Arectangular poster is to contain 162 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?

Let
H > the height of the printed area
W > the width of the printed area
 Then
we know that
HW = 162> H=162/W> equation 1
the area of the poster is given by
A = (H + 4)*(W + 2)> equation 2

I substitute 1 in 2

A=([162/W] + 4)*(W + 2)> A=162+324/W+4W+8
By taking the derivative,
Let A1> derivative of A
we have
A1 = −324/W² + 4
By setting A1 to 0, we have
W²=324/4> W²=81> W=9 in
H=162/W> 162/9> H=18 in

the dimensions of the poster are
(H + 4)> 18+4> 22 in
(W + 2)> 9+2> 11 in

the answer is 
the dimensions of the poster are 11 in x 22 in

Yes you do, so £1973.09 - £1473.50 = £499.59 so tim overpaid £499.59 .hope this ; )