Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

Question

The strength of a beam varies as the product of its breadth and square of its depth. Find the dimensions of the strongest beam which can be cut from a circular log of radius

a

?

Accepted Answer

Let us assume $ABCD$ be the cross-section of the beam that needs to cut from the circular log of radius ‘ $a$ ’.

Let us assume ‘ $b$ ’ be the depth of the rectangle, ‘ $l$ ’ be the length of the rectangle and ‘ $d$ ’ be the diameter of a circular log.

$d = 2 a . . . . . . (i)$

According to the problem, strength of the beam is given by

$S = {lb}^{2} . . . . . . (ii)$

From the $∆ ABC$ , $d^{2} = l^{2} + b^{2}$

From $(i)$ , $(2 a^{2}) = l^{2} + b^{2}$

$b^{2} = 4 a^{2} - l^{2} (iii)$

From (ii) & (iii)

$S = l (4 a^{2} - l^{2}) \Rightarrow 4 {la}^{2} - l^{3}$

We need strength of the beam to be maximum, let us take $S$ as a function of $l$

For maxima and minima,

Differentiate and equate the function to $0$ , we get, $\frac{dS}{dl} = 0$

$4 a^{2} - 3 l^{2} = 0$

Therefore, $l = \frac{2 a}{\sqrt{3}}$ , $b = \frac{2 \sqrt{2} a}{\sqrt{3}}$

Again differentiating the function, we get,

$\frac{d^{2} S}{{dl}^{2}} = - 6 l$

$\frac{d^{2} S}{{dl}^{2}} = - 6 (\frac{2 a}{\sqrt{3}}) = - 4 \sqrt{3} a < 0 maxima$

$S = (\frac{2 a}{\sqrt{3}}) (\frac{2 \sqrt{2} a}{\sqrt{3}})^{2} = \frac{16 a^{2}}{3 \sqrt{3}}$

Dimensions of beam of maximum strength are $(\frac{2 a}{\sqrt{3}}, \frac{2 \sqrt{2} a}{\sqrt{3}})$ .

The strength of a beam varies as the product of its breadth and square of its depth. Find the dimensions of the strongest beam which can be cut from a circular log of radius $a$ ?

Important Questions on Maxima and Minima

Learn and Prepare for any exam you want !