The story the square and the equilateral triangle

Let $x$ be the side length of the equilateral triangle. Therefore, the side length of the square is $\dfrac {12-3x}4$ . The total area of the triangle and square is given by:

$\begin{aligned} A & = A_\triangle + A_\square = \frac {\sqrt 3}4x^2 + \left(\frac {3(4-x)}4\right)^2 \\ & = \frac {\sqrt 3}4x^2 + \frac 9{16} (x^2-8x+16) \\ & = \left(\frac {\sqrt 3}4 + \frac 9{16} \right)x^2-\frac 92 x+9 \\ & = \frac {4\sqrt 3 + 9}{16} x^2-\frac 92 x+9 \\ \frac {dA}{dx} & = \frac {4\sqrt 3 + 9}8 x-\frac 92 \end{aligned}$

$A$ is $\red{\text{minimum}}$ , when $\dfrac {dA}{dx} = 0$ or when:

$\begin{aligned} \frac {4\sqrt 3 + 9}8 x & = \frac 92 \\ \implies x & = \frac {36}{4\sqrt 3+9} \approx \boxed{2.26} \text{ m} \end{aligned}$

Note that $\dfrac {d^2 A}{dx^2} > 0$ for all $x$ , therefore, $A$ is $\red{\text{minimum}}$ when $x \approx 2.26$ .

Let the length of each side of the square be $a$ and of the triangle be $b$ . Then $4a+3b=12\implies a^2+\dfrac{\sqrt 3}{4}b^2= a^2(1+\frac{4\sqrt 3}{9})-\frac{8\sqrt 3}{3}a+4\sqrt 3=\frac{9+4\sqrt 3}{9}(a-\frac{12\sqrt 3}{9+4\sqrt 3})^2+4\sqrt 3-\frac{48}{9+4\sqrt 3}\geq 4\sqrt 3-\frac{48}{9+4\sqrt 3}$ .

So the area is minimum when $a=\frac{12\sqrt 3}{9+4\sqrt 3}$ and $b=\frac{36}{9+4\sqrt 3}\approx \boxed {2.26}$ .

Let x = side length of the square and y = side length of the equilateral triangle

$\text{4x+3y=12}$

$\text{4x=12 - 3y}$

$x=\dfrac{12-3y}{4}$

$A=x^2+\dfrac{\sqrt{3}}{4}y^2$

$A=\left(\dfrac{12-3y}{4}\right)^2+\dfrac{\sqrt{3}}{4}y^2$

$A=\dfrac{144- 72y+9y^2}{16}+\dfrac{\sqrt{3}}{4}y^2$

Take the derivative using the Quotient Rule.

$\dfrac{dA}{dy}=\dfrac{16(-72+18y)-(144- 72y+9y^2)(0)}{16^2}+2\left(\dfrac{\sqrt{3}}{4}\right)(y)$

$=\dfrac{-1152+288y}{256}+\dfrac{\sqrt{3}}{2}y$

Equate the derivative to zero.

$\dfrac{-1152+288y}{256}+\dfrac{\sqrt{3}}{2}y=0$

$\dfrac{-1152+288y}{256}+\dfrac{128}{128} \times \dfrac{\sqrt{3}}{2}y=0$

$\dfrac{-1152+288y}{256}+\dfrac{128\sqrt{3}}{256}y=0$

$-1152+288y+128\sqrt{3}y=0$

$288y+128\sqrt{3}y=1152$

$y \approx \boxed{\text{2.26 meters}}$