Sos help with maximum and minimum

[leprofece]

Between all the cones whose generatrix length given is L, determine the one with the highest volume?

the answer is h= L/sqrt(3) and R = Sqrt(6)L/2

 

[mente oscura]

Re: Sos help whit maximun and minimun

Between all the cones whose generatrix length given is L, determine the one with the highest volume?

the answer is h= L/sqrt(3) and R = Sqrt(6)L/2

Hello.

That solution is not me.

$$L^2=R^2+h^2$$(*)

$$Cone \ volume= V =\frac{\pi R^2 h}{3}$$

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

$$\dfrac{d(V)}{d(h)}=\dfrac{\pi L^2-3 \pi h^2}{3}=0$$

$$\cancel{\pi} L^2=3 \cancel{\pi} h^2 \rightarrow{}h=\dfrac{L}{\sqrt{3}}$$

$$\dfrac{d_2(V)}{d(h)}=- \dfrac{6 \pi h}{3} \rightarrow{} h=\dfrac{L}{\sqrt{3}}= Is \ maximun$$

For (*):

$$R^2=L^2-\dfrac{L^2}{3}$$

$$R=\dfrac{\sqrt{2}L}{\sqrt{3}}=\dfrac{\sqrt{2} \sqrt{3}L}{\sqrt{3} \sqrt{3}}=\dfrac{\sqrt{6}L}{3}$$

Regards.

 

[leprofece]

Re: Sos help whit maximun and minimun

hello.

That solution is not me.

$$l^2=r^2+h^2$$(*)

$$cone \ volume= v =\frac{\pi r^2 h}{3}$$

$$v=\dfrac{\pi (l^2-h^2)h}{3}=\dfrac{\pi h l^2-\pi h^3}{3}$$

$$\dfrac{d(v)}{d(h)}=\dfrac{\pi l^2-3 \pi h^2}{3}=0$$

$$\cancel{\pi} l^2=3 \cancel{\pi} h^2 \rightarrow{}h=\dfrac{l}{\sqrt{3}}$$

$$\dfrac{d_2(v)}{d(h)}=- \dfrac{6 \pi h}{3} \rightarrow{} h=\dfrac{l}{\sqrt{3}}= is \ maximun$$

for (*):

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

$$r=\dfrac{\sqrt{2}l}{\sqrt{3}}=\dfrac{\sqrt{2} \sqrt{3}l}{\sqrt{3} \sqrt{3}}=\dfrac{\sqrt{6}l}{3}$$

regards.

thanks a lot