A wire of length 8 text{ units} is cut into t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the length of the wire used for the square be $x$. Then the length of the wire used for the circle is $8-x$. For the square,the perimeter is $

Vedclass · Accepted Answer

$\frac{16}{\pi+4}$

Vedclass · Answer

(D) Let the length of the wire used for the square be $x$. Then the length of the wire used for the circle is $8-x$. For the square,the perimeter is $4a = x$,so the side $a = \frac{x}{4}$. The area of the square is $A_1 = a^2 = \frac{x^2}{16}$. For the circle,the circumference is $2\pi r = 8-x$,so the radius $r = \frac{8-x}{2\pi}$. The area of the circle is $A_2 = \pi r^2 = \pi \left(\frac{8-x}{2\pi}\right)^2 = \frac{(8-x)^2}{4\pi}$. The total area $A(x) = \frac{x^2}{16} + \frac{(8-x)^2}{4\pi}$. To find the minimum,differentiate with respect to $x$: $A'(x) = \frac{2x}{16} + \frac{2(8-x)(-1)}{4\pi} = \frac{x}{8} - \frac{8-x}{2\pi}$. Set $A'(x) = 0$: $\frac{x}{8} = \frac{8-x}{2\pi} \implies \pi x = 32 - 4x \implies x(\pi+4) = 32 \implies x = \frac{32}{\pi+4}$. Substitute $x$ back into $A(x)$: $A = \frac{1}{16} \left(\frac{32}{\pi+4}\right)^2 + \frac{1}{4\pi} \left(8 - \frac{32}{\pi+4}\right)^2 = \frac{64}{(\pi+4)^2} + \frac{1}{4\pi} \left(\frac{8\pi+32-32}{\pi+4}\right)^2 = \frac{64}{(\pi+4)^2} + \frac{64\pi^2}{4\pi(\pi+4)^2} = \frac{64}{(\pi+4)^2} + \frac{16\pi}{(\pi+4)^2} = \frac{16(4+\pi)}{(\pi+4)^2} = \frac{16}{\pi+4}$.

$A$ wire of length $8 \text{ units}$ is cut into two parts which are bent respectively in the form of a square and a circle. The least value of the sum of the areas so formed is

Similar Questions

Statement-$I$: $e^{\pi} > \pi^e$.
Statement-$II$: The function $f(x) = x^{1/x}$ attains its global maximum at $x = e$.

The point $(0, 5)$ is closest to the curve ${x^2} = 2y$ at

The maximum area of the rectangle that can be inscribed in a circle of radius $r$ is

If the height of a cone of greatest volume that can be inscribed in a sphere of radius $R$ is $kR$,then the ratio of the volume of the cone to the volume of the sphere is

The maximum area of a rectangle that can be inscribed in a circle of radius $2 \text{ unit}$ is (in square unit)

Vedclass Test Series

Exam Paper Generator

Online Exam Module