1. OPTIMIZATION
Alex Teshon
Daffy Durairaj

2. 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
What is Optimization Anyways?

3. Equations You Should Know: Volume
BOX:
LWH H
CONE:
⅓πr²h W h r
CYLINDER: πr²h r h
SPHERE:
¾πr³ r

4. Equations You Should Know Area and Surface Area
Circle:
πr²
Cylinder without top:
πr²+2πrh w
Rectangle:
2l+2w l
Cylinder with a closed top:
2πr²+2πrh

5. Steps of Optimization
BEFORE YOU DO ANYTHING MAKE SURE YOU READ THE PROBLEM CAREFULLY!!!!
1.DRAW THE PICTURE
2.LABEL THE PICTURE WITH YOUR X’S AND Y’S
3. WRITE OUT ANY EQUATIONS YOU WILL NEED
3. SOLVE FOR EITHER Y OR X
5. AFTER SOLVING FOR YOUR VARIABLE, PLUG THE SOLVED VARIABLE BACK INTO THE ORIGINAL EQUATION

6. Even More Steps 6. TAKE THE DERIVATIVE OF THAT EQUATION
7. SET THIS DERIVATIVE EQUAL TO ZERO
8. SOLVE FOR EITHER Y OR X
9. TAKE THE SECOND DERIVATIVE TO DETERMINE IF THIS IS THE MAX OR MIN
10. 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

7. 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?

8. DRAW THE PICTURE!!! x y y
BARN

9. Use the needed equations to solve for either x or y
We know that the perimeter has to be 360 feet so we can plug this into the perimeter equation
P= 360
360= 2y+x
Since we are dealing with area and perimeter we only need 2 equations:
A= xy
P= 2y+x
In this case, we will solve for x
360= 2y+x
X= 360-2y

10. 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
A= xy
X= 360-2y
A=(360-2y)y
A= 360y-2y²
Now that we have our equation, take the derivative
A= 360y-2y²
A’=360-2y

11. SET THE DERIVATIVE EQUAL TO ZERO
A’= 360-4y 4(90-y) = 0 y = 90 ft Now that we have our y value, we can plug this back into the original equation to get our answer!
360= 2y+x
Y= 90
360= 2(90)+x
So to get the maximum are of the fence, y should be 90 ft and x should be 180 ft.
360=180+x
X= 180

12. Take the derivative again
Once 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 max
If A’’>0 then the value is a relative min
-4<0
Therefore this is a relative max

13. 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 ?

14. Let’s See How We Did!
DRAW IT OUT!!!
SA: X²+4XY
V: X²Y y x x

15. SA= x²+4xy
SA= 48
48= x²+4xy
4xy= 48-x²
Y= 48-x²/4xy
Y= 12/x – ¼x
V= x²y
V= x²(12/x – ¼x)
V= 12x²/x – ¼ x³
V= 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

16. 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 ?

17. Let’s See How We Did!
DRAW IT OUT!!! r
SA: πr²+2πrh
V: πr²h h

18. SA= πr²+2πrh
SA= 3π
3π= πr²+2πrh
2πrh= 3π- πr²
h= 3π- πr²/2πr
3/2r- ½r
V= πr²h
V= 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= r≠ 0,-1
h= 3/2r- ½r
h= 3/2(1) – ½(1)
3/2 – 1/2
h= 1
V= π(1)²(1)
V= π ft³

19. 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?

20. Let’s See How We Did!
DRAW IT OUT!!! 8
8-2x 5
5-2x x x x
5-2x
8-2x

21. V=lwh
V= 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, 10/3
X≠ 10/3
X= 1
5-2(1)= 3
8-2(1)= 6
The cut outs will be 1 in on each side and the lengths of the box will be 6x3x1 6 1 1 3 1 3 6

22. 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

23. Solution

24. 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?

25. Solution

26. 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?

27. Solution

28. 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.

29. Getting Ready For The AP Exam
1979 AB3 BC3
Find 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.

30. Let’s See How You Did

31. CONGRATULATIONS!!! You can now efficiently use optimization to find the maximum or minimum amount of stuff can fit in something!!!!

32. Works Cited
plicationsofthederivative/problems_8.html
neDIRECTORY/maxmindirectory/MaxMin.html