Question

A right triangle whose hypotenuse is StartRoot 7 EndRoot7 m long is revolved about one of its legs to generate a right circular cone. Find the​ radius, height, and volume of the cone of greatest volume that can be made this way.

Answer 1

Answer:

The maximum volume of cone is 138.25 m³

$Radius, r=7\sqrt{\dfrac{2}{3}}$ m

$Height, h=\dfrac{7}{\sqrt{3}}$ m

Step-by-step explanation:

A right circular cone whose hypotenuse is $\sqrt{7}$ m

It is revolved about one leg to generate a right circular cone.

Let radius be r m and height be h m

For right angle triangle,

$r^2+h^2=7^2$

$r^2=49-h^2$

Volume of generated cone $=\dfrac{1}{3}\pi r^2h$

$V=\dfrac{1}{3}\pi (49-h^2)h$

Differentiate w.r.t h

$\dfrac{dV}{dh}=\dfrac{1}{3}\pi (49-3h^2)$

For maximum/minimum $\dfrac{dV}{dh}=0$

$\dfrac{1}{3}\pi (49-3h^2)=0$

$h=\dfrac{7}{\sqrt{3}}$

$r^2=49-\dfrac{49}{3}$

$r=7\sqrt{\dfrac{2}{3}}$

Using double derivative test

$\dfrac{d^2V}{dh^2}=\dfrac{1}{3}\pi (-6h)$

At $h=\dfrac{7}{\sqrt{3}}$  

$\dfrac{d^2V}{dh^2}$ so get maximum volume.

Dimension of cone:

$Radius, r=7\sqrt{\dfrac{2}{3}}$ m

$Height, h=\dfrac{7}{\sqrt{3}}$ m

$V=\frac{1}{3}\pi\cdot\left(7\sqrt{\frac{2}{3}}\right)^{2}\cdot\frac{7}{\sqrt{3}}$

The maximum volume of cone is 138.25 m³