Answer to Question #155424 in Calculus for Lance Valderama

Question

A poster must have 32 square inches of printed matter with margins of 4 inches at the top and bottom, and 2 inches at each side. Find the dimensions of the whole poster if its area is minimum.

Answer 1

$\text{let }x,y\text{ length and width of the poster}$

$S = xy$

$S_{prn}=(x-8)(y-4)=32$

$y= 4 +\frac{32}{x-8}$

$S(x)= 4x +\frac{32x}{x-8}$

$\text{find the extremum points where the derivative is 0}$

$S^{'}(x)= 4 +\frac{32(x-8)-32x}{(x-8)^2}=4-\frac{256}{(x-8)^2}$

$4-\frac{256}{(x-8)^2}=0$

$4=\frac{256}{(x-8)^2}$

$(x-8)^2 =64$

$\text{as } x>0 \ x=16$

$S^{'}(10)<0;S^{'}(20)>0$

$\text{then }x=16 \text{ minimum point}$

$y= 4+\frac{32}{16-8}=8$

Answer:16 inches and 8 inches poster sizes