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 :39.294 cm × 9.823 cmExplanation :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,[tex]A = xy\text{ square cm}[/tex]According to the question,A = 386 cm²,[tex]xy = 386[/tex][tex]\implies y = \frac{386}{x}------(1)[/tex]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,[tex]A_1=(x+16)(y+4)[/tex][tex]A_1 =(x+16)(\frac{386}{x}+4)[/tex]Differentiating with respect to x,[tex]\frac{dA_1}{dx}= \frac{4x^2-6176}{x^2}[/tex]Again differentiating with respect to x,[tex]\frac{d^2A_1}{dx^2}=\frac{ 12352}{x^3}[/tex]For maxima or minima,[tex]\frac{4x^2-6176}{x^2}=0[/tex][tex]4x^2-6176=0[/tex][tex]4x^2 = 6176[/tex][tex]x^2 =1544[/tex][tex]x = \pm \sqrt{1544}[/tex][tex]\implies x = 39.294\text{ or } x=-39.294[/tex]Sides can not be negative,Also, at x = 39.294,[tex]\frac{d^2A_1}{dx^2}=\text{ positive}[/tex]i.e. [tex]A_1[/tex] is minimum at x = 39.294,∵ y = [tex]\frac{386}{39.294}[/tex] = 9.823Hence, the dimensions of the poster with the smallest area,39.294 cm × 9.823 cm