SOLUTION: SOLUTION: A carpenter is building a rectangular room with a fixed perimeter of 168 ft. What dimensions would yeild the maximum area? what is the maximum area?

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Question 593762: SOLUTION: A carpenter is building a rectangular room with a fixed perimeter of 168 ft. What dimensions would yeild the maximum area? what is the maximum area?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Start with:
P+=+2%28L%2BW%29 Substitute P = 168.
168+=+2%28L%2BW%29 Divide both sides by 2.
84+=+L%2BW Rewrite as:
L+=+84-W and substitute into;
A+=+L%2AW
A+=+%2884-W%29%2AW Simplify.
A+=+84W-W%5E2 Rewrite this as:
-W%5E2%2B84W-A+=+0 This is a quadratic equation which yields a parabola opening downward when graphed. This is the graph of the area (vertical) versus the width (W) (horizontal). The maximum point (W) on this graph, also known as the vertex, is given by:
W+=+-b%2F2a and a = -1, b = 84.
W+=+-%2884%29%2F2%28-1%29
W+=+42
The dimensions for the maximum area will 42 by 42 which is a square.