since you say the fence is shorter than the wall, I'm not sure what you mean when you say "using part of the side of the store for part of one of the sides." I'll assume the entire side is formed by the store wall. If that's not right, then I'm sure you can make the necessary adjustments.
As with all of these problems, the maximum area is achieved when the fence is divided equally among lengths and widths.
So, with the wall being one length, fence is used for
1 length and 2 widths.
That means that the garden is 250 x 125
To see that this is so, let
x = length
y = width
with 500' of fencing, that means that
x + 2y = 500
or,
x = 500-2y
The area is xy = (500-2y)y
This is just a parabola, with vertex midway between the roots (y=0,250), at y = 125
This is the value noted above.
The owner of a garden supply store wants to construct a fence to enclose a rectangular outdoor storage area adjacent to the store, using part of the side of the store (which is 260 feet long) for part of one of the sides. (See the figure below.) There are 500 feet of fencing available to complete the job. Find the length of the sides parallel to the store and perpendicular that will maximize the total area of the outdoor enclosure.
The size of the parallel fence is shorter than the parallel side of the store
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