Question

A right triangle whose hypotenuse is StartRoot 18 EndRoot18 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:

V = $\frac{1}{3}$π(8 - $\frac{8}{3}$)$\sqrt{\frac{8}{3} }$ = 9.11 $m^{3}$

r = 4$\frac{\sqrt{3} }{3}$

h = $\sqrt{\frac{8}{3} }$

Step-by-step explanation:

Given that the right triangle whose hypotenuse is $\sqrt{8}$

• Let r is the radius of the cone
• Let h is the height of the cone

We know that:

$r^{2} + h^{2} = 8$

<=> $r^{2} = 8 - h^{2}$

The volume of the cone is:

V = π$r^{2} \frac{1}{3} h$

<=> V = π$\frac{1}{3}(8 - h^{2} )h$

Differentiate w.r.t h

$\frac{dV}{dh}$ = π $\frac{1}{3}$ (8 - $3h^{2}$)

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

<=> π $\frac{1}{3}$ (8 - $3h^{2}$)  = 0

<=> $h^{2}$ = $\frac{8}{3}$

<=> h = $\sqrt{\frac{8}{3} }$

=> $r^{2}$ = $\frac{16}{3}$

<=> r = 4$\frac{\sqrt{3} }{3}$

So the volume of the cone is:

V = $\frac{1}{3}$π(8 - $\frac{8}{3}$)$\sqrt{\frac{8}{3} }$ = 9.11 $m^{3}$