# How much wire should be used for the square in order to minimize the total area?A piece of wire 28 m long is cut into two pieces. One piece is bent into a square and the other is bent into an...

How much wire should be used for the square in order to minimize the total area?

A piece of wire 28 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.

(b) How much wire should be used for the square in order to minimize the total area?

_____________m

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### 1 Answer

You should come up with the following notations: x expresses the length of side of square and y expresses the length of side of equilateral triangle.

You need to evaluate the perimeters of square and equilateral triangle such that:

`P square = 4x`

P triangle = `3y`

The problem provides the information that the total length of piece of wire is of 28 m such that:

`28 = 4x + 3y =gt y = (28-4x)/3:`

You need to evaluate the total area of square and triangle such that:

`A = A square + ` A triangle

`A = x^2 + y^2*sqrt3/4`

You need to write the function of area in terms of one variable, hence you may substitute `(28-4x)/3` for y in equation of total area such that:

`A(x) = x^2 + (28-4x)^2*sqrt3/36`

`A(x) = x^2 + 16(7-x)^2*sqrt3/36`

`A(x) = x^2 + 4(7-x)^2*sqrt3/9`

You need to differentiate the function A(x) with respect to x and then you need to solve the equation A'(x)=0 to find how much wire needs to be used for square to minimize the total area.

`A'(x) = 2x- 8sqrt3/9*(7-x)`

You need to solve the equation A'(x)=0 such that:

`2x - 8sqrt3/9*(7-x)= 0`

`2x- 56sqrt3/9 + 8sqrt3/9*x = 0`

`x(2 + 8sqrt3/9) = 56sqrt3/9 =gt x = (56sqrt3/9)/((18+8sqrt3)/9)`

`x = 56sqrt3/(18+8sqrt3) =gt x = 96.994/31.856`

`x = 3.044 m`

`P square = 4*3.044 =gt P square = 12.176 m`

**Hence, evaluating the length of wire to be used for square to minimize the total area yields 12.176 meters.**