Find the length L and width W (with W≤L ) of the rectangle?
Find the length L and width W (with W≤L ) of the rectangle with perimeter 44 that has maximum area, and then find the maximum area.
L =
W =
Maximum area =
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Rick is correct, but the calculation is so simple here it is:
L = 44/2-W = 22-W
Area A = W.(22-W) = 22.W - W²
dA/dW = 0 for a maximum giving 2.W – 22 = 0 or W=11 giving in turn L=11 and A =121
This is a square so each side is 11 and the area is 121. You could solve this with calculus, but it is commonly known that the maximum area is that of a square.