1. Applications of Differentiation
Optimization
Applications of Differentiation

2. ?
What is optimization?
Some of the most important applications of differential calculus are optimization problems, in which we are required to find the optimal (best) way of doing something. Many practical problems require us to minimize a cost or maximize an area or somehow find the best possible outcome of a situation.

3. SA = 2πr² + 2πrh (closed top)
Quick Review of
Helpful Formulas
Cone: V = ⅓πr³
A = πr²
C = 2πr
V = πr²h
SA = 2πr² + 2πrh (closed top)
SA = πr² + 2πrh (open top)
Sphere:
V = 4/3πr³

4. Steps in Solving Optimization Problems
Read the problem.
Draw and label a picture with the given information.
Write 1 equation for every variable.
Solve for either x or y in the equation containing a numerical value.
Substitute x or y into the other equation.
Take the derivative.
Set the derivative equal to 0, and solve for either x or y.
Take the second derivative to determine whether the answer is a minimum or maximum value (< 0 = MAX, > 0 = MIN)
Using the original equation, solve for the remaining variable.

5. Ex. 1 READ THE PROBLEM: WRITE 1 EQUATION FOR EVERY VARIABLE: x = $4
A rectangular field is to be fenced off along the bank of a river, and no fence is required along the river. The material for the fence costs $4 per running foot for the two ends and $6 per running foot for the side parallel to the river. Find the dimensions of the field of largest possible area that can be enclosed with $1,800 worth of fence.
x = $4
WRITE 1 EQUATION FOR
EVERY VARIABLE:
y = $6
C = 2(4x) + 6y
1,800 = 8x + 6y
A = xy x

6. Ex. 1 SOLVE FOR EITHER X OR Y IN THE EQUATION CONTAINING
A NUMERICAL VALUE :
1,800 = 8x + 6y
900 = 4x + 3y
900 – 3y = 4x
225 – ¾y = x
x = $4
SUBSTITUTE EITHER X OR Y INTO THE OTHER EQUATION:
A = xy
A = (225 – ¾y)y
A = 225y – ¾y²
y = $6 x

7. Ex. 1 TAKE THE DERIVATIVE, SET EQUAL TO
ZERO, AND SOLVE FOR EITHER X OR Y:
A = 225y – ¾y²
A’ = 225 – (3/2)y
0 = 225 – (3/2)y
225 = (3/2)y
150 = y
x = $4
TAKE THE 2ND DERIVATIVE TO DETERMINE WHETHER THE ANSWER IS A MIN OR MAX:
In this case, the answer is a maximum value because we were asked to find the dimensions of the field of LARGEST possible area.
y = $6
A’’ = -3/2
-3/2 < 0
MAX x

8. Ex. 1 USING THE ORIGINAL EQUATION, SOLVE FOR THE REMAINING VARIABLE:
y = 150
x = 225 – (3/4)y
x = 225 – (3/4)(150)
x = 225/2 = 112.5
x = $4
FINAL ANSWER
(DIMENSIONS OF THE FIELD):
x = ft
y = 150 ft
y = $6 x

9. WRITE 1 EQUATION FOR EVERY VARIABLE:
Ex. 2 x
16 in. x
READ THE PROBLEM:
An open box with a rectangular base is to be constructed from a rectangular piece of cardboard 16 inches wide and 21 inches long by cutting out a square from each corner and then bending up the sides. Find the size of the corner square which will produce a box having the largest possible volume.
23 in.
WRITE 1 EQUATION FOR EVERY VARIABLE:
V = lwh
V = x (21-2x) (16-2x)
V = 336x – 74x² + 4x³

10. TAKE THE DERIVATIVE, SET EQUAL TO ZERO, AND SOLVE FOR EITHER X OR Y:
Ex. 2 x
16 in. x
TAKE THE DERIVATIVE, SET EQUAL TO ZERO, AND SOLVE FOR EITHER X OR Y:
V = 336x – 74x² + 4x³
V’ = 336 – 148x + 12x²
0 = 4 (3x² - 37x + 84)
0 = 4 (3x – 28) (x – 3)
x = 28/3, x = 3
23 in.
In this case, x cannot equal 28/3 because the length of one side of the paper is only 16 inches. 28/2 multiplied by 2 would be 56/3 or roughly 18.7 inches.

11. (SIZE OF CORNER SQUARE):
Ex. 2 x
16 in. x
2ND DERIVATIVE:
V’ = 336 – 148x + 12x²
V’’ = 24x – 148
V’’ < 0
MAX
23 in.
FINAL ANSWER
(SIZE OF CORNER SQUARE):
x² = 9 in.²

12. Ex. 3
READ THE PROBLEM:
An oil can is to be made in the form of a right
circular cylinder to contain K cubic centimeters.
What dimensions of the can will require the
least amount of material?
WRITE 1 EQUATION FOR EVERY VARIABLE, AND SOLVE FOR h IN THE EQUATION CONTAINING A NUMERICAL VALUE:
V = K cm³
K = πr²h
h = K / πr²
SA = 2πr² + 2πrh
Remember, K is a number!

13. Ex. 3 SUBSTITUTE h INTO THE OTHER EQUATION:
h = K / πr²
SA = 2πr² + 2πr(K / πr²)
SA = 2πr² + (2K / r)
TAKE THE DERIVATIVE, SET EQUAL TO ZERO, AND SOLVE FOR r:
SA’ = 4πr – (2K / r²)
4πr = 2K / r²
4πr³ = 2K
r³ = 2K / 4π
r = (K / 2π) 1/3
2ND DERIVATIVE:
SA’’ = 4π + (4K / r³)
SA’’ > 0
MIN
Remember, K is a number!

14. Ex. 3 USING THE ORIGINAL EQUATION, SOLVE FOR THE REMAINING VARIABLE:
r = (K / 2π) 1/3
h = K / πr²
h = K / π(K/2π)2/3 • (K / 2π) 1/3 / (K / 2π) 1/3
h = 2 (K / 2π) 1/3
FINAL ANSWER
(DIMENSIONS OF THE CAN):
r = (K / 2π) 1/3
h = 2 (K / 2π) 1/3
Remember, K is a number!

15. Now it’s time
to try some
on your own!

16. Try Me #1
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.

17. First, lets draw our picture…
Sol. 1
First, lets draw our picture… X Y 2X

18. A = 2x [6 – x – (1/2)πx] + (1/2)πx²
Sol. 1
P = 2x + 2y + πx
12 = 2x + 2y + πx
y = (12 – 2x – πx) / 2
y = 6 – x – (1/2)πx
A = 2xy + (1/2)πx²
A = 2x [6 – x – (1/2)πx] + (1/2)πx²
A = 12x – 2x² - πx² + (1/2)πx²
A = 12x – 2x² - (1/2)πx²
A’ = 12 – 4x – πx
12 = x (4 + π)
X = 12 / (4 + π) X Y 2X

19. (Dimensions of the Window):
Sol. 1
X = 12 / (4 + π)
A’ = 12 – 4x – πx
A’’ = -4 – π
A’’ < 0
MAX
y = 6 – x – (1/2)πx
y = 6 – [12 / (4 + π)] – (1/2)π[12 / (4 + π)]
y = 24 / (4 + π)
Final Answer
(Dimensions of the Window):
x = 12 / (4 + π) X Y 2X

20. Try Me #2
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 if there is no waste of material, find the dimensions which will minimize the cost.

21. First, lets draw our picture…
Sol. 2
First, lets draw our picture… r h

22. Sol. 2 r h V = 24π V = πr²h 24π = πr²h h = 24 / r²
SA = 3(πr²) + 1(2πrh)
SA = 3πr² + 2πr (24 / r²)
SA = 3πr² + (48π / r)
SA’ = (-48π / r²) + 6πr
6πr = 48π / r²
6πr³ = 48π
r = 2 r h

23. (Dimensions of the Can):
Sol. 2
r = 2
SA’ = (-48π / r²) + 6πr
SA’’ = (96π / r³) + 6π
SA’’ > 0
MIN
h = 24 / r²
h = (24 / 4) = 6
Final Answer
(Dimensions of the Can):
h = 6 r h

24. Try Me #3
A right circular cylinder is inscribed in a right circular cone so that the centerlines of the cylinder and the cone coincide. The cone has a height of 6 and a radius of base 3. Find the volume and the dimensions of the cylinder that has maximum volume.

25. First, lets draw our picture…
Sol. 3
First, lets draw our picture… 6 h 3

26. Sol. 3 6 / 3 = (6 – h) / r h = 6 – 2r V = πr²h V = πr² (6 – 2r )

27. Sol. 3 r = 2 V’ = 12πr – 6πr² V’’ = 12π – 12πr V’’ < 0 MAX
h = 6 – 2r
h = 6 – 2(2)
h = 2
V = π(2)²(2) = 8π units³ 6 r h 3

28. 1972 AB 4 BC 3
A man has 340 yards of fencing for enclosing two separate fields, one of which is to be a rectangle twice as long as it is wide and the other a square. The square field must contain at least 100 square yards and the rectangular one must contain at least 800 square yards.
If x is the width of the rectangular field, what are the maximum and minimum possible values of x?
What is the greatest number of square yards that can be enclosed in the two fields? Justify your answer.

29. 1972 AB 4 BC 3 2x
side length of square =
(340 – 6x) / 4 x

30. 1972 AB 4 BC 3 20 ≤ x ≤ 50 a. Total Fence = 340 yards 2x x 2x² ≥ 800
x ≥ 20 yards
[(340 – 6x) / 4]² ≥ 100
(340 – 6x) / 4 ≥ 10
340 – 6x ≥ 40
-6x ≥ -300
x ≤ 50 yards 2x
side length of square =
(340 – 6x) / 4 x
20 ≤ x ≤ 50

31. 1972 AB 4 BC 3
b. Let A be the total area of the rectangle and the square
A = 2x² + [(340 – 6x) / 4]²
A’ = (17/2)x – 255
0 = (17/2)x – 255
x = 30
A(20) = 3,825
A(30) = 3,400
A(50) = 5,100
Maximum Area = 5,100 yards²
when x = 50 2x
side length of square =
(340 – 6x) / 4 x

32. Works Cited http://etc.usf.edu/clipart/41600/41699/fc_cone_41699.htm
Calculus, James Stewart 5e
© Kelsey Byrd, 2011