The top and bottom margins of a rectangle poster are 2 inches.

Question

Top and bottom margins of a rectangle are 2 inches while the side margins are 1.5 inches. The printed are is fixed at 120 square inches.

FIND THE DIMENSIONS OF THE POSTER WITH THE SMALLEST TOTAL AREA.

Answer 1

Simon

Assume the length of the printed area is l.

Then the width of that area will be 120/l (because the area is fixed at 120).

Considering the area of the margins.

Top and bottom : each 2 (l+3) , total 2l + 6

Left and right (excluding area already included) : each (120/l) x 1.5, total 360/l

So total area of margins A = 2l + 6 + 360/l

You want this to be minimal , so differentiate this with respect to l, and set dA/dl = 0

This gives the value of l for the minimal area, and using this you can find the dimensions needed.

Mike

Answer 2

Let the height of the print area be x and the width be y

xy = 120 or y = 120/x

Then, the height of the poster is x+2+2 = x+4

The width of the poster is y + 1.5 + 1.5 = y+3

We seek to minimize A = (x+4)(y+3) = (x+4)(120/x + 3) = 132 + 3x + 480/x

dA/dx = 3 - 480/x^2

dA/Dx = 0 implies

3 = 480/x^2 or 3x^2 = 480 or x = √160 = 4√10

Then, y = 120/(4√10) = 3√10

The poster has dimensions x+4 by y+3 = 4√10+4 by 3√10+3

height is 4√10+4

width is 3√10+3

The area is 132 + 24√10