As with most such problems, I expect you will find that the maximum area occurs on a square page. But, let's check it out.
If the dimensions are x and y, we have
(x-4)(y-4) = 16
a = xy = x*(16/(x-4) + 4)
da/dx = 4x(x-8)/(x-4)^2
da/dx = 0 when x=8
So, the page is 8x8, with the printed area 4x4, having an area of 16
It would have been a better problem had the margins not been the same. Just sayin'
1) A rectangular page is to contain 16 square inches of print. The page has to have a 2-inch margin on top and at the bottom and a 2-inch margin on each side. Find the dimensions of the page that minimize the amount of paper used.
2) A rectangular garden of area 480 square feet is be surrounded on three sides by a brick wall costing $12 per foot and on one side by a fence costing $8 per foot. Find the dimensions of the garden such that the cost of the materials is minimized.
2 answers
If the fenced side is y and the wall sides are x, then we have
xy = 480
and the cost is
c = 12*2x + 8y = 24x + 8(480/x) = 24x + 3840/x
dc/dx = 24(x^2-160)/x^2
dc/dx = 0 when x = 4√10
So, the garden is 4√10 by 120/√10
xy = 480
and the cost is
c = 12*2x + 8y = 24x + 8(480/x) = 24x + 3840/x
dc/dx = 24(x^2-160)/x^2
dc/dx = 0 when x = 4√10
So, the garden is 4√10 by 120/√10
2 answers