Presentation on theme: "AP Calculus Review: Optimization"— Presentation transcript:
1AP Calculus Review: Optimization Created by Sarah Tuggle
2Basic Steps to Solve Optimization Problems Read the problemDraw a picture to help you visualize the situationWrite 1 equation for each variable. (You should have the same number of equations as variables)
3Basic Steps to Solve Optimization Problems 4. Take the equation with the number so you can solve for one of the variables5. Substitute the variable you found into one of the other equations6. Take the derivative and set the equation equal to 0
4Basic Steps to Solve Optimization Problems 7. Take the second derivative to check your answer (if it’s <0 max, >0 min)8. Solve for the other variables needed to answer the question
5Example Problem 1. Read the Problem Find the area of the largest rectangle with a perimeter of 200 feet.
62. Draw a picture that represents the scenario yx
73. Write an equation for each variable 3. Write an equation for each variable. There are two variables therefore there should be two equations.A= x+ y200= 2x+ 2y
84. Take the equation with a number in it and solve for one of the variables. 100= x+ yx= 100-y
95. Substitute that variable into the other equation and solve for the other variable. A= xy x= (100-y)A= (100-y)yA= 100y -y²
106. Take the derivative and set it equal to 0 to solve for the variable A′= y0= 100-2yy= 50
117. Take the second derivative to check your answer (<0 max, >0 min) A′′= -2 < 0 max
128. Solve for the other variables needed to answer the question 200= 2x+ 100x= 50A= 50(50)A= 2500 sq ft
13Try This!!!A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that it’s width is 4 meters and volume is 36 cubic meters. If building the tank cost $10 per sq meter for the base and $5 per sq meter for the sides, what is the cost of the least expensive tank?
15Write an equation for each variable 36= 4xyC= 10(4x)+ 5(8y)+ 5(2xy)C= 40x+ 40y+ 10xy
16Take the equation with the variable and solve for one of the variables 36= 4xy9= xyy= 9x
17Substitute the variable you solved for into the other equation C= 40x+ 40(9)+ 10x(9)x xC= 40xx
18Take the derivative and set it equal to 0 to solve for the variable C′= /x²0= /x²40= 360/x²40x²= 360x²= 9x= 3
19Take the second derivative (< 0 max, > 0 min) C′′= 720/x³ > 0 min
20Solve for the other variables needed to answer the question x= 336= 4(3)yy= 3C= 40(3)+ 40(3)+ 10(3)(3)cost= $330
21Here’s another problem to try! A closed box with a square base is to have a volume of 2000 cubic in. The material for the top and bottom of the box is to cost $3 per sq in, and the material for the sides is to cost $1.50 per sq in. If the cost of the material is to be the least, find the dimensions of the box.