A rectangular poster is to contain 578 square inches of print. The margins at the top and bottom of the poster are to be 2 inches, and the margins on the left and right are to be 1 inch. What should the dimensions of the poster be so that the least amount of poster is used?

19 by 38

Step-by-step explanation:

The area of the poster, A=xy

Margin at both up and bottom is 2inches, at the right and left is 1 inch

Then we can say

(x-4)(y-2)= 578

Let us make y subject of the formula

(y-2)= 578/(x-4)

y=[ 578/(x-4)] + 2

If we substitute into the area A equation we have

A= x { [ 578/(x-4)] + 2}

A= 2x + (578x)/(x - 4)

If we differentiate we have

A'(x)= [2+578(x-4)+578x ]/ (x-4)^2

=[ 2(x-4)^2 + 578(x -4) + 578x ] / (x-4)^2

=[ 2(x^2 - 8x +16) + 578(x-4) + 578x ] /(x-4)^2

If we simplify this we have

=[ 2x^2 -16x +32+578x -578x - 2312]/ (x-4)^2

=( 2x^2 -16x - 2312) / (x-4)^2

At A'(x)= 0

= ( x^2 - 8x - 2312) / (x-4)^ =0

If we divide through by 2 we have

x^2 - 8x - 1156= 0

Solving the quadratic eqn( CHECK THE ATTACHMENT)

X= 38 or -30 then we choose the positive one, then x= 38

Then from area of the poster

A= (x-4)(y-2)= 578

Substitute 38 as value of x

(38-4)(y-2)= 578

34(y-2)=578

34y-68=578

34y=646

y=19

Hence dimensions of the poster is 19 by 38

CHECK THE ATTACHMENT FOR QUADRATIC SOLUTION