# A 10 m wire is cut into 2 parts and bent to form a rectangle with sides x and y and a square. Find the dimensions x and y that make the total area maximum.

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I suggest you to use the following notation: x denotes the length and y denotes the width.

You know the total dimension of the wire such that you may use the perimeter formula such that:

`P = 2(x+y) =gt 10 = 2(x+y) =gt x + y = 5`

Since the problem is a maximum optimization problem, you need to find what is the constraint equation and what is optimization equation. You know the total length of the wire, hence the constraint equation is the perimeter equation. Since the optimization equation is the area equation, use the perimeter equation to solve for y variable.

Hence you should express the area of rectangle with respect to one of the two dimensions defining the rectangle. You need to remember the formula of the area of rectangle.

`A = x*y`

`y = 5 - x`

Plugging `y = 5-x` in the formula of area yields:

`A(x) = x*(5-x)=gt A(x) = 5x - x^2`

You need to differentiate the area equation above with respect to x:

`A'(x) = 5 - 2x`

If A'(x) is maximum, then `A'(x) = 0 =gt 5 - 2x = 0 =gt x = 5/2`

`y = 5 - 5/2 =gt 5/2`

**Hence, area is maximum when the dimensions x and y are equal and the rectangle becomes a square: x = y = `5/2` m.**