# AP Calculus Review: Optimization

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AP Calculus Review: Optimization
Created by Sarah Tuggle

Basic Steps to Solve Optimization Problems
Read the problem Draw a picture to help you visualize the situation Write 1 equation for each variable. (You should have the same number of equations as variables)

Basic Steps to Solve Optimization Problems
4. Take the equation with the number so you can solve for one of the variables 5. Substitute the variable you found into one of the other equations 6. Take the derivative and set the equation equal to 0

Basic 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

Example Problem 1. Read the Problem
Find the area of the largest rectangle with a perimeter of 200 feet.

2. Draw a picture that represents the scenario
y x

3. 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+ y 200= 2x+ 2y

4. Take the equation with a number in it and solve for one of the variables.
100= x+ y x= 100-y

5. Substitute that variable into the other equation and solve for the other variable.
A= xy x= (100-y) A= (100-y)y A= 100y -y²

6. Take the derivative and set it equal to 0 to solve for the variable
A′= y 0= 100-2y y= 50

7. Take the second derivative to check your answer (<0 max, >0 min)
A′′= -2 < 0 max

8. Solve for the other variables needed to answer the question
200= 2x+ 100 x= 50 A= 50(50) A= 2500 sq ft

Try 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?

Draw a picture x 4 y

Write an equation for each variable
36= 4xy C= 10(4x)+ 5(8y)+ 5(2xy) C= 40x+ 40y+ 10xy

Take the equation with the variable and solve for one of the variables
36= 4xy 9= xy y= 9 x

Substitute the variable you solved for into the other equation
C= 40x+ 40(9)+ 10x(9) x x C= 40x x

Take the derivative and set it equal to 0 to solve for the variable
C′= /x² 0= /x² 40= 360/x² 40x²= 360 x²= 9 x= 3

Take the second derivative (< 0 max, > 0 min)
C′′= 720/x³ > 0 min

Solve for the other variables needed to answer the question
x= 3 36= 4(3)y y= 3 C= 40(3)+ 40(3)+ 10(3)(3) cost= \$330

Here’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.

Draw a picture x x y

Write an equation for each variable
2000= x²y A= 3(2x²)+ 1.5(4xy) A= 6x²+ 6xy

Take the equation with the number and solve for one of the variables
2000= x²y y= 2000

Substitute the variable you solved for into the other equation
A= 6x²+ 6x(2000) A= 6x² x

Take the derivative and set it equal to 0 to solve for the variable
A′= 12x /x² 0= 12x /x² 12x³=12000 x³=1000 x=10

Take the second derivative (<0 max, >0 min)
A′′= >0 min

Solve for the other variables needed to answer the question
y= 2000/10² y= 20 Dimensions: 10in x 10in x 20in

Here’s something a little more difficult
A window is in the shape of a rectangle surmounted by a semicircle. Find the dimensions when the perimeter is 12 meters and the area is as large as possible.

12= 2y + x+ ½πx A=xy+ ⅛πx² 12- x- ½πx= 2y 6- ½x- ¼πx= y A=x(6- ½x- ¼πx)+ ⅛πx² A=6x- ½x²- ⅛πx² A′=6- x- ¼πx

0=6- x- ¼πx 6=x+ ¼πx 6=x(1+ ¼π) x= 24/(4+π) A′′-1- ¼π <0 max y=6- 12/(4+π)- 6π/(4+π) y= 12/(4+π)

Bibliography Youse, Bevan K., and F. Lane Hardy. Calculus with Analytical Geometry Reprint. New York: Holt, Rinehart and Winston, Print. © Sarah Tuggle 2011

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