SOLUTION: A piece of wire 57 inches long is cut into two pieces, and two rectangles are formed from the pieces. If you want the sum of the areas of the two rectangles to be as large as poss

Question 261760: A piece of wire 57 inches long is cut into two pieces, and two rectangles are formed from the pieces. If you want the sum of the areas of the two rectangles to be as large as possible, how long should you make the pieces?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

For any given perimeter, the rectangle with the greatest area is a square with sides that measure the perimeter divided by 4. This can be proven rather easily in general, so I'm just going to ask you to take it on faith at this point.

Given that fact, the maximum area of the sum of the two rectangles described would be achieved when each of the rectangles was a square, one of which has sides that measure some portion of 57 inches, let's call that measure

, and the other has sides that measure the balance of the 57 inches, or

.

The area of the first square must then be

and the area of the second square must be

. The sum of the areas is then:

Such function having a domain of

. Note that the domain for this function necessarily excludes the endpoints because if either

you would only have one rectangle and the problem specifies that there are two rectangles.

Now we come to the real problem here. You cannot solve this problem. If you expand the binomial above and collect like terms the result is:

Note that this is a quadratic which would have a graph that is a portion of a parabola. I say a portion of a parabola because of the aforementioned restrictions on the domain of the function. Since the lead coefficient is positive, the parabola opens upward. Hence the function value at the vertex of the parabola is a minimum and the function, with the domain restrictions only has a maximum AT the excluded endpoints. The point is, no matter how close to zero you make

or how close to 57 you make

, someone can always come along and get closer on either end. Hence the length of the pieces for maximum area of the two rectangles cannot be specified as long as the problem insists on two rectangles.

John

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