A farmer has 54m of fencing with which to build two animal pens with a common side. one pen is rectangular; the other is square. if the area of the pens is to be maximized, what are their dimensions?

Answer 1

let the side of the square be x
let the sides of the rectangle be x by y

so 5x + 2y = 54
y = (54-5x)/2

area = x^2 + xy
= x^2 + x(54-5x)/2
= x^2 + 27x - (5/2)x^2

d(area)/dx = 2x + 27 - 5x
= 0 for a max of area
3x = 27
x = 9
y = (54 - 45)/2 = 4.5
the square is 9 by 9, and the rectangle is 9 by 4.5

Answer 2

bur