# find the dimensions of the largest rectangle whose perimeter is 4000 feet. (enter the dimensions from smallest to largest.) side: side: Let us assume the dimensions as follws.

Length = xft

Width = yft

perimeter`(S) = x+x+y+y = 2(x+y)`

`2(x+y) = 4000`

` y = 2000-x`

Area of the rectangle` (A) = x*y`

`A = x(2000-x) = 2000x-x^2`

When the area is maximum or minimum then...

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Let us assume the dimensions as follws.

Length = xft

Width = yft

perimeter`(S) = x+x+y+y = 2(x+y)`

`2(x+y) = 4000`

` y = 2000-x`

Area of the rectangle` (A) = x*y`

`A = x(2000-x) = 2000x-x^2`

When the area is maximum or minimum then `(dA)/dx = 0`

`(dA)/dx = 2000-2x`

When `(dA)/dx = 0` ;

`2000-2x = 0`

`x = 1000`

If A is a maximum then `(d^2A)/(dx^2) < 0` at x = 1000

`(d^2A)/(dx^2) = -2 < 0`

`So (d^2A)/(dx^2) < 0`

This means A has a maximum.

The area will be maximum when;

x = 1000ft

y = 1000ft

At maximum area the rectangle will become a square.

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