A wire of length 2 units is cut into two parts, which are bent respectively to form a square of side $x$…

A wire of length 2 units is cut into two parts, which are bent respectively to form a square of side $x$ units and a circle of radius of r units. If the sum of the areas of square and the circle so formed is minimum, then
  1. $2 x=(\pi+4) \mathrm{r}$
  2. $(4-\pi) x=\pi \mathrm{r}$
  3. $x=2 r$
  4. $2 x=\mathrm{r}$

Solution

Perimeter of the square $=4 x$ Perimeter of the circle $=2 \pi \mathrm{r}$ $\begin{array}{ll} \therefore & 4 x+2 \pi r=2 \\ \therefore & 2 x+\pi r=1 \Rightarrow r=\frac{1-2 x}{\pi} \end{array}...(i)$
Sum of the areas $(\mathrm{A})=x^2+\pi \mathrm{r}^2$ $\therefore \quad \mathrm{A}=x^2+\pi\left(\frac{1-2 x}{\pi}\right)^2$ ...[From (i) $\therefore \quad \mathrm{A}=x^2+\frac{1}{\pi}(1-2 x)^2$ Differentiating A w.r.t. $x$, we get $\begin{aligned} & \frac{\mathrm{dA}}{\mathrm{~d} x}=2 x+\frac{2}{\pi}(1-2 x)(-2), \frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{~d} x^2}=2+\frac{8}{\pi} \gt 0 \\ & \frac{\mathrm{dA}}{\mathrm{~d} x}=0 \Rightarrow 2 x-\frac{4}{\pi}+\frac{8 x}{\pi}=0 \\ & \Rightarrow(2 \pi+8) x=4 \\ & \Rightarrow(\pi+4) x=2 \\ & \Rightarrow x=\frac{2}{\pi+4} \end{aligned}$ $\therefore \quad$ Area is minimum when $x=\frac{2}{\pi+4}$ Substituting $x=\frac{2}{\pi+4}$ in equation (i), we get $\begin{aligned} & \mathrm{r}=\frac{1}{\pi+4} \\ & \Rightarrow x=2 \mathrm{r} \end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 1)

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