Answer to Question #349932 in Calculus for Mark

Question

. A poster must have 32 sq.in. of printed materials with margins of 4 inches each at the top and 2 inches at each side. Find the dimensions of the whole poster if its area is maximum.

Answer 1

Let $x=$ the length of printed matter, $y=$the width of printed matter.

Given $xy=32.$

The length of a poster will be $x+2(2),$ the width of a poster will be $y+4.$

The area of a poster is $A=(x+4)(y+4).$

$xy=32=>y=\dfrac{32}{x}$

Substitute

$A=A(x)=(x+4)(\dfrac{32}{x}+4), x>0$ $A(x)=32+4x+\dfrac{128}{x}+16$

Find the derivative with respect to $x$

$A'(x)=(48+4x+\dfrac{128}{x})'=4-\dfrac{128}{x^2}$

Find the critical number(s)

$A'(x)=0=>4-\dfrac{128}{x^2}=0$ $x^2=32$ $x_1=-4\sqrt{2}, x_2=4\sqrt{2}$

If $0<x<4\sqrt{2}, A'(x)>0, A(x)$ increases.

If $x>4\sqrt{2}, A'(x)<0, A(x)$ decreases.

The function $A(x)$ has a local maximum at $x=4\sqrt{2}.$

Since the function $A(x)$ has the only extremum for $x>0,$ then the function $A(x)$ has the absolute maximum at $x=4\sqrt{2}$ for $x>0.$

$y=\dfrac{32}{4\sqrt{2}}=4\sqrt{2}$

The poster is $(4\sqrt{2}+4)\ inches\times (4\sqrt{2}+4)\ inches.$