# 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?

##### 2 Answers

For maximum area, all of the wire should be used to construct the square.

The minimum total area is obtained when

#### Explanation:

Let's say that

Then the length of the side of the square

On the other hand, the side of the equilateral triangle will then be

Hence the total area is

To extremize this, we find the derivatives

and

Thus, the extremum is at

Since

Thus the minimum area of

This is the only extremum for

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.

Minimum Area occurs is we use

Maximum Area occurs is we use

#### 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

The total perimeter is that of

# 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,

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

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

At a critical point we have

# :. 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

# {: (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

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