Posted by **kayla** on Sunday, October 26, 2008 at 8:48pm.

a farmer with 10,000 meters of fencing wants to enclose a rectangular field and then divide it into two plots with a fence parallel to one of the sides. What is the largest area that can be enclose?

- math -
**Damon**, Sunday, October 26, 2008 at 9:07pm
3 w + 2 L = 10,000

so

w = (10,000 -2 L)/3

A = w L = (10,000/3)L -(2/3)L^2

that is a parabola (quadratic)

we could find the vertex or we could use calculus

first find the vertex

A is 0 when L = 0

A is 0 when (2/3)L =10,000/3

or when L = 5000

so vertex is halfway between

L = 2500

then

w = (10,000 - 5,000) /3 = 5,000/3 = 1667

Now with calculus

dA /dL = w dL/dL + l dw/dL = w + L dw/dL

when dA/dL = 0 we have an extreme

w = -L dw/dL

but w = 10,000/3 - (2/3) L

so

dw/dL = -2/3

so

w = (2/3) L

so

2 w = 10,000/3

w = 5000/3

then L = 10,000/4 = 2,500

which we knew from the parabola anyway

- math -
**dudu**, Saturday, October 17, 2015 at 11:18am
a farmers wants fence 250 h/a of land,

how much meters are in there?

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