SOLUTION: A carpenter is building a rectangular room with a fixed perimeterof 112 ft. What dementions would yeild the maximum area? What is the maximum area? The lenght that would yeild t

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Question 334210: A carpenter is building a rectangular room with a fixed perimeterof 112 ft.
What dementions would yeild the maximum area? What is the maximum area?
The lenght that would yeild the maximum are is __?__ft
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Let's solve this one in general, that is for any given perimeter.

Let w represent the width of the field.

Let l represent the length of the field.

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:








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 available fencing. Therefore, the shape must be a square, and the area is the width squared.

Calculus Solution:

The area function is continuous and twice differentiable across 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.









Therefore the maximum area is obtained when



And the shape is therefore a square.

And that maximum area is:



John

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