# CALCULUS OPTIMIZATION PROBLEMS

To find the maximum and minimum turning points of y = f(x), we need to find x such that f'(x) = 0

Step 1 :

Draw a large, clear diagram for the situation.

Step 2 :

Construct the equation with the variable to be maximized or minimized as the subject of the formula in terms of the single variable x.

Step 3 :

Find the first derivative and equate to 0 to get the critical number.

Step 4 :

Apply the value that we have received as critical number in the second derivative.

• If f''(x) > 0 at x = a, then at x = a the function will reach its minimum
• If f''(x) < 0 at x = a, then at x = a the function will reach its maximum.

Step 5 :

To find the maximum or minimum value, we have to apply the value of x in the original function.

Problem 1 :

If a business employs x workers to manufacture furniture then the profit made by the business given by

P(x) = -3x3 + 6084x - 5000 dollars per week

a) How many employees should the business have to maximize the profit ?

b) What is the maximum profit ?

Solution :

P(x) = -3x3 + 6084x - 5000

P'(x) = -3(3x2) + 6084(1) - 0

P'(x) = -9x2 + 6084

P'(x) = 0

-9x2 + 6084 = 0

-9x2 = -6084

x2 = 6084/9

x2 = 676

x = 26

Applying the value of x in the second derivative.

P''(x) = -9(2x) + 0

P''(x) = -18x

Applying x = 26

P''(26) = -18(26) < 0 maximum at x = 26

a) So, the number of employees should have in the business is 26.

b) Maximum profit,

P(26) = -3(26)3 + 6084(26) - 5000

= -3(17576) + 158184 - 5000

= -52728  + 158184 - 5000

= 100456

Problem 2 :

Two numbers have the sum of 10. What is the minimum value that the sum of their cubes could be ?

Solution :

Let x and 10 - x be two numbers,

Let the required function be f(x).

f(x) = x3 + (10 - x)3

f'(x) = 3x2 + 3(10 - x)2 (-1)

f'(x) = 3x2 - 3(10 - x)2

f'(x) = 0

3x2 - 3(10 - x)2 = 0

x= (10 - x)2

x = 10 - x

x + x = 10

2x = 10

x = 5

Applying the value of x in the second derivative.

f''(x) = 6x + 6(10 - x)

= 6(5) + 6(10 - 5) > 0

Minimum at x = 5.

Minimum value = x3 + (10 - x)3

= 53 + (10 - 5)3

= 125 + 125

= 250

Problem 3 :

What is the least possible value of the sum of a positive number and its reciprocal ?

Solution :

Least value means minimum value. Let x be the required number.

Let f(x) be the required function and it will be

f(x) = x + 1/x

f'(x) = 1 - 1/x2

1 - 1/x2 = 0

1 = 1/x2

x2 = 1

x = 1 and -1

Since it is positive number -1 to be ignored.

Applying x = 1 in the second derivative.

f'(x) = 1 - 1/x2

f''(x) = 0 + 1/x3

f''(1) = 1/1==> 1

To find the least value, we can apply x = 1 in the function f(x).

f(x) = 1 + 1/1

= 2

So, the least value is 2.

Problem 4 :

A rectangle sits on the x-axis under the graph of y = 9 - x2

a) Find the coordinate of P.

b) Write down a formula for the area of the rectangle in terms of x.

c)  Find the maximum possible area of the rectangle.

Solution :

a) The coordinate P is (x, 9 - x2)

b)  Area of the shaded region.

Area of the rectangle = length x width

length = x + x ==> 2x

width = 9 - x2

= 2x (9 - x2)

A(x) = 18x - 2x3

c) Maximum area :

A(x) = 18x - 2x3

A'(x) = 18(1) - 2(3x2)

= 18 - 6x2

A'(x) = 0

18 - 6x2 = 0

6x2 = 18

x2 = 3

x = 3 and -3

A' (x) = 18 - 6x2

A'' (x) = 0 - 12x

= -12x

 A'' (x) = -12xWhen x = √3A'' (√3) = -12√3 < 0Maximum A'' (x) = -12xWhen x = -√3A'' (√3) = -12(-√3) > 0Minimum

We will get maximum possible area when x = 3

Finding area of the rectangle :

A(3) = 183 - 2(3)3

= 183 - 63

= 123

So, the maximum area is 123 square units.

Problem 4 :

A beam with a rectangular cross section is to be cut from a log of diameter of 1 m.

The strength of the beam varies in proportion to the width and to the square of the depth.

a) If the beam is x m and y m deep, write down the equation connecting x and y.

b) The strength S of the beam is given by width times the square of depth. Write down the formula for S in terms of x only.

c)  Find the dimensions of the strongest possible beam that can be cut from the log.

Solution :

a) By connecting points shown above, the red line is the diameter of the sphere. Using Pythagorean theorem,

x+ y2 = 12

x+ y2 = 1

b) Strength =  xy2

From x+ y2 = 1, y2 = 1 - x2

Applying the value of y2, we get

Strength = x (1 - x2)

c)  Strongest means maximum.

f(x) = x (1 - x2)

f'(x) = x(0 - 2x) + (1 - x2)(1)

= -2x2 + 1 - x2

f'(x) = -3x2 + 1

f'(x) = 0

-3x2 + 1 = 0

x2 = 1/3

x = 1/√3

x = 0.577

Applying the value of x in second derivative, we get

f'(x) = -6x + 0

= -6x

f''(0.577) = -6(0.577)

= -3.46 < 0 maximum

y= 1 - x2

= 1 - (0.577)2

= 1 - 0.3329

y= 0.6670

y = 0.816

So, the dimension is 0.577 m and 0.816 m.

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