# 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?

May 31, 2018

For maximum area, all of the wire should be used to construct the square.
The minimum total area is obtained when $\frac{44}{4 + 3 \sqrt{3}} \setminus \text{m"~~4.78\ "m}$ 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 $= \frac{x}{4}$ m and its area $= {x}^{2} / 16 \setminus {\text{m}}^{2}$

On the other hand, the side of the equilateral triangle will then be $a = \frac{11 - x}{3} \setminus \text{m}$ and its area will be $\frac{\sqrt{3}}{4} {a}^{2} = \frac{\sqrt{3}}{36} {\left(11 - x\right)}^{2} \setminus {\text{m}}^{2}$

Hence the total area is $A \setminus {\text{m}}^{2}$ where

$A = {x}^{2} / 16 + \frac{\sqrt{3}}{36} {\left(11 - x\right)}^{2}$

To extremize this, we find the derivatives

$\frac{\mathrm{dA}}{\mathrm{dx}} = \frac{x}{8} - \frac{\sqrt{3}}{18} \left(11 - x\right)$

and

$\frac{{d}^{2} A}{\mathrm{dx}} ^ 2 = \frac{1}{8} + \frac{\sqrt{3}}{18} > 0$

Thus, the extremum is at

$\frac{\mathrm{dA}}{\mathrm{dx}} = 0 \implies$
$\frac{x}{11 - x} = \frac{4 \sqrt{3}}{9} = \frac{4}{3 \sqrt{3}} \implies$
$\frac{x}{11} = \frac{4}{4 + 3 \sqrt{3}} \implies$
$x = \frac{44}{4 + 3 \sqrt{3}} \approx 4.78$

Since $\frac{{d}^{2} A}{\mathrm{dx}} ^ 2 > 0$ this is a minimum.

Thus the minimum area of $\approx 3.29 \setminus {\text{m}}^{2}$ will be obtained when
$4.78 \setminus \text{m}$ is used for the square.

This is the only extremum for $x \in \left(0 , 11\right)$ 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.

May 31, 2018

Minimum Area occurs is we use $12 \sqrt{3} - 16 \approx 4.786609 \setminus m$ for the square.

Maximum Area occurs is we use $11 \setminus m$ 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 $A \left(t , l\right)$, 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)

${P}_{s} + {P}_{t} = 11 \implies 4 l + 3 t = 11$
$\therefore t = 11$

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

$A = {A}_{s} + {A}_{t}$

$\setminus \setminus \setminus = \left(l\right) \left(l\right) + \frac{1}{2} \left(t\right) \left(t\right) \sin {60}^{o}$

$\setminus \setminus \setminus = {l}^{2} + \frac{1}{2} {t}^{2} \setminus \frac{\sqrt{3}}{2}$

$\setminus \setminus \setminus = {l}^{2} + \frac{\sqrt{3}}{4} {t}^{2}$

$\setminus \setminus \setminus = {l}^{2} + \frac{\sqrt{3}}{4} {\left(\frac{11 - 4 l}{3}\right)}^{2}$

$\setminus \setminus \setminus = {l}^{2} + \frac{\sqrt{3}}{4} {\left(11 - 4 l\right)}^{2} / 9$

$\setminus \setminus \setminus = {l}^{2} + \frac{\sqrt{3}}{36} {\left(11 - 4 l\right)}^{2}$

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

$\frac{\mathrm{dA}}{\mathrm{dl}} = 2 l + 2 \frac{\sqrt{3}}{36} \left(11 - 4 l\right) \left(- 4\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus = 2 l - \frac{2 \sqrt{3}}{9} \left(11 - 4 l\right)$

At a critical point we have $\frac{\mathrm{dA}}{\mathrm{dl}} = 0 \implies 2 l - \frac{2 \sqrt{3}}{9} \left(11 - 4 l\right) = 0$

$\therefore l - \frac{\sqrt{3}}{9} \left(11 - 4 l\right) = 0$

$\therefore 9 l - \sqrt{3} \left(11 - 4 l\right) = 0$

$\therefore 9 l - 11 \sqrt{3} + 4 \sqrt{3} l = 0$

$\therefore \left(9 + 4 \sqrt{3}\right) l = 11 \sqrt{3}$

$\therefore l = \frac{11 \sqrt{3}}{9 + 4 \sqrt{3}}$

$\therefore l = \frac{11 \sqrt{3}}{9 + 4 \sqrt{3}} \cdot \frac{9 - 4 \sqrt{3}}{9 - 4 \sqrt{3}}$

$\therefore l = \frac{99 \sqrt{3} - 44 \cdot 3}{81 - 16 \cdot 3}$

$\therefore l = \frac{99 \sqrt{3} - 132}{81 - 48}$

$\therefore l = \frac{99 \sqrt{3} - 132}{33}$

$\therefore l = 3 \sqrt{3} - 4 \approx 1.196152$

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 $y = A \left(l\right)$ (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