 4𝜋 𝑟 3 =2000 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝑟 3 = 500 𝜋 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 𝐴′=4𝜋𝑟−2000 𝑟 −2 4𝜋𝑟= 2000 𝑟 2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝑟 = 3 500 𝜋 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 4𝜋𝑟= 2000 𝑟 2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝑟 = 3 500 𝜋 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴 ′′ =4𝜋+ 4000 𝑟 3 4𝜋𝑟= 2000 𝑟 2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝑟 = 3 500 𝜋 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴 ′′ =4𝜋+ 4000 𝑟 3 4𝜋𝑟= 2000 𝑟 2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 𝐴 ′′ 3 500 𝜋 =+ 2. 𝑟 = 3 500 𝜋 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴 ′′ =4𝜋+ 4000 𝑟 3 4𝜋𝑟= 2000 𝑟 2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 𝐴 ′′ 3 500 𝜋 =+ 2. 𝑟 = 3 500 𝜋 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. 𝐴 ′′ =+ means local MIN The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴 ′′ =4𝜋+ 4000 𝑟 3 4𝜋𝑟= 2000 𝑟 2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 𝐴 ′′ 3 500 𝜋 =+ 2. 𝑟 = 3 500 𝜋 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. 𝐴 ′′ =+ means local MIN The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0 So, 𝑟= 3 500 𝜋 is a local MIN.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴 ′′ =4𝜋+ 4000 𝑟 3 4𝜋𝑟= 2000 𝑟 2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 𝐴 ′′ 3 500 𝜋 =+ 2. 𝑟 = 3 500 𝜋 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. Using an I/D chart you can verify that 𝑟= 3 500 𝜋 also gives rise to an absolute minimum. The associated height ℎ is ℎ= 1000 𝜋 3 500 𝜋 2 So, 𝑟≈5.42 𝑐𝑚 , ℎ≈10.84 𝑐𝑚 𝐴 ′′ =+ means local MIN The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0 So, 𝑟= 3 500 𝜋 is a local MIN.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴 ′′ =4𝜋+ 4000 𝑟 3 4𝜋𝑟= 2000 𝑟 2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 𝐴 ′′ 3 500 𝜋 =+ 2. 𝑟 = 3 500 𝜋 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. Using an I/D chart you can verify that 𝑟= 3 500 𝜋 also gives rise to an absolute minimum. The associated height ℎ is ℎ= 1000 𝜋 3 500 𝜋 2 So, 𝑟≈5.42 𝑐𝑚 , ℎ≈10.84 𝑐𝑚 𝐴 ′′ =+ means local MIN The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0 So, 𝑟= 3 500 𝜋 is a local MIN.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴 ′′ =4𝜋+ 4000 𝑟 3 4𝜋𝑟= 2000 𝑟 2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 𝐴 ′′ 3 500 𝜋 =+ 2. 𝑟 = 3 500 𝜋 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. Using an I/D chart you can verify that 𝑟= 3 500 𝜋 also gives rise to an absolute minimum. The associated height ℎ is ℎ= 1000 𝜋 3 500 𝜋 2 So, 𝑟≈5.42 𝑐𝑚 , ℎ≈10.84 𝑐𝑚 𝐴 ′′ =+ means local MIN The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0 So, 𝑟= 3 500 𝜋 is a local MIN.
Optimization Problems 1. Understand the Problem 3. Use Variables 4. Write an equation 5. Simplify, if needed 6. Find Domain 7. Find Extrema 2. Draw a Diagram A cylindrical can is to be made to hold a liter (1000 𝑐𝑚 3 ) of oil. Find the dimensions to minimize the cost of the metal to manufacture the can. In simple terms, we must find the cylinder size that will MINIMIZE the cost of making the can …and material cost relates to the surface area of the cylinder. 7. 𝐴=2𝜋 𝑟 2 +2000 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴 ′′ =4𝜋+ 4000 𝑟 3 4𝜋𝑟= 2000 𝑟 2 𝐴′=4𝜋𝑟− 2000 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 𝐴 ′′ 3 500 𝜋 =+ 2. 𝑟 = 3 500 𝜋 𝐴 ′′ =4𝜋+ 4000 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. Using an I/D chart you can verify that 𝑟= 3 500 𝜋 also gives rise to an absolute minimum. The associated height ℎ is ℎ= 1000 𝜋 3 500 𝜋 2 So, 𝑟≈5.42 𝑐𝑚 , ℎ≈10.84 𝑐𝑚 𝐴 ′′ =+ means local MIN The volume is 1000 𝑐𝑚 3 𝑉=1000=𝜋 𝑟 2 ℎ ℎ= 1000 𝜋 𝑟 2 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟. 6. 𝐴=2𝜋 𝑟 2 + 2000 𝑟 ,𝑟>0 So, 𝑟= 3 500 𝜋 is a local MIN.
Optimization Problems This next theorem sums up what we just did when using the I/C chart just now.
Optimization Problems This next theorem sums up what we just did when using the I/C chart just now. FIRST DERIVATIVE TEST FOR ABSOLUTE EXTREME VALUES Suppose that 𝑐 is a critical number of a continuous function 𝑓 defined on an interval. If 𝑓 ′ 𝑥 >0 for all 𝑥<𝑐 and 𝑓 ′ 𝑥 <0 for all 𝑥>𝑐, then 𝑓(𝑐) is the absolute maximum value of 𝑓. (b) If 𝑓 ′ 𝑥 <0 for all 𝑥<𝑐 and 𝑓 ′ 𝑥 >0 for all 𝑥>𝑐, then 𝑓(𝑐) is the absolute minimum value of 𝑓.
Optimization Problems This next theorem sums up what we just did when using the I/C chart just now. FIRST DERIVATIVE TEST FOR ABSOLUTE EXTREME VALUES Suppose that 𝑐 is a critical number of a continuous function 𝑓 defined on an interval. If 𝑓 ′ 𝑥 >0 for all 𝑥<𝑐 and 𝑓 ′ 𝑥 <0 for all 𝑥>𝑐, then 𝑓(𝑐) is the absolute maximum value of 𝑓. (b) If 𝑓 ′ 𝑥 <0 for all 𝑥<𝑐 and 𝑓 ′ 𝑥 >0 for all 𝑥>𝑐, then 𝑓(𝑐) is the absolute minimum value of 𝑓.
Optimization Problems This next theorem sums up what we just did when using the I/C chart just now. FIRST DERIVATIVE TEST FOR ABSOLUTE EXTREME VALUES Suppose that 𝑐 is a critical number of a continuous function 𝑓 defined on an interval. If 𝑓 ′ 𝑥 >0 for all 𝑥<𝑐 and 𝑓 ′ 𝑥 <0 for all 𝑥>𝑐, then 𝑓(𝑐) is the absolute maximum value of 𝑓. (b) If 𝑓 ′ 𝑥 <0 for all 𝑥<𝑐 and 𝑓 ′ 𝑥 >0 for all 𝑥>𝑐, then 𝑓(𝑐) is the absolute minimum value of 𝑓.
Optimization Problems Find the point on the parabola y2 = 2x that is closest to the point (1, 4).
Optimization Problems Find the point on the parabola y2 = 2x that is closest to the point (1, 4).
Optimization Problems Find the point on the parabola y2 = 2x that is closest to the point (1, 4). The distance between the point (1, 4) and the point (x, y) is
Optimization Problems Find the point on the parabola y2 = 2x that is closest to the point (1, 4). The distance between the point (1, 4) and the point (x, y) is We know that 𝑦 2 =2𝑥 so 𝑥= 𝑦 2 2 .
Optimization Problems Find the point on the parabola y2 = 2x that is closest to the point (1, 4). The distance between the point (1, 4) and the point (x, y) is We know that 𝑦 2 =2𝑥 so 𝑥= 𝑦 2 2 .
Optimization Problems Find the point on the parabola y2 = 2x that is closest to the point (1, 4). The distance between the point (1, 4) and the point (x, y) is We know that 𝑦 2 =2𝑥 so 𝑥= 𝑦 2 2 . It is easier to minimize 𝑑 2 than 𝑑 so let’s do that instead.
Optimization Problems Find the point on the parabola y2 = 2x that is closest to the point (1, 4). The distance between the point (1, 4) and the point (x, y) is We know that 𝑦 2 =2𝑥 so 𝑥= 𝑦 2 2 . It is easier to minimize 𝑑 2 than 𝑑 so let’s do that instead.
Optimization Problems Find the point on the parabola y2 = 2x that is closest to the point (1, 4). The distance between the point (1, 4) and the point (x, y) is We know that 𝑦 2 =2𝑥 so 𝑥= 𝑦 2 2 . It is easier to minimize 𝑑 2 than 𝑑 so let’s do that instead.
Optimization Problems Find the point on the parabola y2 = 2x that is closest to the point (1, 4). The distance between the point (1, 4) and the point (x, y) is We know that 𝑦 2 =2𝑥 so 𝑥= 𝑦 2 2 . It is easier to minimize 𝑑 2 than 𝑑 so let’s do that instead.
Optimization Problems Find the point on the parabola y2 = 2x that is closest to the point (1, 4). The distance between the point (1, 4) and the point (x, y) is We know that 𝑦 2 =2𝑥 so 𝑥= 𝑦 2 2 . It is easier to minimize 𝑑 2 than 𝑑 so let’s do that instead. =0
Optimization Problems Find the point on the parabola y2 = 2x that is closest to the point (1, 4). The distance between the point (1, 4) and the point (x, y) is We know that 𝑦 2 =2𝑥 so 𝑥= 𝑦 2 2 . It is easier to minimize 𝑑 2 than 𝑑 so let’s do that instead. So, 𝑓 ′ 𝑦 =0 when 𝑦=2. =0
Optimization Problems Find the point on the parabola y2 = 2x that is closest to the point (1, 4). The distance between the point (1, 4) and the point (x, y) is We know that 𝑦 2 =2𝑥 so 𝑥= 𝑦 2 2 . It is easier to minimize 𝑑 2 than 𝑑 so let’s do that instead. So, 𝑓 ′ 𝑦 =0 when 𝑦=2. =0 It’s easy enough to use the 1st Derivative Test for Absolute Extrema to prove 𝑦=2 is a minimum but it is clear from the graph that there is no max distance from (1,4).
Optimization Problems Find the point on the parabola y2 = 2x that is closest to the point (1, 4). The distance between the point (1, 4) and the point (x, y) is If 𝑦=2, then x=2. We know that 𝑦 2 =2𝑥 so 𝑥= 𝑦 2 2 . It is easier to minimize 𝑑 2 than 𝑑 so let’s do that instead. So, 𝑓 ′ 𝑦 =0 when 𝑦=2. =0 It’s easy enough to use the 1st Derivative Test for Absolute Extrema to prove 𝑦=2 is a minimum but it is clear from the graph that there is no max distance from (1,4).
Optimization Problems Find the point on the parabola y2 = 2x that is closest to the point (1, 4). The distance between the point (1, 4) and the point (x, y) is If 𝑦=2, then x=2. We know that 𝑦 2 =2𝑥 so 𝑥= 𝑦 2 2 . It is easier to minimize 𝑑 2 than 𝑑 so let’s do that instead. So, 𝑓 ′ 𝑦 =0 when 𝑦=2. =0 It’s easy enough to use the 1st Derivative Test for Absolute Extrema to prove 𝑦=2 is a minimum but it is clear from the graph that there is no max distance from (1,4). So the point on 𝑦 2 =2𝑥 that is closest to the point (1,4) is (2,2).