# What is the maximum area that he can enclose with his materials? A farmer has 160 m of fence to enclose a rectangular area against a straight river. He only needs to fence in three sides. What...

What is the maximum area that he can enclose with his materials?

A farmer has 160 m of fence to enclose a rectangular area against a straight river. He only needs to fence in three sides. What is the maximum area that he can enclose with his materials?

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Let the dimensions for the rectangle be w and l.

Then we have 2w+l=160 so l=160-2w. (We only have 2w+l, not 2w+2l since one of teh sides is covered by the river.)

We want to maximize the area. The area is given by A=lw. Substituting for l we get A=w(160-2w).

To find the maximum, we take the first derivative and find the critical points.

`A=160w-2w^2 ==> (dA)/(dw)=160-4w` . The derivative exists everywhere, so the only critical points are when `(dA)/(dt)=0` :

160-4w=0 ==> w=40.

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**The area is maximized when the dimensions are 40x80, the 2 widths are 40m and the single length is 80m. The area is `3200m^2` .**

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