Question

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

Answer 1

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

Then $xy=32.$

Solve for $y$

$$y=\dfrac{32}{x}, x>0$$

The total area of the poster is

$$A=(x+4)(y+4)$$

Substitute

$$A=A(x)=(x+4)(\dfrac{32}{x}+4)$$

$$=32+\dfrac{128}{x}+4x+16=48+\dfrac{128}{x}+4x$$

Find the first derivative with respect to $x$

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

Find the critical number(s)

$$A'(x)=0=>-\dfrac{128}{x^2}+4=0$$

$$x=-4\sqrt{2}\ or\ x=4\sqrt{2}$$

We consider $x>0.$ Then

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

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

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

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

The area of the poster has no maximum.

The area of the poster will be minimum, when the poster will be the square with side $4\sqrt{2}$ inches.