# math

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?

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1. 3 w + 2 L = 10,000
so
w = (10,000 -2 L)/3

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

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posted by Damon
2. a farmers wants fence 250 h/a of land,
how much meters are in there?

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posted by dudu

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