# A piece of wire 6 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.A piece of wire 6 m long is cut into two pieces. One piece is...

A piece of wire 6 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.

A piece of wire 6 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.

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

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

sciencesolve | Teacher | (Level 3) Educator Emeritus

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a)The problem provides you the information that the piece of wire is cut into two pieces, a square and equilateral triangle.

You should come up with the substitutions for the perimeters of the equilateral triangle and perimeter of square such that:

x = perimeter of square

y = perimeter of equilateral triangle

The total length of 6 meters may be evaluated using the perimeters of equilateral triangle and the square such that:

`6 = x+y`

Notice that the length of the side of square is of  `x/4 ` and the length of the side of equilateral triangle is of `y/3` .

You need to express the area of square such that:

A square = `(x/4)^2`  => A square = `(x^2)/16`

You need to express the area of equilateral triangle such that:

A triangle = ((y^2)/9)*(sqrt3/4)

You need to maximize the total area, hence you should differentiate the function:

`A = (x^2)/16 + ((y^2)/9)*(sqrt3/4)`

You need to write the total area in terms of one variable, hence you may use the equation x+y= 6=>x=6-y.

`A(y) = ((6-y)^2)/16 + ((y^2)/9)*(sqrt3/4)`

You need to differentiate the function with respect to y such that:

`A'(y) = -2(6-y)/16 + (2y/9)*(sqrt3/4)`

`A'(y) = -(6-y)/8 + (y/9)*(sqrt3/2)`

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

`-(6-y)/8 + (y/9)*(sqrt3/2) = 0`

`-9(6-y)+ 4sqrt3*y = 0 =gt -54 + 9y + 4sqrt3*y = 0`

`y(9+4sqrt3) = 54 =gt y = 54/(9+4sqrt3)`

Notice that the second derivative of the function is positive, hence the function reaches its minimum at `y = 54/(9+4sqrt3).`

You may find the maximum area at `x=0`  and `x=6`  (endpoints of wire) such that:

`A = (6^2)/16 + ((0^2)/9)*(sqrt3/4)`

`A = 36/16 =gt A = 9/4`

`A = (0^2)/16 + ((6^2)/9)*(sqrt3/4)`

`A = sqrt3`

Hence, you should use the following lengths for the square for the total area to be maximum: `x=0 mor x=6 m` .

b) The function A(y) reaches its minimum at `y = 54/(9+4sqrt3), ` hence, you may evaluate the length of the square such that:

`x = 6 - 54/(9+4sqrt3) =gt x = (54 + 24sqrt3 - 54)/(9+4sqrt3)`

`x = (24sqrt3)/(9+4sqrt3)`

Hence, you should use the following lengths of wire to minimize the total area such that `x = (24sqrt3)/(9+4sqrt3) m`  and `y = 54/(9+4sqrt3) m.`

thez | Student, Grade 11 | (Level 1) eNoter

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Well the let us assume that length of wire used for square is x and for the triangle is 6-x

then the area of the two shapes are going to be:

Square: (x/4)^2

Triangle: 1/2*a*b*sinC : ((sqrt3)/2)*(6-x)^2

and the total area will equal to the sum of these two equatons.

Then use differentiation to find the turning points of the curve which will be the max/min value of the curve.

Below is a graph using Wolfram showing the results

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