Hello!
Unfortunately you have no drawings attached, but I can envision.
I suppose all three corrals has one common dimension, denote its length as Then there are 4 walls with this length and 2 walls with another length, denote it as
So we have to use (ft) of fence while the enclosed area to maximize is
Express from the equation,
and substitute it into the formula for the area:
Of course and
so
The graph of this function is a parabola branches down. It has a maximum at a point (ft). The corresponding
(ft).
The answer: the bounding walls will have a length of 500 ft and the separation walls will have a length of 250 ft.
I believe you can now solve this problem for corrals with a different mutual arrangement.
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