Of all the closed cylindrical cans (right cir | vedclass

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

Question

(A) Let $r$ be the radius and $h$ be the height of the cylinder.Given volume $V = \pi r^2 h = 100 	ext{ cm}^3$.Thus,$h = \frac{100}{\pi r^2}$.The sur

Vedclass · Accepted Answer

Radius $= \left(\frac{50}{\pi}ight)^{1/3} 	ext{ cm}$,Height $= 2\left(\frac{50}{\pi}ight)^{1/3} 	ext{ cm}$

Vedclass · Answer

(A) Let $r$ be the radius and $h$ be the height of the cylinder.Given volume $V = \pi r^2 h = 100 	ext{ cm}^3$.Thus,$h = \frac{100}{\pi r^2}$.The surface area $S$ is given by $S = 2\pi r^2 + 2\pi rh$.Substituting $h$,we get $S = 2\pi r^2 + 2\pi r \left(\frac{100}{\pi r^2}ight) = 2\pi r^2 + \frac{200}{r}$.Differentiating with respect to $r$,we get $\frac{dS}{dr} = 4\pi r - \frac{200}{r^2}$.Setting $\frac{dS}{dr} = 0$,we have $4\pi r = \frac{200}{r^2} \Rightarrow r^3 = \frac{50}{\pi} \Rightarrow r = \left(\frac{50}{\pi}ight)^{1/3}$.Checking the second derivative,$\frac{d^2S}{dr^2} = 4\pi + \frac{400}{r^3}$. Since $r > 0$,$\frac{d^2S}{dr^2} > 0$,so $S$ is minimum at $r = \left(\frac{50}{\pi}ight)^{1/3}$.Then $h = \frac{100}{\pi (50/\pi)^{2/3}} = \frac{100}{\pi} \cdot \frac{\pi^{2/3}}{50^{2/3}} = 2 \cdot \frac{50}{\pi^{1/3} \cdot 50^{2/3}} = 2 \left(\frac{50}{\pi}ight)^{1/3}$.Thus,the dimensions are $r = \left(\frac{50}{\pi}ight)^{1/3} 	ext{ cm}$ and $h = 2\left(\frac{50}{\pi}ight)^{1/3} 	ext{ cm}$.

Of all the closed cylindrical cans (right circular) with a given volume of $100 \text{ cm}^3$,find the dimensions of the can that has the minimum surface area.

Similar Questions

For a steamer,the consumption of petrol (per hour) varies as the cube of its speed (in $km/hr$). If the speed of the current is steady at $C \, km/hr$,then the most economical speed of the steamer going against the current will be ........... $C$.

Find the local maximum and local minimum values of the function given by $f(x) = x^{3} - 6x^{2} + 9x + 15$.

Slope of the tangent to the curve $y=2 e^x \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)$ where $0 \leq x \leq 2 \pi$ is minimum at $x=$

The function $f(x) = 2x^3 - 15x^2 + 36x + 4$ is maximum at $x=$ ......

$A$ solid hemisphere is mounted on a solid cylinder,both having equal radii. If the whole solid is to have a fixed surface area and the maximum possible volume,then the ratio of the height of the cylinder to the common radius is

Vedclass Test Series

Exam Paper Generator

Online Exam Module