Question

A piece of wire 36cm long is cut into two, one part being bent in the shape of an equilateral triangle and the other in the form of circle. Find the lengths of these two pieces of wire if the sum of the areas of these two figures is to be minimum.

Answer 1

Solution:

Let length of one piece be x, then the length of the other piece will be 36-x. Let from first piece we make the square, then $\mathrm{x}=4 \mathrm{y} \quad \Longrightarrow \mathrm{y}=\frac{\mathrm{x}}{4}$ , where y is the side of the square $\ldots . .(\mathrm{i})$

From the second piece of length (36-x) we make an equilateral triangle, then side of the equilateral triangle $=\left(\frac{36-\mathrm{x}}{3}\right)$

Now combined area of the two $=\mathrm{A}=\left(\frac{\mathrm{x}}{4}\right)^{2}+\frac{\sqrt{3}}{4}\left(\frac{36-\mathrm{x}}{3}\right)^{2}$

Differentiating with respect to \mathrm{x}, we have

$\begin{aligned} &\Longrightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{2 \mathrm{x}}{4} \cdot \frac{1}{4}+\frac{\sqrt{3}}{4} \cdot 2 \cdot \frac{(36-\mathrm{x})}{3} \cdot\left(-\frac{1}{3}\right) \\ &\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{\mathrm{x}}{8}-\frac{\sqrt{3}}{18}(36-\mathrm{x}) \end{aligned}$

For maximum/minimum, we have $\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=0$

$\begin{aligned} &\Rightarrow \frac{x}{8}-\frac{\sqrt{3}}{18}(36-x)=0 \\ &\Rightarrow \frac{x}{8}=\frac{\sqrt{3}}{18} \cdot(36-x) \quad \Rightarrow \frac{x}{8}=2 \sqrt{3}-\frac{1}{6 \sqrt{3}} x \end{aligned}$

$\begin{aligned} &\Longrightarrow \mathrm{x}\left(\frac{1}{8}+\frac{1}{6 \sqrt{3}}\right)=2 \sqrt{3} \quad \Longrightarrow \mathrm{x}\left(\frac{3 \sqrt{3}+4}{24 \sqrt{3}}\right)=2 \sqrt{3} \\ &\Rightarrow \mathrm{x}(4+3 \sqrt{3})=144 \quad \Longrightarrow \mathrm{x}=\frac{144}{4+3 \sqrt{3}} \end{aligned}$

Thus, length of one piece is $\mathrm{x}=\frac{144}{4+3 \sqrt{3}}$ and the length of other piece is

$\begin{aligned} &36-\frac{144}{(4+3 \sqrt{3})}=\frac{144+108 \sqrt{3}-144}{(4+3 \sqrt{3})} \\ &=\frac{108 \sqrt{3}}{(4+3 \sqrt{3})} \mathrm{cm} \end{aligned}$