SOLUTION: what is the largest rectangular region that can be enclosed with fencing on all four sides? a rectangular region can be enclosed with on side against the house and fencing on al

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Question 254482: what is the largest rectangular region that can be enclosed with fencing on all four sides?
a rectangular region can be enclosed with on side against the house and fencing on all three sides. what are the dimensions of the largest rectangular region that can be enclosed with fencing on three sides?

Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
If there are no restrictions, we have the following equations:
(i) P+=+2L+%2B+2W
(ii) A+=+LW
solve (i) for L and we get
(iii) L+=+%28P-2W%29%2F2
substitute this in for (ii) to get
(iv) A+=+%28%28P-2W%29%2F2%29%2AW = -1W%5E2+%2B+PW%2F2
the area is a parabola opening down, we can find the vertex using -b/2a as
-b/2a = -(p/2)/(-2) = p/4
and then put p/4 into the original equation to get
A = -(p/4)^2 + p(p/4)/2 = -p^2/16 + p^2/8 = p^2/16.
But this is the same thing as
A+=+%28p%2F4%29%5E2 which is a square.
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If there is a house on one side and fencing on the other three sides, we get:
v) P+=+L+%2B+2W
(vi) A+=+LW
solve (i) for L and we get
(iii) L+=+P-2W
substitute this in for (ii) to get
(iv) A = (P-2W)*W = -2W^2 + PW
the area is a parabola opening down, we can find the vertex using -b/2a as
-b/2a = -(p)/(-4) = p/4
and then put p/4 into the (iv) to get
A = -2(p/4)^2 + p(p/4) = -p^2/8 + p^2/4 = p^2/8.
But this is the same thing as
A+=+%281%2F2%29%2A%28p%2F2%29%5E2