I hav 150m of wire 2 build around a rectngular lot.If a 10m openin s left,wat wud b d dimensions 4 max. area?

Question

Please use the quadratic function in solving. Sorry I've not specified yesterday that it was a rectangular lot. Please do help me,it's my project please.

Answer 1

Well, u can make believe u have 160 m of wire since u allow 4 an opening. So the perimeter is 160m = 2 L +2W
Area = L*W
U can substitute in for either L or W from the perimeter eqtn. Then take derivitive dA/dL or dA/dW, which ever is the surviving variable. Set that = to 0 and find L or W.
OR
U can be lazy and realize the area is square, so 160=4L and A= (40^2)=1600 m^2

Answer 2

For a given perimeter for a rectangle the area will be maximum it is a square.
The perimeter is 4 x 40 = 160 after leaving a gap of 10 m the wire needed is 150 m
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Total length = 2 a + 2b -10 = 150
2 a + 2b = 160 implies a = [160 - 2b] /2
Area A = ab = [160 - 2b] b /2 = [160 b - 2b^2] /2 = 80b - b^2
Differentiating
A'(b) = 80 - 2b.
A'(b) = 0 implies 2b = 80 => b = 40 => a = 40
A" = negative and hence A is the maximum at a = 40m.
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Or from
A = 80b - b^2
b^2 -80b + A = 0
b = 80 ± √ [6400 - 4A] / 2
b = 40 ± √ [1600 - A]
For maximum value and for real roots
1600 = A and hence b = 40.m
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