Answer to Question #129848 in Calculus for Hetisani Sewela

Question

You are designing a poster to contain 50 cm2 of printing with margins of 4 cm each at the top and bottom and 2 cm at each side. What overall dimensions will minimize the amount of paper used?

Answer 1

Let the paper size be $X$ inches in length and $Y$ inches in width.

the length of the printed space would be $X-8$ inches and width would be $Y-4$ inches. Print area would thus be $(X-8)(Y-4)=50$.

From this $Y=4+\frac {50} {X-8}=\frac{4X+18}{X-8}$

Also from the same equation on simplifying, it is $XY-8X+32=50$.

Since the area of the paper of size $X$ inches by $Y$ inches is $XY$, let it be denoted as $A$. Thus

$A-8Y-4X=18$ Or $A=8Y+4X+18$

$A=\frac{32X+144}{X-8}+4X+18$ . For minimum paper size $\frac{dA}{dX}$ must be $=0$ , hence,

$\frac{dA}{dX}=\frac{32(X-8)-(32X+144)}{(X-8)^2}+4=0$

$\frac{-400}{(X-8)^2}+4=0$

$(X-8)^2=100$

$X-8=10$

$X=18$ , hence $Y=4+\frac{50}{18-8}=9$

Dimension of minimum paper size would be 18 inches by 9 inches.