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

Answer 1

let the width of the whole thing be x m
and let the length be y m

your sketch should show 3 widths and 2 lengths
where 3x + 2y = 54
y = (54-3x)/2 = 27 - (3/2)x

area = xy
x(27 - (3/2)x)
= 27x - (3/2)x^2

d(area)/dx = 27 - 3x
= 0 for a max/min of area
3x = 27
x = 9
y = (54-27)/2 = 13.5

the whole field is 13.5 m by 9 m, with a 9 m divider as shown in your diagram.