Question

A piece of wire 5 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? m (b) How much wire should be used for the square in order to minimize the total area?

Answer 1

a) For maximum area, all the wire should be used to make the square.

b) For minimum area, 2.17 should be used for square and 2.83 should be used for the equilateral triangle.

Explanation:

Given : A piece of wire 5 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.

To find :

(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?

Solution :

A piece of wire 5 m long is cut into two pieces.

One piece is bent into a square let the amount cut be 'x'.

The other is bent into an equilateral triangle let the amount left be '5-x'.

Side length of the square is
`(x)/(4)`

So, The area of the square is
`A=((x)/(4))^2=(x^2)/(16)`

Side length of the equilateral triangle is
`(5-x)/(3)`

So, The area of the equilateral triangle is
`A=(\sqrt3)/(4)((5-x)/(3))^2=(\sqrt3(5-x)^2)/(36)`

Total area = Area of square +Area of equilateral triangle

`A(x)=(x^2)/(16)+(\sqrt3(5-x)^2)/(36)`

Differentiate w.r.t x,

`A'(x)=(x)/(8)-(\sqrt3(5-x))/(18)`

To find critical point put A'(x)=0

`(x)/(8)-(\sqrt3(5-x))/(18)=0`

`(18x-8\sqrt3(5-x))/(18* 8)=0`

`18x-8\sqrt3(5-x)=0`

`18x-40\sqrt3+8\sqrt3x=0`

`x(18+8\sqrt3)=40\sqrt3`

`x=(40\sqrt3)/(18+8\sqrt3)`

`x\approx 2.17`

Now, The domain of x is [0,5]

So, End point are 0 and 5.

Substitute x=0,2.17,5 in the total area,

`A(0)=(0^2)/(16)+(\sqrt3(5-0)^2)/(36)`

`A(0)=0+(\sqrt3* 25)/(36)`

`A(0)\approx 1.20`

`A(2.17)=((2.17)^2)/(16)+(\sqrt3(5-2.17)^2)/(36)`

`A(2.17)\approx 0.67`

`A(5)=(5^2)/(16)+(\sqrt3(5-5)^2)/(36)`

`A(5)\approx 1.56`

a) For maximum area, all the wire should be used to make the square.

b) For minimum area, 2.17 should be used for square and 5-2.17=2.83 should be used for the equilateral triangle.