Let the radius and height of a right circular | vedclass

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

Question

(A) The volume of a right circular cylinder is given by $V = \pi r^2 h$. Given the relation $r^2 + h = 6$,we can express $h$ as $h = 6 - r^2$. Substit

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}}$

Vedclass · Answer

(A) The volume of a right circular cylinder is given by $V = \pi r^2 h$. Given the relation $r^2 + h = 6$,we can express $h$ as $h = 6 - r^2$. Substituting this into the volume formula: $V = \pi r^2 (6 - r^2) = \pi (6r^2 - r^4)$. To find the maximum volume,we differentiate $V$ with respect to $r$: $\frac{dV}{dr} = \pi (12r - 4r^3)$. Setting $\frac{dV}{dr} = 0$,we get $12r - 4r^3 = 0$,which implies $4r(3 - r^2) = 0$. Since $r > 0$,we have $r^2 = 3$,so $r = \sqrt{3}$. Substituting $r^2 = 3$ into the relation $h = 6 - r^2$,we get $h = 6 - 3 = 3$. Therefore,the ratio $\frac{r}{h} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.

Let the radius and height of a right circular cylinder be related as $r^2 + h = 6$. If the volume of the cylinder is maximum,then the value of $\frac{r}{h}$ is:

Similar Questions

If $y = a \log |x| + b x^2 + x$ has its extremum values at $x = -1$ and $x = 2$,then

Observe the statements given below :
Assertion $(A)$ : $f(x)=x e^{-x}$ has the maximum at $x=1$
Reason $(R)$ : $f^{\prime}(1)=0$ and $f^{\prime \prime}(1) < 0$
Which of the following is correct?

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

The ordinates of the points on the curve $y = \tan^{-1}(\sin \sqrt{x})$,$0 \leq x \leq 8\pi^2$,at which the tangent is parallel to the $x$-axis are

The number of distinct real roots of the equation $3x^{4} + 4x^{3} - 12x^{2} + 4 = 0$ is ..... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module