math
posted by kayla on .
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?

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 
a farmers wants fence 250 h/a of land,
how much meters are in there?