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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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) (Show Source):
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
