OPTIMIZATION PROBLEMS

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.

GENERAL PROCEDURE FOR SOLVING OPTIMIZATION PROBLEMS

GENERAL PROCEDURE FOR SOLVING OPTIMIZATION PROBLEMS Understand the Problem: Read problem carefully until clearly understood.

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?

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.

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.

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.

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.

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.

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.

Let’s illustrate the procedure.

Understand the Problem: Read problem carefully until clearly understood. ASK…What’s the unknown? What quantities are given? What conditions are given?

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.

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.

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.

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.

FOCUS ON HOMEWORK

CONTINUE HERE NEXT TIME

What do these two rectangular solids (boxes) have in common? 40 10 10 10 5 5

What do these two rectangular solids (boxes) have in common? VOLUME = L x W x H 40 10 10 10 5 5

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

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?

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!

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 = 850 square feet of material to construct the one on the left.

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 = 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.

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 = 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.

Let’s illustrate the general problem-solving procedure with another example.

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.

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.

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 𝑐𝑚 3 5. 𝐴 is a function of 𝑟 & ℎ so we simplify by writing ℎ in terms of 𝑟.

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 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

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 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.

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 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 𝜋 𝑟 2 =2𝜋 𝑟 𝑟

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 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 𝑟

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 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 𝑟 ,𝑟>0

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 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 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 𝑟 ,𝑟>0

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 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 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 𝑟 ,𝑟>0

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 𝑟 −1 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴′=4𝜋𝑟− 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 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 𝑟 ,𝑟>0

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 𝑟 −1 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴′=4𝜋𝑟− 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴 ′′ =4𝜋 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 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 𝑟 ,𝑟>0

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 𝑟 −1 critical # Set 𝐴′=0 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴′=4𝜋𝑟− 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴 ′′ =4𝜋 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 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 𝑟 ,𝑟>0

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 𝑟 −1 critical # Set 𝐴′=0 𝐴′=4𝜋𝑟−2000 𝑟 −2 4𝜋𝑟− 𝑟 2 =0 𝐴′=4𝜋𝑟− 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴 ′′ =4𝜋 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 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 𝑟 ,𝑟>0

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 𝑟 −1 critical # Set 𝐴′=0 𝐴′=4𝜋𝑟−2000 𝑟 −2 4𝜋𝑟= 𝑟 2 𝐴′=4𝜋𝑟− 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝐴 ′′ =4𝜋 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 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 𝑟 ,𝑟>0

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 𝑟 −1 critical # Set 𝐴′=0 𝐴′=4𝜋𝑟−2000 𝑟 −2 4𝜋𝑟= 𝑟 2 𝐴′=4𝜋𝑟− 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 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 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 𝑟 ,𝑟>0

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 𝑟 −1 critical # Set 𝐴′=0 𝐴′=4𝜋𝑟−2000 𝑟 −2 4𝜋𝑟= 𝑟 2 𝐴′=4𝜋𝑟− 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 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝑟 3 = 500 𝜋 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 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 𝑟 ,𝑟>0

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 𝑟 −1 critical # Set 𝐴′=0 𝐴′=4𝜋𝑟−2000 𝑟 −2 4𝜋𝑟= 𝑟 2 𝐴′=4𝜋𝑟− 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝑟 = 𝜋 𝐴 ′′ =4𝜋 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 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 𝑟 ,𝑟>0

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 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 4𝜋𝑟= 𝑟 2 𝐴′=4𝜋𝑟− 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝑟 = 𝜋 𝐴 ′′ =4𝜋 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 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 𝑟 ,𝑟>0

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 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴 ′′ =4𝜋 𝑟 3 4𝜋𝑟= 𝑟 2 𝐴′=4𝜋𝑟− 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 2. 𝑟 = 𝜋 𝐴 ′′ =4𝜋 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 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 𝑟 ,𝑟>0

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 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴 ′′ =4𝜋 𝑟 3 4𝜋𝑟= 𝑟 2 𝐴′=4𝜋𝑟− 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 𝐴 ′′ 𝜋 =+ 2. 𝑟 = 𝜋 𝐴 ′′ =4𝜋 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. The volume is 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 𝑟 ,𝑟>0

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 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴 ′′ =4𝜋 𝑟 3 4𝜋𝑟= 𝑟 2 𝐴′=4𝜋𝑟− 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 𝐴 ′′ 𝜋 =+ 2. 𝑟 = 𝜋 𝐴 ′′ =4𝜋 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. 𝐴 ′′ =+ means local MIN The volume is 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 𝑟 ,𝑟>0

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 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴 ′′ =4𝜋 𝑟 3 4𝜋𝑟= 𝑟 2 𝐴′=4𝜋𝑟− 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 𝐴 ′′ 𝜋 =+ 2. 𝑟 = 𝜋 𝐴 ′′ =4𝜋 𝑟 3 𝐴=2𝜋 𝑟 2 +2𝜋𝑟ℎ 𝐴=𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 ℎ=ℎ𝑒𝑖𝑔ℎ𝑡 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠 3. 𝐴 ′′ =+ means local MIN The volume is 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 𝑟 ,𝑟>0 So, 𝑟= 𝜋 is a local MIN.

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 𝑟 −1 critical # Set 𝐴′=0 Using 2nd Der Test 𝐴′=4𝜋𝑟−2000 𝑟 −2 𝐴 ′′ =4𝜋 𝑟 3 4𝜋𝑟= 𝑟 2 𝐴′=4𝜋𝑟− 𝑟 2 4. To ‘see’ the surface area, think of disassembling it into 2 circles and a rectangle. 𝐴 ′′ 𝜋 =+ 2. 𝑟 = 𝜋 𝐴 ′′ =4𝜋 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= 𝜋 also gives rise to an absolute minimum. The associated height ℎ is ℎ= 1000 𝜋 𝜋 2 So, 𝑟≈5.42 𝑐𝑚 , ℎ≈10.84 𝑐𝑚 𝐴 ′′ =+ means local MIN The volume is 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 𝑟 ,𝑟>0 So, 𝑟= 𝜋 is a local MIN.

This next theorem sums up what we just did when using the I/C chart just now.

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.

Find the point on the parabola y2 = 2x that is closest to the point (1, 4).

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

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

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 It is easier to minimize 𝑑 2 than 𝑑 so let’s do that instead.

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 It is easier to minimize 𝑑 2 than 𝑑 so let’s do that instead. =0

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 It is easier to minimize 𝑑 2 than 𝑑 so let’s do that instead. So, 𝑓 ′ 𝑦 =0 when 𝑦=2. =0

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 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).

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 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).

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 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).

OPTIMIZATION PROBLEMS

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