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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?

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sdmatera
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)