# 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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### 1 Answer

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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