Farmer ed has 500meters of fencing, and wants to enclose a rectangular plot that borders on a river. if farmer ed does not fence the side along the river, find the length and width of the plot that will maximize the area. what is the largest area that can be enclosed?

The length is 125 and the width is 250 for a maximum area of 3125.

Use the equations A=L*w and 500=2L+w (the second one represents perimeter)

The second one can be rewritten as w=500-2L and this can be plugged into the first to get A=L(500-2L), which can be rewritten as 500L-2L^2

You then take the derivative to get A'=500-4L and set this equal to 0 to get a maximum area. 
500=4L
L=125 and plug this into the original to get W=250 and A=3125

Hope this helps

Perimeter=500
P=2(l+w)
250=l+w
l=250-w

Area=l*w
A=w*(250-w)
A=250w-w^2
A=-w^2+250w (parabolic eqn)

The vertex of a parabola is the point (h, k), where h = –b/2a .  In this case:
h=-250/2*-1
h=125

To find the "k" part of the vertex, all I do is plug 125 in w
k= -w^2+250w (parabolic eqn)
k=- (125)^2+250*125
k=15625

My points from this equation are (W, A) — that is, I plug in a width and figure out the area — so the "h" is the maximizing width and the "k" is the maximum area. So the answer is:

15625 is maximum area

Now u can calculate L(length)

And i oop, gurlll he not gonna have no more left .. anna ou