Question

A farmer wants to enclose a rectangular field along a river on three sides. If 4,400 feet of fencing is to be used, what dimensions will maximize the enclosed area?

Answer 1

F(x,y,L) = xy-L(2x+y-4400)

Fx = y-2L = 0; y = 2L
Fy = x - L = 0; x = L
FL = 2x +y -440

2L +2L -440 = 0

4L = 440

L = 110

x = 110
y = 220

Answer 2

the the side opposite the river is half of the fencing, the other two sides use 1/4 of the fence each

Answer 3

you can bet your bottom dollar so when you see this on a test go right to the answer. the amount of fence is a red herring.
if there was no river make a square

Answer 4

@amistre you have a function of 3 variables! wow

Answer 5

I do :) L is spose to be lamda tho

Answer 6

oh, you actually lost me entirely.

Answer 7

I read the section on LaGrange multipliers the other night and have been dying to use it ;)

Answer 8

\[f(x,y)\text{ ; function to max/min}\]\[g(x,y)=0\text{ ; constraint function}\]
\[F(x,y,\lambda)=f(x,y)-\lambda\ g(x,y)\text{ ; partial derivatives to 0}\]

Answer 9

put P = perimeter of fence, x = length of side opposite river so other two sides are
\[\frac{P-x}{2}\] and area is
\[A(x)=\frac{x(P-x)}{2}=\frac{P}{2}x -\frac{x^2}{2}\] a parabola facing down whose vertex is at
\[\frac{-b}{2a}=\frac{P}{2}\]
so opposite side is half the perimeter and other two are one -fourth

Answer 10

it has something to do with lining up the gradients of each function and making them equal

Answer 11

you gotta be kidding!

Answer 12

simple algebra does this, not calculus, but i am impressed for sure!

Answer 13

Satellite73- how much would that be in feet