SOLUTION: A rancher has 400 feet of fencing to put around a rectangular field and then subdivide the field into 3 identical smaller rectangular plots by placing two fences parallel to one of

Question 1038991: A rancher has 400 feet of fencing to put around a rectangular field and then subdivide the field into 3 identical smaller rectangular plots by placing two fences parallel to one of the field's shorter sides. Find the dimensions that maximize the enclosed area. Write your answers as fractions reduced to lowest terms.
--please help me!
Found 2 solutions by addingup, solver91311:
Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
So we have two long sides, and four short sides that divide the field into three identical rectangular plots.
Length of each shorter side: x
Length of each longer side: 1/2(400-4x) = 200-2x
Total area:
A(x) = x(200-2x) = 200x-2x^2
A'(x) = 200-4x
A' = 0 when x = 200/4 = 50
So your 2 long sides are 100 each and your 4 short sides are 50 each
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Algebra Method

Negative lead coefficient so parabola opens down. Hence, vertex is a maximum.

The -coordinate of the vertex of a parabola is given by . So:

Overall dimensions are 100 by 50. Short side of each small rectangle is

Calculus Method

Set the first derivative equal to zero and solve for the coordinates of any extreme points.

Since the second derivative is negative for all in the domain of , is the -coordinate of a maximum.

As above:

Overall dimensions are 100 by 50. Short side of each small rectangle is

John

My calculator said it, I believe it, that settles it

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