# Question #40c26

Feb 12, 2017

Minimum area of $2.901$ occurs when triangle length is $2.321$ and square length is $0.760$

#### Explanation:

Let us set up the following variables:

$\left\{\begin{matrix}t & \text{Length of a side of the triangle" \\ s & "Length of a side of the square" \\ A & "Total Area}\end{matrix}\right.$

Our aim is to find $A \left(t , s\right)$, 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 $3$ sides of a triangle and $4$ side of a square; we are told that this perimeter is $10$

$\setminus \setminus \setminus 10 = 3 t + 4 s$
$\therefore s = \frac{1}{4} \left(10 - 3 t\right)$

And the total Area is that of an equilibrium triangle and a square:

$A = \left(\frac{1}{2}\right) \left(t\right) \left(t\right) \sin {60}^{o} + \left(s\right) \left(s\right)$
$\setminus \setminus \setminus = \left(\frac{1}{2}\right) {t}^{2} \left(\frac{1}{2} \sqrt{3}\right) + {s}^{2}$
$\setminus \setminus \setminus = \frac{1}{4} \sqrt{3} \setminus {t}^{2} + \frac{1}{16} {\left(10 - 3 t\right)}^{2}$

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

$\frac{\mathrm{dA}}{\mathrm{dt}} = \left(2\right) \left(\frac{1}{4} \sqrt{3}\right) t + \frac{1}{16} \left(2\right) \left(10 - 3 t\right) \left(- 3\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} \sqrt{3} \setminus t - \frac{3}{8} \left(10 - 3 t\right)$

At a critical point we have $\frac{\mathrm{dA}}{\mathrm{dt}} = 0 \implies$

$\frac{1}{2} \sqrt{3} \setminus t - \frac{3}{8} \left(10 - 3 t\right) = 0$
$\therefore \frac{1}{2} \sqrt{3} \setminus t - \frac{15}{4} + \frac{3}{4} t = 0$
$\therefore \left(\frac{1}{2} \sqrt{3} + \frac{3}{4}\right) t = \frac{15}{4}$
$\therefore t = \frac{15}{4 \left(\frac{1}{2} \sqrt{3} + \frac{3}{4}\right)}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{15}{2 \sqrt{3} + 3}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 10 \sqrt{3} - 15$
$\setminus \setminus \setminus \setminus \setminus \setminus \approx 2.3205080 \ldots$

With this value of $t$ we have:

$A = \frac{1}{4} \sqrt{3} {\left(10 \sqrt{3} - 15\right)}^{2} + \frac{1}{16} {\left(10 - 3 \left(10 \sqrt{3} - 15\right)\right)}^{2}$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{2125}{16} - 75 \sqrt{3}$
$\setminus \setminus \setminus \setminus \approx 2.908689 \ldots$

And:

$s = \frac{1}{4} \left(10 - 3 t\right)$
$\setminus \setminus = \frac{1}{4} \left(10 - 3 \left(10 \sqrt{3} - 15\right)\right)$
$\setminus \setminus = \frac{55}{4} - \frac{15}{2} \sqrt{3}$
$\setminus \setminus \approx 0.75961894 \ldots$

We can visually verify that this corresponds to a maximum by looking at the graph of $y = A \left(t\right)$:

graph{1/4sqrt(3) x^2 + 1/16(10-3x) ^2 [-5, 10, -20, 20]}

So minimum area of $2.901$ occurs when triangle length is $2.321$ and square length is $0.760$