What is the maximum volume of a cone?

[Karol]

Homework Statement

Homework Equations

First derivative=maxima/minima/vertical tangent/rising/falling
Volume of a cone: ##~\displaystyle V=\frac{\pi}{3}r^2h##

The Attempt at a Solution

$$L=a^2+b^2~\rightarrow b^2=L-a^2$$
$$V=\frac{\pi}{3}a(L-a^2)$$
$$V'=\frac{\pi}{3}(L-3a^2),~V'=0:~a=\sqrt\frac{L}{3}$$
$$V=\frac{\pi}{3}\left[ \sqrt\frac{L}{3}\left( L-\frac{L}{3} \right) \right]$$
The answer should be:
$$V=\frac{2\pi L^3}{9\sqrt{3}}$$

 

[Abhishek kumar]

Homework Statement

View attachment 214806

Homework Equations

First derivative=maxima/minima/vertical tangent/rising/falling
Volume of a cone: ##~\displaystyle V=\frac{\pi}{3}r^2h##

The Attempt at a Solution

$$L=a^2+b^2~\rightarrow b^2=L-a^2$$
$$V=\frac{\pi}{3}a(L-a^2)$$
$$V'=\frac{\pi}{3}(L-3a^2),~V'=0:~a=\sqrt\frac{L}{3}$$
$$V=\frac{\pi}{3}\left[ \sqrt\frac{L}{3}\left( L-\frac{L}{3} \right) \right]$$
The answer should be:
$$V=\frac{2\pi L^3}{9\sqrt{3}}$$

If L is hypotenuse then L^2=a^2+b^2

 

[Karol]

$$L^2=a^2+b^2~\rightarrow b^2=L^2-a^2$$
$$V=\frac{\pi}{3}a(L^2-a^2)$$
$$V'=\frac{\pi}{3}(L^2-3a^2),~V'=0:~a=\frac{L}{\sqrt{3}}$$
$$V=\frac{\pi}{3}\left[ \frac{L}{\sqrt{3}} \left( L^2-\frac{L^2}{3} \right) \right]=\frac{2\pi L^3}{9\sqrt{3}}$$
Thank you Kumar

 

[Abhishek kumar]

$$L^2=a^2+b^2~\rightarrow b^2=L^2-a^2$$
$$V=\frac{\pi}{3}a(L^2-a^2)$$
$$V'=\frac{\pi}{3}(L^2-3a^2),~V'=0:~a=\frac{L}{\sqrt{3}}$$
$$V=\frac{\pi}{3}\left[ \frac{L}{\sqrt{3}} \left( L^2-\frac{L^2}{3} \right) \right]=\frac{2\pi L^3}{27\sqrt{3}}$$
Almost

$$L^2=a^2+b^2~\rightarrow b^2=L^2-a^2$$
$$V=\frac{\pi}{3}a(L^2-a^2)$$
$$V'=\frac{\pi}{3}(L^2-3a^2),~V'=0:~a=\frac{L}{\sqrt{3}}$$
$$V=\frac{\pi}{3}\left[ \frac{L}{\sqrt{3}} \left( L^2-\frac{L^2}{3} \right) \right]=\frac{2\pi L^3}{27\sqrt{3}}$$
Almost

You have done a mistake in last line