Maxima and Minima Examples

Example 1: Find the local maxima and local minima for the function y = x³ – 3x + 2

Solution:

y = x³ – 3x + 2

Find first order derivative

Differentiating y

y’ = (d / dx) [x³ – 3x + 2]

⇒ y’ = (d / dx) x³ – (d / dx) (3x) + (d / dx) 2

⇒ y’ = 3x² – 3 + 0

⇒ y’ = 3x² – 3

Now equate y’ = 0, to find the critical points

y’ = 0

⇒ 3x² – 3 = 0

⇒ 3x² = 3

⇒ x² = 1

⇒ x = 1 or x = -1

The critical points are x = 1 and x = -1

Now we will find second derivative to check the critical point is maxima or minima.

y” = (d / dx) [3x² – 3]

⇒ y” = (d / dx) [3x²] – (d /dx) [3]

⇒ y” = 6x – 0

⇒ y” = 6x

Now we will put the values of x and find whether y” is greater than 0 or less than 0.

At x = 1

y” = 6(1) = 6

Since, y” > 0 x = 1 is the minima of y

At x = -1

y” = 6(-1) = -6

Since, y” < 0 x = -1 is the maxima of y

The local maxima and minima of y are x = -1 and x = 1 respectively.

Example 2: Find the extremum of the function f(x) = -3x² + 4x + 7 and the extremum value.

Solution:

y =-3x² + 4x + 7

Find first order derivative

Differentiating y

y’ = (d / dx) [-3x² + 4x + 7]

⇒ y’ = (d / dx) (-3x²) – (d / dx) (4x) + (d / dx) 7

⇒ y’ = -6x – 4 + 0

⇒ y’ = -6x – 4

Now equate y’ = 0, to find the critical points

y’ = 0

⇒ -6x – 4 = 0

⇒ -6x = 4

⇒ x = -2 / 3

The critical point is x = -2/3

Now we will find second derivative to check the critical point is maxima or minima.

y” = (d / dx) [-6x – 4]

⇒ y” = (d / dx) [-6x] – (d /dx) [4]

⇒ y” = -6 – 0

⇒ y” = – 6

Since, y” < 0 x = -2 / 3 is the maxima of y and the maximum value is obtained by putting x = -2/3 in y

The local maxima of y is x = -2/3. (extremum)

The maximum value of y = -3(-2/3)² + 4(-2/3) + 7 = – 2/ 3 – 8 / 3 + 7 = 10 /3 (extremum value)

Example 3: Find the maximum height when a stone is thrown at any time t and height is given by h = -10t² + 20t + 8.

Solution:

To find the maximum height we will differentiate h = -10t² + 20t + 8

h’ = (d /dx) [-10t² + 20t + 8]

⇒ h’ = (d/dx) -10t² + (d /dx) 20t + (d/ dx) 8

⇒ h’ = -20t + 20 + 0

⇒ h’ = -20t + 20

To find the time at which height is maximum we find h’ = 0

⇒ h’ = 0

⇒ -20t + 20 = 0

⇒ -20t = -20

⇒ t = 1

Now we will find second derivative of h

h” = -20 < 0

Therefore, height is maximum at t = 1

putting value of t in h = -10t² + 20t + 8

h = -10(1)² + 20(1) + 8

⇒ h = -10 + 20 + 8

⇒ h = 18

The maximum height is 18 unit.

Example 4: Find the value of the function (x – 1)(x – 2)² at its minima.

Solution:

Let y = (x – 1)(x – 2)²

⇒ y = (x – 1) (x² + 4 – 4x)

⇒ y = x³ + 4x – 4x² – x² – 4 + 4x

⇒ y = x³ – 5x² + 8x – 4

Differentiating y

y’ = (d /dx) [x³ – 5x² + 8x – 4]

⇒ y’ = 3x² – 10x + 8

Now we will equate y’ = 0 to find the critical points

y’ = 0

3x² – 10x + 8 = 0

⇒ 3x² – 6x – 4x + 8 = 0

⇒ 3x (x – 2) – 4 (x – 2) = 0

⇒ (x – 2)(3x – 4) = 0

⇒ x = 2 or x = 4/3

We will find second derivative to check point is maxima or minima

y” = (d / dx) [3x² – 10x + 8]

⇒ y” = 6x – 10

putting x = 2, y” = 12 – 10 = 2 > 0

So, x = 2 is minima point

putting x = 4 / 3, y” = 8 – 10 = -2 < 0

So, x = 4/3 is maxima point

In the question we have to find minima and value of y at its minima

At x = 2

⇒ y = (x – 1)(x – 2)²

⇒ y = (2 – 1)(2 – 2)²

⇒ y = 1

The value of y at its minima is 1.

Example 5: Find the minimum value of the function 6e^3x + 4e^-3x

Solution:

Let y = 6e^3x + 4e^-3x

Find first order derivative

Differentiating y

y’ = (d / dx) [6e^3x + 4e^-3x]

⇒ y’ = (d / dx) 6e^3x + (d / dx) 4e^-3x

⇒ y’ = 18e^3x – 12e^-3x

Now equate y’ = 0, to find the critical points

y’ = 0

⇒ 18e^3x – 12e^-3x = 0

⇒ 6[3e^3x – 2e^-3x] = 0

⇒ 3e^3x – 2e^-3x = 0

⇒ 3e^3x = 2e^-3x

⇒ e^6x = 2/3

vx = log(2/3)^1/6

The critical points are x = log(2/3)^1/6

Now we will find second derivative to check the critical point is maxima or minima.

y” = (d / dx) [18e^3x – 12e^-3x]

⇒ y” = 54 e^3x + 36 e^-3x

Now we will put the values of x and find whether y” is greater than 0 or less than 0.

At x = log(2/3)^1/6

Since, y” > 0 x = log(2/3)^1/6 is the minima of y

The minima of y is x = log(2/3)^1/6

The minimum value of y is at x = log(2/3)^1/6

Putting value of x in y

⇒ y = 6e^3x + 4e^-3x

⇒ y = 6e^3log(2/3)1/6 + 4e^{-3log(2/3)1/6}

⇒ y = 6e^{(1/2)log(2/3)} + 4e^{(-1/2)log(2/3)}

⇒ y = 6e^(1/2)e^log(2/3) + 4e^(-1/2)e^log(2/3)

⇒ y = (2/3) 6e^(1/2) + (2/3) 4e^(-1/2)

⇒ y = 4e^(1/2) + (8/3)e^(-1/2)

Maxima and Minima in Calculus is an important application of derivatives. The Maxima and Minima of a function are the points that give the maximum and minimum values of the function within the given range. Maxima and minima are called the extremum points of a function.

This article explores the concept of maxima and minima. In addition to details about maxima and minima, we will also cover the types of maxima and minima, properties of Maxima and Minima, provide examples of maxima and minima, and discuss applications of Maxima and Minima.