Maximum and minimum problems??? ?

Question

A rectangular field is to be put up at a water area with 240 m of fencing material. No fencing material is required for the side nearest the water area. Fencing will only be used for the 3 other sides and 2 fences for dividing the field into four pens of the same size.

You need to find the dimensions of the field so that the enclosed area is as large as possible.

I need solutions and the final answer :)

thanks!

Answer 1

Assume an area A of length l and width w with one 'l' bounded by the water.  Divide this area up into four with one internal fence of length l and one of w.

This means that you need two fences of length l and three of length w

2l + 3w = 240   ...(1)

also

A =lw

l = A/w ....(2)

Substitute this value in (1)

2A/w + 3w = 240

multiply by w

2A + 3w^2 = 240w

A = 120w - 1.5w^2

dA/dw = 120 - 3w

For maximum this = 0

120 - 3w = 0

w = 40 metres

Substitute in (1)

2l + 120 = 240

l= 60 metres

Of course, if all four pens need access to the water's edge, then you need to divide the area up differently,

so that

l + 5w = 240   ...(3)

A/w + 5w =240

A = 240w - 5w^2

dA/dw = 240 - 10w = 0

w = 24 metres

l+120 =240

l =120 metres

Which, interestingly, is a better deal in terms of area.

Answer 2

If there are three sides and the two sections of fence that divide the pens are the same length as the sides then:

240 = 5x

that means  240/5 = x

x = 48