The owner of a used tire store wants to construct a fence

Question

The owner of a used tire 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 460 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.

Length of parallel side(s) =

Length of perpendicular sides =

Answer 1

Assuming that the side of the store which is 260 feet is long enough to be the side of the rectangle, then there will be 3 lengths cut from the 460 foot length. Then

The 3 lengths cut from the wire will be WLW and L will be the length used by the side of the building.

The area formed by the 3 sides of the wire A = LWW = LW²

The wire Perimeter P = L + 2W = 460' using all of the wire fencing.

A = (460 - 2W)W² = 460W² - 2W³

Taking the derivative set to 0 to find the critical points,

A' = 920W - 6W²

A' = 0 = W(920 - 6W)

W = 0 or 153.33 feet which is 1/3 of the 460 feet fence wire.

Note. This says that the shape should be a square which is the figure that proves the maximum area.

Then the parallel side and perpendicular sides are each 153.33 feet,

and the side of the store used is also 153.33 feet

Another way to arrive at these figures is knowing that a SQUARE provides the maximum area from a given length of wire, than any rectangle given that all of the wire is used to make the perimeter.

Then you don't have to know Calculus to get the maximum size, you can work like a carpenter or handyman who just knows these things from experience...