You have 196 feet of fencing to enclose a rectangular region. Find the Dimensions of the rectangle that maximizes the enclosed area.
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Let us say the lengths are X and Y
So 2(X+Y) gives the length of the fencing.
`2(X+Y) = 196`
`X+Y = 98`
`Y = 98-X`
If the area is A;
`A = X*Y`
`A = X*(98-X)`
`A = 98X-X^2`
When area is maximum/minimum then `(dA)/(dX) = 0`
`(dA)/(dX) = 98-2X`
When `(dA)/(dX) = 0;`
`98-2X = 0`
`X = 49`
If we have maximum at `X = 49 then (d^2A)/(dX^2)` at that point will be negative.
`(d^2A)/(dX^2) = -2 < 0` (negative)
So we have maximum for area at X = 49.
When X = 49 then Y = 98-49 = 49.
So the dimensions of the rectangle is 49ft which means it becomes a square.
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