Solved Examples on Maxima and Minima

Introduction to Maxima and Minima

As the name suggests, it is finding the maximum and minimum value of a given function. In this article, we come across Solved Examples of Maxima and Minima.

For instance, from the values of the maximum and minimum speed of a train, an engineer will be able to decide on the materials required to withstand the speed to manufacture brakes for the train to run smoothly. The maximum and minimum value of thyroid in the bloodstream enables the doctor to prescribe the appropriate medicine for the patients to bring down the thyroid levels.

Maxima and Minima – Top 12 Must-Do Important JEE Questions

The extreme value of the function is the maxima or minima. If a function f (x) is defined on x, then based on the x – interval, the function attains an extremum termed as “global” or “local” extremum.

The types of maxima and minima are as follows:

• Local Maxima

The local maxima of a function can be defined as any point in the domain of the function, whose value exceeds the value of the local maxima. It is the maximum value when compared to other points which are nearby.

• Global Maxima

The global maxima is the one which doesn’t have any point in the domain of the function, whose value is greater than the value of the global maxima.

What Is a Stationary Point on a Curve?

A point on the curve whose derivative vanishes is a stationary point. A point (y₀, f (y₀)) can be called a stationary point of f (y) if [df / dy]_y=y0=0.

The types of stationary points are:

• Local Maxima
• Local Minimas
• Inflection Points

Also, Read:

• Limits of Functions
• Differentiation

Maxima and Minima Solved Examples

Example 1: What is the value of the function (x − 1) (x − 2)² at its maxima?

Solution:

Given f (x) = (x − 1) (x − 2)²

f (x) = (x − 1) (x² + 4 −4x);

f (x) = (x³ − 5x² + 8x −4)

Now, f′(x) = 3x² − 10x + 8, f′(x) = 0

3x² − 10x + 8 = 0

(3x − 4) (x − 2) = 0

x = 4 / 3, 2

Now, f′′(x) = 6x − 10

f′′(4 / 3) = 6 × [4 / 3] −10 < 0

f′′(2) = 12 − 10 > 0

Hence, at x = 4 / 3, the function will occupy the maximum value.

∴ The maximum value = f (4 / 3) = 4 / 27.

Example 2: Find the maximum value of function x³ − 12x² + 36x + 17 in the interval [1, 10].

Solution:

Let f (x) = x³ − 12x² + 36x + 17

∴ f′ (x) = 3x² − 24x + 36 = 0 at x = 2, 6

Again f′′(x) = 6x − 24 is negative at x = 2

So that f (6) = 17, f (2) = 49

At the endpoints = f (1) = 42, f (10) = 177

So, that f (x) has its maximum value of 177.

Example 3: Check whether the function x² logx in the interval (1, e) has a point of maximum or minimum?

Solution:

Let f (x) = x² logx

f′(x) = 2x logx + x and

f′′(x) = 2 (1 + logx) + 1

Now f′′(1) = 3 + 2 log_e1 and f′′(e) = 3 + 2 log_ee

f(x) has local minimum at 1 / √e, but x lies only in interval (1, e) so that y₂ = √x has not extremum in (1, e).

Hence, neither a point of maximum nor minimum.

Example 4: The minimum value of |x| + |x + (1/2)| + |x − 3| + |x − (5/2)| is ____________.

Solution:

f (x) = |x| + |x + (1/2)| + |x − 3| + |x − (5/2)|

= e^x [1 + (x − 2) logx / x³] for x ≤ −12

= −2x + 6, for −12 ≤ x ≤ 0

dy / dx = 0 ⇒ x =−1 for 0 ≤ x ≤ 5 / 2

=2x + 1, for 5 / 2 ≤ x ≤ 3 =4x − 5, for x ≥ 3

The minimum value of the function is 6 from the graph.

Example 5: What is the local maximum value of the function logx / x?

Solution:

Let f (x) = logx / x ⇒ f′ (x) = 1/ x² − logx / x²

For maximum or minimum value of f (x), f′(x) = 0

f′(x) = [1 − log_ex] / x² = 0 or [1 − log_ex] / x² = 0

∴ log_e x = 1 or x = e, which lie in (0, ∞).

For x = e, d²y / dx² = −1 / e³, which is negative.

Hence, y is maximum at x = e and its maximum value = log e / e = 1 / e.

Example 6: Calculate the adjacent sides of a rectangle with a given perimeter of 100 cm and enclose the maximum area.

Solution:

2x + 2y =100 ⇒ x + y = 50 ..(i)

Let the area of the rectangle be A,

∴ A = x y ⇒ y = A / x

Put in (i), we have x + A / x = 50 ⇒ A = 50x − x²

dA / dx = 50 − 2x

For maximum area dA / dx = 0

∴ 50 − 2x = 0 ⇒ x = 25 and y = 25

Hence, the adjacent sides are 25 and 25 cm.

Example 7: If P = (1, 1), Q = (3, 2), and R is a point on the x-axis  then the value of PR + RQ will be minimum at               

A) (5 / 3, 0)

B) (1 / 3, 0)

C) (3, 0)

D) (1, 0)

Solution: 

Let the coordinate of R be (x, 0).                   

Given P (1, 1) and Q (3, 2).

For the minimum value of PR + RQ, 

Squaring both sides, 

Also 1 < x < 3.

Example 8: The minimum value of 4e^2x + 9e^-2x is 

A) 11

B) 12

C) 10

D) 14

Solution:

Let

Therefore,

Put

Again,

Now, 

Hence, the minimum value = 12.

Example 9: The function f(x) = x^-x, x ∈ R, attains a maximum value at x =     

A) 2

B) 3

C) 1/e

D) 1

Solution:

Differentiating w.r.t. x,  

Put dy/dx = 0.

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