SOLUTION: A fourth-grade class decides to enclose a rectangular garden, using the side of the school as one side of the rectangle. What is the maximum area that the class can enclose with 3

Question 33263: A fourth-grade class decides to enclose a rectangular garden, using the side of the school as one side of the rectangle. What is the maximum area that the class can enclose with 32 ft of fence? What should the dimensions of the garden be in order to yield this area?
Found 2 solutions by mukhopadhyay, venugopalramana:
Answer by mukhopadhyay(490) (Show Source): You can put this solution on YOUR website!
Total length of the fence = Sum of three sides of the garden;
Sum of three sides of the garden =length+width+width;
If length of the fence = x ft, the width of the fence would be = (32-x)/2;
The area of the fenced portion = Length*WIdth = x(32-x)/2;
If A(x) is the function for area, A(x) = x(32-x)/2 = -x^2/2 + 16x;
A(x) is a parabola, opening downward; the maximum value is attained at its vertex; x-coordinate of Vertex of A(x) = -b/2a = 16/1 = 16;
So, the length and width of the fence to maximize the area is 16 ft and 8 ft respectively.

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
TOTAL FENCING=32=L+L+B+B=2(L+B)
L+B=16.......L=16-B
AREA=LB=B(16-B)=16B-B^2=-(B^2-16B)=-(B^2-2B*8+8^2-8^2)=64-(B-8)^2
(B-8)^2BENIG A PERFRC SQUARE IS ALWAYS POSITIVE.SINCE IT IS HAVING A MINUS SIGN INFRONT AREA WILL BE MAXIMUM WHEN THIS IS ZERO..THAT IS B-8=0..OR...B=8
SO L=8 AND B=8 IS THE SOLUTION WHICH GIVES MAXIMUM AREA OF 64
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