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 the set of all values of $a$,for which the equation $5x^3 - 15x - a = 0$ has three distinct real roots,is the interval $(\alpha, \beta)$,then $\beta - 2\alpha$ is equal to . . . . . . .

Find the maximum profit that a company can make,if the profit function is given by $p(x) = 41 - 72x - 18x^{2}$. (in $units$)

The maximum value of $x^4 y^4$ given the constraint $a^2 x^4 + b^2 y^4 = c^6$ is

The difference between the absolute maximum and absolute minimum values of the function $f(x)=2x^3-15x^2+36x-30$ on the interval $[-1, 4]$ is:

The minimum value of $ax + by$ where $xy = c^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module