find the dimensions of the largest rectangle whose perimeter is 4000 feet. (enter the dimensions from smallest to largest.) side: side:

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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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.

 

 

Sources:

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