The top and bottom margins of a poster are 4 cm and the side margins are each 5 cm. If the area of printed material on the poster is fixed at 388 square centimeters, find the dimensions of the poster with the smallest area.

Answer:The smallest poster has dimension 25.6 cm by 32.05 cm.Step-by-step explanation:Let "x" and "y" be the length and the width of the poster.The margin of a poster are 4 cm and the side margins are 5 cm.The length of the print = x - 2(4) = x - 8The width of the print = y - 2(5) = y - 10The area of the print = (x- 8)(y -10)The area of the print is given as 388 square inches.(x-8)(y -10) = 388From this let's find y.y -10 = [tex]\frac{388}{(x - 8)}[/tex]y = [tex]\frac{388}{x - 8} + 10[/tex] -------------------(1)The area of the poster = xyNow replace y by [tex]\frac{388}{x - 8} + 10[/tex], we getThe area of the poster = x ([tex]\frac{388}{x - 8} + 10[/tex])= [tex]10x + \frac{388x}{x - 8}[/tex]To minimizing the area of the poster, take the derivative.A'(x) =  [tex]10 + 388(\frac{-8}{(x-8)^{2} } )[/tex]A'(x) = [tex]10 - \frac{3104}{(x-8)^2}[/tex]Now set the derivative equal to zero and find the critical point.A'(x) = 0 [tex]10 - \frac{3104}{(x-8)^2}[/tex] = 0[tex]10 = \frac{3104}{(x-8)^2}[/tex][tex](x - 8)^2 = \frac{3104}{10}[/tex][tex](x - 8)^2 = 310.4[/tex]Taking square root on both sides, we getx - 8 = 17.6x = 17.6 + 8x = 25.6So, x = 25.6 cm takes the minimum.Now let's find y.Plug in x = 25.6 cm in equation (1)y =  [tex]\frac{388}{25.6 - 8} + 10[/tex] y = 22.05 + 10y = 32.05Therefore, the smallest poster has dimension 25.6 cm by 32.05 cm.