math help?

(total fencing available) - (fencing already used) = 2000 - 2x = length of the third side
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A = L*W
= x*(2000 - 2x)
= 2000x - 2x^2
A(x) = 2000x - 2x^2

find x that maximizes A(x) and evaluate A(x)

500,000 m^2

(If you know calculus take the derivative and set it to zero)
So what I did:
A'(x) = 2000 - 4x
2000 - 4x = 0
x = -2000/-4
x = 500

Plug it back into the equation
A(x) = 2000x - 2x^2
A(x) = 2000*500m - 2*(500m)^2 = 500,000 m^2

I doubt she knows calc if she is stubbed on a simple word problem.

The Calculus free solution

l field length, h is width, A is area
A=lh
l+2h=2000 so l=2000-2h

A=lh=(2000-2h)h=2000h-2h^2

2h^2-2000h+A=0 divide by 2

h^2-1000h+(A/2)=o

This is a quadratic eqn. use [-b+/-sqrt(b^2-4ac)]/(2a)
a=1, b=-1000, c=A/2 and you get one root when sqrt(b^2-4ac)]=0

set (b^2-4ac)=0
(1000)^2-4(1)(A/2)=1,000,000-2A=0 so A=500,000
or
h^2-1000h+(A/2)=o
becomes h^2-1000h+250,000=o
and
h^2-1000h+250,000=(h-500)^2=o

so h=500m & l=1000m should maximize area

check
less 50
h'=450 l'=1100 A'=495,000 less than 500,000

more 50
h'=550 l'=900 A'=495,000 less than 500,000

so max is h=500, l=1000 A=500,000

Rap