Name: Rates of Change
Uploaded: 2019-07-19
Duration: 48 min 10 s
Description: In this video, let us know the average rate of change and instantaneous rate of change. Learn more about this concept in the video.

Question

A window is in the form of a rectangle surmounted by a semicircle. If the perimeter be

30 m

, find the dimensions so that the greatest possible amount of light may be admitted.

Accepted Answer

Given,

Let $x$ be the length, $y$ be the breadth of the rectangle and $r$ be the radius of the semicircle. Then perimeter of the window.

Then perimeter of the window $= x + 2 y + πr,$ where $x = 2 r$

This is given to be $m$

$∴ 2 r + 2 y + πr = 30$

$∴ 2 y = 30 - (π + 2) r$

$∴ y = 15 - \frac{(π + 2) r}{2} . . . . . . . . . . . . . (1)$

The greatest possible amount of light may be admitted if the area of the window is maximum. Let $A$ be the area of the window.

Then $A = x y + \frac{{πr}^{2}}{2}$

$A = 2 y r + \frac{{πr}^{2}}{2} . . . . [∵ x = 2 r]$

$A = 2 r [15 - \frac{(π + 2) r}{2}] + \frac{{πr}^{2}}{2} . . . . [By (1)]$

$A = 30 r - (π + 2) r^{2} + \frac{{πr}^{2}}{2}$

$A = 30 r - (π + 2 - \frac{π}{2}) r^{2}$

$∴ A = 30 r - (\frac{π + 4}{2}) r^{2}$

Differentiating w.r.t. $r$ , we get

$∴ \frac{d A}{d r} = \frac{d}{d r} [30 r - (\frac{π + 4}{2}) r^{2}]$

$∴ \frac{d A}{d r} = 30 \times 1 - (\frac{π + 4}{2}) \times 2 r$

$∴ \frac{d A}{d r} = 30 - (π + 4) r$

Again differentiating w.r.t. $r$ , we get

$∴ \frac{d^{2} A}{d r^{2}} = \frac{d}{d r} [30 - (π + 4) r]$

$∴ \frac{d^{2} A}{d r^{2}} = 0 - (π + 4) \times 1$

$∴ \frac{d^{2} A}{d r^{2}} = - (π + 4)$

For maximum $A, \frac{d A}{d r} = 0$

$∴ 30 - (π + 4) r = 0$

$∴ r = \frac{30}{π + 4}$

And, ${(\frac{d^{2} A}{d r^{2}})}_{a t r = \frac{30}{π + 4}} = - (π + 4) < 0$

$∴ A$ is maximum when $r = \frac{30}{π + 4}$

When $r = \frac{30}{π + 4}$

$y = 15 - \frac{(π + 2)}{2} \times \frac{30}{π + 4} . . . [by (1)]$

$y = \frac{30 π + 120 - 30 π - 60}{2 (π + 4)}$

$y = \frac{30}{π + 4}$

Hence, the required dimensions of the window are as follows:

Length of rectangle $= (\frac{60}{π + 4}) meters$

Breadth of rectangle $= (\frac{30}{π + 4}) meters$ and

Radius of the semicircle $= (\frac{30}{π + 4}) meters$ .

A window is in the form of a rectangle surmounted by a semicircle. If the perimeter be $30 m$ , find the dimensions so that the greatest possible amount of light may be admitted.

Important Questions on Applications of Derivatives

Learn and Prepare for any exam you want !

A window is in the form of a rectangle surmounted by a semicircle. If the perimeter be 30 m, find the dimensions so that the greatest possible amount of light may be admitted.

Important Questions on Applications of Derivatives

Learn and Prepare for any exam you want !

A window is in the form of a rectangle surmounted by a semicircle. If the perimeter be $30 m$ , find the dimensions so that the greatest possible amount of light may be admitted.