Question

Johnny is designing a rectangular poster to contain 24in^2 of printing with a 3-in margin at the top and bottom and a 2-in margin at each side. what overall dimensions will minimize the amount of paper used?

Answer 1

Let the paper size be $y$ inches in length and $x$ inches in width.

The length of the printed space would be $(y-2\cdot3)$ inches and width would be $(x-2\cdot2)$ inches.

Johnny is designing a rectangular poster to contain $24\ in^2$ of printing 

$$(x-4)\cdot(y-6)=24$$

Solve for $y$

$$y={24\over x-4}+6$$

Since the area of the paper of size $x$ inches by $y$ inches is $xy,$ let it be denoted as A

$$A=x\cdot y$$

Then

$$A=A(x)=x({24\over x-4}+6), x>4$$

Find the first derivative with respect to $x$

$$A'(x)=({24x\over x-4}+6x))'={24\over x-4} -{24x\over (x-4)^2}+6$$

Find the critical number(s)

$$A'(x)=0=>{24\over x-4} -{24x\over (x-4)^2}+6=0$$

$$4(x-4)-4x+(x-4)^2=0$$

$$4x-16-4x+x^2-8x+16=0$$

$$x(x-8)=0$$

Critical numbers: $x=0, x=8$

First derivative test

$If \ x<0, A'(x)>0, A(x)\ increases.$

$If \ 0<x<8, A'(x)<0, A(x)\ decreases.$

$If \ x>8, A'(x)>0, A(x)\ increases.$

The function $A(x)$ has the local maximum at $x=0.$

The function $A(x)$ has the local minimum at $x=8.$

We consider $x>4.$ Hence the function $A(x)$ has the absolute minimum for $x>4$ at $x=8.$

Find the length

$$y={24\over 8-4}+6=12$$

We need to use the paper $8\ in\times 12 \ in.$