Question 1491: find the length and width of a rectangle having a perimeter of 200 meteres that will maximize its area.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Let L& W be the length & width of the rectangle.
The perimeter = 2(L+W) = 200, so L+W =100 or W = 100 -L
Its area = LW = L(100 -L) = 100 L - L^2
= - (L^2 - 100 L + (100/2)^2) + (100/2)^2 [Complete square]
= 2500 - (L -50)^2
We see that when L= 50,the area has the maximum value 2500.
Also, when L = 50,W = 100 -50 = 50.
Answer: when L= 50,and W = 50, we attain the maximum area 2500.
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