A wire of length 28 , m is to be cut into two | vedclass

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

Question

(A) Let the length of the wire used for the square be $l \, m$. Then the length of the wire used for the circle is $(28-l) \, m$.The side of the squar

Vedclass · Accepted Answer

Square: $\frac{112}{\pi+4} \, m$,Circle: $\frac{28\pi}{\pi+4} \, m$

Vedclass · Answer

(A) Let the length of the wire used for the square be $l \, m$. Then the length of the wire used for the circle is $(28-l) \, m$.The side of the square is $s = \frac{l}{4}$. The area of the square is $A_s = s^2 = \frac{l^2}{16}$.For the circle,the circumference is $2\pi r = 28-l$,so the radius is $r = \frac{28-l}{2\pi}$. The area of the circle is $A_c = \pi r^2 = \pi \left( \frac{28-l}{2\pi} \right)^2 = \frac{(28-l)^2}{4\pi}$.The total area $A = A_s + A_c = \frac{l^2}{16} + \frac{(28-l)^2}{4\pi}$.To find the minimum,we differentiate $A$ with respect to $l$: $\frac{dA}{dl} = \frac{2l}{16} + \frac{2(28-l)(-1)}{4\pi} = \frac{l}{8} - \frac{28-l}{2\pi}$.Setting $\frac{dA}{dl} = 0$,we get $\frac{l}{8} = \frac{28-l}{2\pi} \Rightarrow \pi l = 4(28-l) \Rightarrow \pi l = 112 - 4l \Rightarrow l(\pi+4) = 112 \Rightarrow l = \frac{112}{\pi+4}$.The second derivative $\frac{d^2A}{dl^2} = \frac{1}{8} + \frac{1}{2\pi} > 0$,confirming a minimum.The length for the square is $\frac{112}{\pi+4} \, m$ and the length for the circle is $28 - \frac{112}{\pi+4} = \frac{28\pi+112-112}{\pi+4} = \frac{28\pi}{\pi+4} \, m$.

$A$ wire of length $28 \, m$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Similar Questions

The least value of $a$ for which the equation $\frac{4}{\sin x} + \frac{1}{1 - \sin x} = a$ has at least one solution on the interval $(0, \pi/2)$ is:

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

Let $f(x) = 1 + 2x^2 + 2^2x^4 + \dots + 2^{10}x^{20}$. Then $f(x)$ has:

The maximum value of $xy$ subject to $x+y=7$ is

Consider all rectangles lying in the region $\left\{( x , y ) \in R \times R : 0 \leq x \leq \frac{\pi}{2} \text{ and } 0 \leq y \leq 2 \sin (2 x )\right\}$ and having one side on the $x$-axis. The area of the rectangle which has the maximum perimeter among all such rectangles is

Vedclass Test Series

Exam Paper Generator

Online Exam Module