 # 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 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.

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, 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

## 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 maximum value.

∴ 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 end points = f (1) = 42, f (10) = 177

So, that f (x) has its maximum value as 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 loge1 and f′′(e) = 3 + 2 logee

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|

= ex [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 − logex] / x2 = 0 or [1 − logex] / x2 = 0

∴loge 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 as 100 cm and enclosing the maximum area.

Solution:

2x + 2y =100 ⇒ x + y = 50 ..(i)

Let area of 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, adjacent sides are 25 and 25 cm.

Example 7: If P = (1, 1), Q = (3, 2) and R is a point on 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 co-ordinate of R (x, 0).

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

$PR+RQ=\sqrt{{{(x-1)}^{2}}+{{(0-1)}^{2}}}+\sqrt{{{(x-3)}^{2}}+{{(0-2)}^{2}}}\\ = \sqrt{{{x}^{2}}-2x+2}+\sqrt{{{x}^{2}}-6x+13}$

For minimum value of PR + RQ,

$\frac{d}{dx}(PR+RQ)=0 \\ \frac{d}{dx}(\sqrt{{{x}^{2}}-2x+2})+\frac{d}{dx}(\sqrt{{{x}^{2}}-6x+13})=0 \\ \frac{(x-1)}{\sqrt{{{x}^{2}}-2x+2}}=-\frac{(x-3)}{\sqrt{{{x}^{2}}-6x+13}}$

Squaring both sides,

$\frac{{{(x-1)}^{2}}}{({{x}^{2}}-2x+2)}=\frac{{{(x-3)}^{2}}}{{{x}^{2}}-6x+13} \\ 3{{x}^{2}}-2x-5=0\\ (3x-5)\,(x+1)=0, \\ x=\frac{5}{3},\,-1$

Also 1 < x < 3.

$R=(5/3,\,0)$

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

A) 11

B) 12

C) 10

D) 14

Solution:

Let $f(x)=4{{e}^{2x}}+9{{e}^{-2x}}$

Therefore, ${f}'(x)=8{{e}^{2x}}-18{{e}^{-2x}}$

Put ${f}'(x)=0\Rightarrow 8{{e}^{2x}}-18{{e}^{-2x}}=0 \\ {{e}^{2x}}=3/2\Rightarrow x=\log {{(3/2)}^{1/2}}$

Again, ${f}”(x)=16{{e}^{2x}}+36{{e}^{-2x}}>0$

Now,

$f(\log {{(3/2)}^{1/2}})=4{{e}^{2.(\log {{(3/2)}^{1/2}})}}+9{{e}^{-2(\log {{(3/2)}^{1/2}})}} \\= 4\times \frac{3}{2}+9\times \frac{2}{3} = 6+6\\=12$

Hence minimum value = 12.

Example 9: The function $f(x)={{x}^{-x}},\,(x\,\in \,R)$ attains a maximum value at x =

A) 2

B) 3

C) 1/e

D) 1

Solution:

$f(x)=y={{x}^{-x}} \\ \log y=-\,x\log x$

Differentiating w.r.t. x,

$\frac{1}{y}.\frac{dy}{dx}=-\left[ x.\frac{1}{x}+\log x \right] \\ \frac{1}{y}.\frac{dy}{dx}=-[1+\log x]\\ \frac{dy}{dx}=-{{x}^{-x}}[1+\log x]\\ \frac{dy}{dx}={{x}^{-x}}\left[ \log \frac{1}{x}-1 \right]$

Put $\frac{dy}{dx}=0\\ {{\log }_{e}}\frac{1}{x}={{\log }_{e}}e\\ \frac{1}{x}=e\Rightarrow x=\frac{1}{e}$