Question 1014500: Karl proposed a project to put up a rectangular garden whose lot perimeter is 36 meters. He was soliciting suggestions for feasible dimensions of the lot. what will i suggest if i want a maximum lot area?
Found 2 solutions by macston, FrankM:
Answer by macston(5194)   (Show Source):
You can put this solution on YOUR website! P=2(L+W)
36m=2(L+W)
18m=L+W
18-W=L
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Area=LW
Area=(18-W)(W)
Area=18W-W^2
Maximum occurs when first derivative=0
Area=f(W)
f(W)=-W^2+18W
f'(W)=-2W+18 (first derivative)
-2W+18=0
-2W=-18
W=9
The maximum area occurs when width is 9 meters.
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L=18-W=18-9=9
ANSWER:The maximum area occurs when length and width are both 9 meters.
Answer by FrankM(1040)  (Show Source):
You can put this solution on YOUR website! 1/2 perimeter is 18 meters.
Since L+W = 18, we can substitute and say L = 18-W and the area, A = (W)(18-W)
Distributing, we get A= 18W-W^2 which is a parabola, with a shape that opens down. Now, the X intercepts are 0 and 18, which is a fence with no space in between. The vertex, or maximum area occurs at 9, the midpoint between these zeros.
At 9 x 9 you get an area of 81,
From this, you should learn that the maximum area you can get given a perimeter of a rectangle is to form a square.
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