HOW TO FIND MAXIMUM AND MINIMUM POINTS USING DIFFERENTIATION

The following steps would be useful to find the maximum and minimum value of a function using first and second derivatives.

Examples 1-2 : Find the maximum and minimum points of the following functions.

Example 1 :

2x³ - 3x² - 12x + 5

Solution :

Let f(x) = 2x³ - 3x² - 12x + 5.

f'(x) = 2(3x²) - 3(2x) - 12(1) + 0

f'(x) = 6x² - 6x - 12

Equating f'(x) to zero,

f'(x) = 0

 6x² - 6x - 12 = 0

Divide both sides by 6.

x² - x - 2 = 0

(x - 2)((x + 1) = 0

x - 2 = 0  or  x + 1 = 0

x = 2  or  x = -1

Find the second derivative :

f'(x) = 6x² - 6x - 12

f"(x) = 6(2x) - 6(1) - 0

f"(x) = 12 x - 6

Substitute x = 2 in f"(x).

f"(2) = 12(2) - 6

= 24 - 6

f"(2) = 18 > 0 Minimum

To find the minimum value, substitute x = 2 in f(x).

f(x) = 2x³ - 3 x² - 12 x + 5

f (2) = 2(2)³ - 3(2)² - 12(2) + 5

  = 2(8) - 3(4) - 24 + 5

= 16 - 12 - 24 + 5

= 21 - 36

= -15

Substitute x = -1 in f"(x).

f"(-1) = 12(-1) -6

  = -12 - 6

f"(-1) = -18 > 0 Maximum

To find the maximum value, substitute x = -1 in f(x).

f(x) = 2x³ - 3x² - 12 x + 5

f(-1) = 2(-1)³ - 3(-1)² - 12(-1) + 5

= 2(-1) - 3(1) + 12 + 5

= -2 - 3 + 12 + 5

= -5 + 17

= 12

Therefore, 

maximum point = (-1, 12)

minimum point = (2, 15)

Example 2 :

Find the maximum and minimum value of the function

x³ - 3x² - 9x + 12

Solution :

Let f(x) = x³ - 3x² - 9 x + 12.

f'(x) = 3x² - 3(2x) - 9(1) + 0

f'(x) = 3x² - 6x - 9

f'(x) = 0

3x² - 6x - 9 = 0

Divide both sides by 3.

x² - 2x - 3 = 0

(x + 1)(x - 3) = 0

x + 1 = 0  or  x - 3 = 0

x = -1  or  x = 3

Find the second derivative :

f'(x) = 3x² - 6x - 9

f"(x) = 3(2x) - 6(1) - 0

f"(x) = 6x - 6

Substitute x = -1 in f"(x).

f"(-1) = 6(-1) - 6

= -6 - 6

f"(-1) = -12 < 0 ----> f(x) is maximum

To find the maximum value, substitute x = -1 in f(x).

f(x) = x³ - 3x² - 9 x + 12

f(-1) = (-1)³ - 3(-1)² - 9(-1) + 12

= -1 - 3(1) + 9 + 12

= -1 - 3 + 9 + 12

= -4 + 21

= 17

Substitute x = 3 in f"(x).

f"(3) = 6(3) - 6

  = 18 - 6

 f"(3) = 12 > 0 ----> f(x) is minimum

To find the minimum value, substitute x = 3 in f(x).

f(x) = x³ - 3x² - 9 x + 12

f(3) = 3³ - 3(3)² - 9(3) + 12

= 27 - 3(9) - 27 + 12

= 27 - 27 - 27 + 12

= -27 + 12 

= -15

Therefore, 

maximum point = (-1, 17)

minimum point = (3, -15)

Example 3 :

The sum of two numbers is 80. Find the largest possible product.

Solution :

Let x and y be the two numbers.

x + y = 80

y = 80 - x ----(1)

Product of the two numbers :

= xy

= x(80 - x)

= 80x - x²

Let P(x) be the product of the two numbers.

P(x) = 80x - x²

Find the first derivative of P(x).

P'(x) = 80 - 2x

Find the second derivative of P(x).

P"(x) = -2

Equate the first derivative of P(x) to zero and solve for x, which is called critical number.

P'(x) = 0

80 - 2x = 0

-2x = -80

x = 40

Substitute x = 40 into P"(x) = -2.

P"(40) = -2 < 0

So, P(x) is maximum at x = 40.

Subsitute x = 40 into (1).

y = 80 - 40

y = 40

When the two numbers are 40 and 40, their product is minimum.

The largest possible product of the two numbers :

= 40 ⋅ 40

= 1600

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