© Alex TeshonDaffy Durairaj

OPTIMIZATION

What is Optimization Anyways?

The term optimization means to optimize something, or use something at its best. This refers, both in real life and in Calculus, to the maximum or minimum value of something.

When you optimize, you try to find the maximum or minimum value required for the given problem

Equations You Should Know:Volume

BOX:LWH

L

H

W

CYLINDER: πr²h

r

h

CONE:⅓πr²h

r

h

SPHERE:¾πr³

r

Equations You Should KnowArea and Surface Area

Surface Area

Area

Cylinder without top:πr²+2πrh

Cylinder with a closed top:2πr²+2πrh

Circle: πr²

Rectangle: 2l+2w

w

l

Steps of Optimization

BEFORE YOU DO ANYTHING MAKE SURE YOU READ THE PROBLEM CAREFULLY!!!!

1.DRAW THE PICTURE2.LABEL THE PICTURE WITH YOUR X’S AND

Y’S3. WRITE OUT ANY EQUATIONS YOU WILL

NEED3. SOLVE FOR EITHER Y OR X5. AFTER SOLVING FOR YOUR VARIABLE, PLUG

THE SOLVED VARIABLE BACK INTO THE ORIGINAL EQUATION

Even More Steps

6. TAKE THE DERIVATIVE OF THAT EQUATION7. SET THIS DERIVATIVE EQUAL TO ZERO8. SOLVE FOR EITHER Y OR X9. TAKE THE SECOND DERIVATIVE TO DETERMINE

IF THIS IS THE MAX OR MIN10. PLUG THE SOLVED FOR VARIABLE BACK INTO

THE EQUATION TO SOLVE FOR THE OTHER VARIABLE

11.PLUG BOTH THE SOLVED VARIABLES BACK INTO THE EQUATION TO GET THE FINAL ANSWER

Lets Try This Out!A shepherd wishes to build a rectangular

fenced area against the side of a barn. He has 360 feet of fencing material, and only needs to use it on three sides of the enclosure, since the wall of the barn will provide the last side. What dimensions should the shepherd choose to maximize the area of the enclosure?

DRAW THE PICTURE!!!

x

y y

BARN

Use the needed equations to solve for either x or y

Since we are dealing with area and perimeter we only need 2 equations:A= xy

P= 2y+x

We know that the perimeter has to be 360 feet so we can plug this into the perimeter equation

P= 360360= 2y+x

In this case, we will solve for x

360= 2y+xX= 360-2y

Plug it back in and get the derivative

Now that we have solved for x, we can plug this into the other equation to solve for our other variable

X= 360-2yA=(360-2y)y

A= xy

A= 360y-2y²

Now that we have our equation, take the derivative

A= 360y-2y²

A’=360-2y

SET THE DERIVATIVE EQUAL TO ZERO

A’= 360-4y4(90-y) = 0y = 90 ft

Now that we have our y value, we can plug this back into the original equation to get

our answer!

360= 2y+xY= 90360= 2(90)+x

360=180+xX= 180

So to get the maximum are of the fence, y should be 90 ft and x should be 180 ft.

Take the derivative againOnce you find your value to plug in, take

the second derivative to find if what you have is a relative maximum or minimum

A’= 360-4y A’’= -4

if A’’<0 then the value is a relative maxIf A’’>0 then the value is a relative min

-4<0Therefore this is a

relative max

PROBLEM #1

An open rectangular box with square base is to be made from 48 ft.2 of material. What dimensions will result in a box with the largest possible volume ?

Let’s See How We Did!

DRAW IT OUT!!!

x

y

x

SA: X²+4XYV: X²Y

SA= x²+4xySA= 4848= x²+4xy

4xy= 48-x²Y= 48-x²/4xy

Y= 12/x – ¼x

V= x²y

V= x²(12/x – ¼x)

V= 12x²/x – ¼ x³V= 12x- ¼x³

V’= 12- ¼ x² ¾(16-x²) ¾(x-4)(x+4)

X= 4 x≠0, -4

Y= 12/(4) – (4)/4 3-1 Y= 2

V= x²y V= (4)²(2)

V= 32 ft³

PROBLEM #2

A container in the shape of a right circular cylinder with no top has surface area 3 ft.2 What height h and base radius r will maximize the volume of the cylinder ?

Let’s See How We Did!

DRAW IT OUT!!!

r

h SA: πr²+2πrhV: πr²h

SA= πr²+2πrhSA= 3π3π= πr²+2πrh

2πrh= 3π- πr²h= 3π- πr²/2πr 3/2r- ½r

V= πr²hV= πr²(3r/2 – ½r)

V= 3πr²/2r - πr³/2 3πr/2- πr³/2

V’= 3π/2 - 3πr²/2

3π/2(1-r²) 3π/2(1-r)(1+r)

r= 1 r≠ 0,-1

h= 3/2r- ½rh= 3/2(1) – ½(1) 3/2 – 1/2 h= 1

V= π(1)²(1)

V= π ft³

PROBLEM #3

A piece of sheet metal is rectangular, 5ft wide and 8ft long. Congruent squares are to be cut from its corners. The resulting piece of metal is to be folded and welded to form an open top box. How should this be done to get a box of largest possible volume?

Let’s See How We Did!

DRAW IT OUT!!!

x

x

x

8-2x

5-2x

8

5

8-2x

5-2x

V=lwhV= x(8-2x)(5-2x) (8x-2x²)(5-2x) 4x³-26x²-40x

V’=12x²-52x-40 4(3x²-13x-10) 4(3x-10)(x-1)

X= 1, 10/3X≠ 10/3

X= 1

5-2(1)= 38-2(1)= 6

The cut outs will be 1 in on each side and the lengths of the box will be 6x3x1

1

6

13

6

1

3

NOW IT’S YOUR TURN! Find the maximum volume of a right

cylinder that can be inscribed in a cone of altitude 12 inches and a base radius 4 inches if the axes of the cylinder and cone coincide

SOLUTION

HERE, TRY ANOTHER ONE A rectangular plot of land containing

216 square meters is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. What dimensions of the outer rectangle require the smallest total length for the two fences?

SOLUTION

3RD TIMES THE CHARM A printed page has 1 inch margins at

the top and .75 on the sides. If the area of the printed paper is to be 48 square inches, what should the dimensions be?

SOLUTION

If You Want Even More Practice TRY THESE PROBLEMS!

A window is in the shape of a rectangle surmounted by a semicircle. Find the dimensions when the perimeter is 24 meters and the area is as large as possible

A container with a rectangular base, rectangular sides, and no top is to have a volume of 2 cubic meters. The width of the base is to be 1 meter. When cut to size, material costs $10 per square meter for the base and $5 per square meter for the sides. What is the cost of the least expensive container?

A circular cylindrical container, open at the top and having a capacity of 24π cubic inches, is to be manufactured. If the cost of the material used for the bottom of the container is three times that used for the curved part and there is no waste of material, find the dimensions which will minimize the cost.

Getting Ready For The AP Exam

1979 AB3 BC3Find the maximum volume of a box

that can be made by cutting out squares from the corners of an 8 inch by 15 inch rectangular sheet of cardboard and folding up the sides.

Let’s See How You Did

CONGRATULATIONS!!!You can now efficiently use

optimization to find the maximum or minimum

amount of stuff can fit in something!!!!

http://www.sparknotes.com/math/calcab/applicationsofthederivative/problems_8.html

http://www.qcalculus.com/cal08.htm http://www.math.ucdavis.edu/~kouba/CalcO

neDIRECTORY/maxmindirectory/MaxMin.html

Works Cited