Question

Suppose that 320 feet of fencing are available to enclose a rectangular field and that one side of the field must be given double fencing. What are the dimensions of the rectangle that will yield the largest possible area?

it has to be explained in a polynomial equation finding the maximum area.

Answer 1

Let w be width of the rectangle.
3w+2h=320 <=>
h = -3w/2+160
A(w) = w*h = w(-3w/2+160)

solve maximum value for A(w)

Answer 2

w is width. h is height. a is area.
4w + 2h = 320
a = w*h

4w + 2h = 320
-> 4w = 320-2h
-> w = 80 - h/2

so
a= (80 - h/2)*h
=80h - h^2/2

Answer 3

Ah. Zuige is correct. It's 3w. Not 4.

Answer 4

The trick is always setting up the problem. You are dealing with perimeter and one side (doesn't matter which) is doubled up. So, normal perimeter would be P=2l+2w. But since one of the sides is doubled, you'd have
\[P=320=2l+3w\]\[l=160-(3/2)w\]Area remains the same \[A=lw\]\[A=[160-(3/2)w]w\]\[A=-(3/2)w^{2}+160w\]
Find the vertex to find the maximum point.

Answer 5

As Bender said. Know how to maximize that equation?