Question

The top and bottom margins of a poster are 2 cm and the side margins are each 8 cm. If the area of printed material on the poster is fixed at 386 square centimeters, find the dimensions of the poster with the smallest area.

Answer 1

Answer :

39.294 cm × 9.823 cm

Explanation :

Let x be the length ( in cm ) of the printed material and y be the width ( in cm ) of the printed material,

So, the area of the printed material,

$A = xy\text{ square cm}$

According to the question,

A = 386 cm²,

$xy = 386$

$\implies y = \frac{386}{x}------(1)$

Since, the printed area is obtained in the margin of 2 cm from both top and bottom and margin of 8 cm in sides,

Thus, length of the poster = (x + 16) cm and width = (y + 4)

So, the area of the poster,

$A_1=(x+16)(y+4)$

$A_1 =(x+16)(\frac{386}{x}+4)$

Differentiating with respect to x,

$\frac{dA_1}{dx}= \frac{4x^2-6176}{x^2}$

Again differentiating with respect to x,

$\frac{d^2A_1}{dx^2}=\frac{ 12352}{x^3}$

For maxima or minima,

$\frac{4x^2-6176}{x^2}=0$

$4x^2-6176=0$

$4x^2 = 6176$

$x^2 =1544$

$x = \pm \sqrt{1544}$

$\implies x = 39.294\text{ or } x=-39.294$

Sides can not be negative,

Also, at x = 39.294,

$\frac{d^2A_1}{dx^2}=\text{ positive}$

i.e. $A_1$ is minimum at x = 39.294,

∵ y = $\frac{386}{39.294}$ = 9.823

Hence, the dimensions of the poster with the smallest area,

39.294 cm × 9.823 cm