Question

A piece of wire 11 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How much wire should be used for the square in order to maximize and minimize the total area?

Answer 1

For maximum area, all of the wire should be used to construct the square.
The minimum total area is obtained when `(44)/(4+3sqrt 3)\ "m"~~4.78\ "m"` is used for the square.

Explanation:

Let's say that `x` m is used for the square.

Then the length of the side of the square `= x/4` m and its area `=x^2/16\ "m"^2`

On the other hand, the side of the equilateral triangle will then be `a= (11-x)/3\ "m"` and its area will be `sqrt 3/4a^2 = sqrt3/36(11-x)^2\ "m"^2`

Hence the total area is `A\ "m"^2` where

`A = x^2/16+ sqrt3/36(11-x)^2`

To extremize this, we find the derivatives

`(dA)/dx = x/8-sqrt 3/18 (11-x)`

and

`(d^2A)/dx^2 = 1/8+sqrt3/18>0`

Thus, the extremum is at

`(dA)/dx = 0 implies`
`x/(11-x) = (4sqrt 3)/9 = 4/(3sqrt 3)implies`
`x/11 = (4)/(4+3sqrt 3)implies`
`x = (44)/(4+3sqrt 3)~~4.78`

Since `(d^2A)/dx^2 >0` this is a minimum.

Thus the minimum area of `~~3.29\ "m"^2` will be obtained when
`4.78\ "m"` is used for the square.

This is the only extremum for `x in (0,11)` and so the maximum occurs at a boundary of the allowed values. It is easy to see that the absolute maximum is for `x= 11`.

This is understandable, because for a given perimeter, a square has more area than a triangle - hence to get the largest area we need to use all of the wire to construct the square.

Answer 2

Minimum Area occurs is we use `12sqrt(3)-16 ~~ 4.786609 \ m` for the square.

Maximum Area occurs is we use `11 \ m` for the square.

Explanation:

Let us set up the following variables:

# { (t, "Length of a side of the triangle",), (l, "Length of a side of the square",), (P_t, "Perimeter of a side of the triangle",=4t), (P_s, "Perimeter of a side of the square",=4s), (A_t, "Area of the triangle",), (A_s, "Area of the square",), (A, "Total Area",=A_t+A_s) :} #

Our aim is to find `A(t,l)`, as a function of a single variable and to maximize the total area, `A`, wrt that variable (It won't matter which variable we do this with as we will get the same result). ie we want a critical point of `A` wrt the variable.

The total perimeter is that of `4` sides of a square and the `3` sides of the triangle; we are told that this perimeter is `11` (m)

`P_s + P_t =11 => 4l+3t = 11`
`:. t = 11`

Similarly, the total Area is that of a square and the triangle:

`A = A_s+A_t`

`\ \ \ = (l)(l) + 1/2(t)(t)sin60^o`

`\ \ \ = l^2 + 1/2t^2 \ sqrt(3)/2`

`\ \ \ = l^2 + sqrt(3)/4 t^2`

`\ \ \ = l^2 + sqrt(3)/4 ((11-4l)/3)^2`

`\ \ \ = l^2 + sqrt(3)/4 (11-4l)^2/9`

`\ \ \ = l^2 + sqrt(3)/36 (11-4l)^2`

We now have the Area, `A`, as a function of a single variable `l`, so differentiating wrt `t` we get:

`(dA)/(dl) = 2l + 2sqrt(3)/36 (11-4l)(-4)`

`\ \ \ \ \ \ = 2l -(2sqrt(3))/9 (11-4l)`

At a critical point we have `(dA)/(dl) =0 => 2l -(2sqrt(3))/9 (11-4l) = 0`

`:. l - sqrt(3)/9 (11-4l) = 0`

`:. 9l - sqrt(3) (11-4l) = 0`

`:. 9l - 11sqrt(3) +4sqrt(3)l = 0`

`:. (9 +4sqrt(3))l = 11sqrt(3)`

`:. l = (11sqrt(3))/(9 +4sqrt(3))`

`:. l = (11sqrt(3))/(9 +4sqrt(3))* (9 -4sqrt(3))/(9 -4sqrt(3))`

`:. l = (99sqrt(3)-44*3)/(81-16*3)`

`:. l = (99sqrt(3)-132)/(81-48)`

`:. l = (99sqrt(3)-132)/(33)`

`:. l = 3sqrt(3)-4 ~~ 1.196152`

With this value of `l` we have, for the square, and triangle:

# {: (P_s, = 4l, = 12sqrt(3)-16, ~~ 4.786609 \ m), (P_t, = 11-4l, = 27-12sqrt(3), ~~ 6.215390 \ m) :} #

Similarly, if sought, we could compute:

# {: (A_s, = l^2, = 43-24sqrt(3), ~~ 1.430780 \ m^2), (A_t, = sqrt(3)/36 (11-4l)^2, = (129sqrt(3))/4-54, ~~ 1.858638 \ m^2) :} #

We can visually verify that this corresponds to a minimum by looking at the graph of `y=A(l)` (we could also perform a second derivative test)

graph{x^2 + sqrt(3)/36*(11-4x)^2 [-10, 10, -5, 20]}

It should also be intuitive (and from the graph) that there is no maximum bound on the area, thus to maximize the area we should use all of the wire