SOLUTION: Good afternoon The perimeter of a rectangle is 50 yds. What are the dimensions that will produce the maximum area of such a rectangle? Thanks :D

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Question 867777: Good afternoon
The perimeter of a rectangle is 50 yds. What are the dimensions that will produce the maximum area of such a rectangle?
Thanks :D
Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

The maximum area for a rectangle of a given perimeter is when the the rectangle is a square with side measure .

Let represent the perimeter of some arbitrary rectangle. Then since , we can say that the width is and the length is

Since area is length times width, we can write an expression for area as a function of width, thus:

which we recognize from algebra as a quadratic in .

The graph of such a quadratic using as the independent variable is a parabola opening downward. Hence the value of the function at the vertex is the maximum value of the function. From algebra we know that the horizontal coordinate of the vertex of a parabola described by the equation:

is given by:

So for the area function developed above:

If you achieve the maximum area when the width is one-fourth of the perimeter, then the length must also be one-fourth of the perimeter. That is, a square with side length one-fourth of the perimeter as claimed.

John

My calculator said it, I believe it, that settles it