SOLUTION: I really appreciate the help i'm getting from this website! Question: A rectangle has perimeter P. Find the maximum possible area of the rectangle. Thank you so so so so so so much
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Question 454480: I really appreciate the help i'm getting from this website! Question: A rectangle has perimeter P. Find the maximum possible area of the rectangle. Thank you so so so so so so much for your time and effort!
Found 2 solutions by rfer, solver91311:
Answer by rfer(16322) (Show Source): You can put this solution on YOUR website!
A square has the maximum area
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Let P represent the fixed perimeter.
Let w represent the width of the rectangle.
Let l represent the length of the rectangle.
The perimeter of a rectangle is:
So
The area of a rectangle is the length times the width so a function for the area in terms of the width is:
 = \left(\frac{P - 2w}{2}\right)w)
 = \frac{Pw - 2w^2}{2})
 = \frac{P}{2}w - w^2)
Algebra Solution:
The area function is a parabola, opening downward, with vertex at:
Since the parabola opens downward, the vertex represents a maximum value of the area function. The value of the width that gives this maximum value is one-fourth of the given perimeter. Therefore, the shape must be a square, and the area is the width squared.
Calculus Solution:
The area function is continuous and twice differentiable over its domain, therefore there will be a local extrema wherever the first derivative is equal to zero and that extreme point will be a maximum if the second derivative is negative at that point.
Therefore the maximum area is obtained when
And that maximum area is:
 = \frac{P}{2}\left(\frac{P}{4}\right) - \left(\frac{P}{4}\right)^2 = \left(\frac{P}{4}\right)^2 = w^2)
John
My calculator said it, I believe it, that settles it
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