Question 758375
I need to find the optimum perimeter for a rectangle with the area of 242cm squared.
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You can find the minimum perimeter, not sure what optiumum would be.
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Pick a perimeter and find the max area: P = 100
P = 2L + 2W = 100
L + W = 50
L = 50 - W
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Area = L*W = W*(50 - W)
Area = 50W - W^2
W^2 - 50W + Area = 0
It's a quadratic in W
Ther vertex is on the Axis of Symmetry which is
W = -b/2a
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W = 50/2 = 25
L = 25
---> Max area of a rectangle of given perimeter is a square.
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For Area = 242
L = W = sqrt(242)
= {{{11*sqrt(2)}}} cm