A wire of length 20 m is to be cut into two | vedclass

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

Question

(D) Let the wire be cut into two pieces of length $x$ and $20-x$.The side of the square is $s_1 = \frac{x}{4}$,so the area of the square is $A_1 = \le

Vedclass · Accepted Answer

$\frac{10}{3+2 \sqrt{3}}$

Vedclass · Answer

(D) Let the wire be cut into two pieces of length $x$ and $20-x$.The side of the square is $s_1 = \frac{x}{4}$,so the area of the square is $A_1 = \left(\frac{x}{4}ight)^2 = \frac{x^2}{16}$.The side of the regular hexagon is $s_2 = \frac{20-x}{6}$,so the area of the regular hexagon is $A_2 = 6 	imes \frac{\sqrt{3}}{4} \left(\frac{20-x}{6}ight)^2 = \frac{3\sqrt{3}}{2} \frac{(20-x)^2}{36} = \frac{\sqrt{3}}{24} (20-x)^2$.The total area is $A(x) = \frac{x^2}{16} + \frac{\sqrt{3}}{24} (20-x)^2$.To find the minimum area,we differentiate $A(x)$ with respect to $x$ and set it to $0$:$A'(x) = \frac{2x}{16} + \frac{\sqrt{3}}{24} 	imes 2(20-x)(-1) = \frac{x}{8} - \frac{\sqrt{3}}{12}(20-x) = 0$.Multiplying by $24$ gives $3x - 2\sqrt{3}(20-x) = 0$,which simplifies to $3x - 40\sqrt{3} + 2\sqrt{3}x = 0$.Thus,$x(3 + 2\sqrt{3}) = 40\sqrt{3}$,so $x = \frac{40\sqrt{3}}{3 + 2\sqrt{3}}$.The side of the hexagon is $s_2 = \frac{20-x}{6} = \frac{1}{6} \left( 20 - \frac{40\sqrt{3}}{3 + 2\sqrt{3}} ight)$.$s_2 = \frac{1}{6} \left( \frac{60 + 40\sqrt{3} - 40\sqrt{3}}{3 + 2\sqrt{3}} ight) = \frac{1}{6} \left( \frac{60}{3 + 2\sqrt{3}} ight) = \frac{10}{3 + 2\sqrt{3}}$.

$A$ wire of length $20 \ m$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in $meters$) of the hexagon,so that the combined area of the square and the hexagon is minimum,is:

Similar Questions

If $x+y=6, x \geqslant 0, y \geqslant 0$,then the maximum value of $x^2 y$ is

In a regular triangular prism,the distance from the centre of one base to one of the vertices of the other base is $l$. Find the altitude of the prism for which the volume is greatest.

In an isosceles trapezium,the length of one of the parallel sides and the lengths of the non-parallel sides are all equal to $30$. In order to maximize the area of the trapezium,the smallest angle should be:

If $m$ is the minimum value of $k$ for which the function $f(x) = x\sqrt{kx - x^2}$ is increasing in the interval $[0, 3]$ and $M$ is the maximum value of $f$ in $[0, 3]$ when $k = m$,then the ordered pair $(m, M)$ is equal to

The function $f(x) = x^3 + ax^2 + bx + c$ where $a^2 \leq 3b$ has:

Vedclass Test Series

Exam Paper Generator

Online Exam Module