Question

A piece of wire 26 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?

Answer 1

For the maximize the total area, the all wire i.e. 26 m should be used for the square.

Explanation:

Length of the wire = 26 m

Let Amount of wire cut for square = x

Amount of wire cut for triangle = 26 - x

Side of the square =
`(x)/(4)`

Area of the square =
`(x^(2) )/(16)` ------ (1)

Side of the triangle is given by

`a = (26 - x)/(3)`

Side of the triangle is
`a = (26 - x)/(3)`

Area of the triangle is given by

`A = (√(3) )/(4) a^(2)`

Area of the triangle is

`A =(√(3) )/(36) (26 - x)^(2)` ------- (2)

Now the total area = Area of square + Area of triangle

The total area =
`(x^(2) )/(16) + (√(3) )/(36) (26 - x)^(2)` ------- (3)

Differentiate above equation with respect to x we get

`A' = (x)/(8) - (√(3) )/(18) (26 - x)`

Take
`A' = 0`

`(x)/(8) - (√(3) )/(18) (26 - x) = 0` ------- (4)

By solving the above equation we get

x = 11.31 m

Again take
`A''` by differentiating equation (4)

`(x)/(8) + (√(3) )/(18) (26)`

Which is greater than zero. so the value x = 11.31 m gives the area minimum.

Thus for the maximize the total area, the all wire i.e. 26 m should be used for the square.