SOLUTION: find the dimensions of a rectangle "a" with the greatest area whose perimetter is 30 feet
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Question 183949: find the dimensions of a rectangle "a" with the greatest area whose perimetter is 30 feet
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
find the dimensions of a rectangle "a" with the greatest area whose perimetter is 30 feet
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Perimeter = 2(L + W)
30 + 2(L+W)
L+W = 15
W = 15-L
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Area = LW
Substitute to get:
Area = L(15-L)
Area = -L^2 + 15L
That is a quadratic with a = -1,b = 15
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Maximum area occurs when L = -b/2a = -15/(-2) = 15/2
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Solve for W when L = (15/2)
W = 15 - (15/2) = 15/2
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The Width and the Length are both 15/2 ft.
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Cheers,
Stan H.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The perimeter of a rectangle is given by:
The area of a rectangle is given by:
Substituting from the perimeter equation:
 = w\left(\frac{P}{2} - w\right)\ \ \Rightarrow\ \ A(w) = \left(\frac{wP}{2} - w^2\right))
This function graphs to a parabola opening downward meaning that the vertex is a maximum. The maximum value of the function, hence the maximum area, is where the value of the first derivative is equal to zero:
 = A'(w) = -2w + \frac{P}{2})
Set equal to zero:
Hence, the maximum area rectangle for a given perimeter is a square with sides of length one-fourth of the perimeter.
John
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