SOLUTION: An architect is designing an atrium for a hotel. The atrium is to be rectangular with a perimeter of 720ft of brass piping. What dimensions will maximize the area of the atrium?

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Question 191814: An architect is designing an atrium for a hotel. The atrium is to be rectangular with a perimeter of 720ft of brass piping. What dimensions will maximize the area of the atrium?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
The perimeter of a rectangle is given by:
P+=+2%28L%2BW%29 and this is equal to 720. so...
2%28L%2BW%29+=+720 Divide both sides by 2 to get:
L%2BW+=+360 which can be written as:
L+=+360-W
The area of a rectangle is given by:
A+=+L%2AW Substitute L+=+360-W to get:
A+=+%28360-W%29%2AW Simplify.
A+=+360W-W%5E2 Now you have a quadratic equation:
-W%5E2%2B360W+=+A When graphed, this will show a parabola opening downwards so there will be a maximum point on the curve and this occurs at the vertex of the parabola. To find the value of W that corresponds to this point, use W+=+%28-b%29%2F2a where a = -1 and b = 360.
W+=+%28-360%29%2F2%28-1%29
W+=+180
So the maximum area is obtained when W = 180ft. and L = 180ft.
This should come as no surprise as it is well-known that the maximum area of a rectangle enclosed by a given perimeter is, in fact a square.
Check:
P+=+2%28L%2BW%29 Substitute L = 180 and W = 180
P+=+2%28180%2B180%29
P+=+2%28360%29
P+=+720 ...the given perimeter.