Question

A rancher has 5100 ft of fencing to enclose a rectangular area bordering a river. He wants to divide the enclosure into two equal areas. If no fencing is used along the river, what is the length of the center partition that will yield the maximum area?

Answer 1

We have a perimeter `3l+w=5100` so an area `A=lw=l(5100-3l)`. Completing the square `A=-3(l-850)^2 + text{constant}` so a maximum when the center partition is `l=850.`

Explanation:

Instead of the rectangle having a perimeter `2l+2w` the perimeter fencing used in this problem is `3l + w`. That's one less width and the center partition an extra length.

Maximize `A=lw` subject to `3l+w=5100.`

`w= 5100 - 3l`

`A = lw = l(5100 -3l) = -3l^2 + 5100l`

`A = -3 (l^2 - 1700 l)`

`A = -3(l^2 - 1700 l + (1700/2)^2 ) + 3 (1700/2)^2`

`A = -3(l - 850)^2+ 3 (1700/2)^2`

That's clearly maximized when `l=850.`