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leprofece
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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
the answer is h= L/sqrt(3) and R = Sqrt(6)L/2
leprofece said: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 said:hello.
That solution is not me.
[tex]l^2=r^2+h^2[/tex](*)
[tex]cone \ volume= v =\frac{\pi r^2 h}{3}[/tex]
[tex]v=\dfrac{\pi (l^2-h^2)h}{3}=\dfrac{\pi h l^2-\pi h^3}{3}[/tex]
[tex]\dfrac{d(v)}{d(h)}=\dfrac{\pi l^2-3 \pi h^2}{3}=0[/tex]
[tex]\cancel{\pi} l^2=3 \cancel{\pi} h^2 \rightarrow{}h=\dfrac{l}{\sqrt{3}}[/tex]
[tex]\dfrac{d_2(v)}{d(h)}=- \dfrac{6 \pi h}{3} \rightarrow{} h=\dfrac{l}{\sqrt{3}}= is \ maximun[/tex]
for (*):
[tex]r^2=l^2-\dfrac{l^2}{3}[/tex]
[tex]r=\dfrac{\sqrt{2}l}{\sqrt{3}}=\dfrac{\sqrt{2} \sqrt{3}l}{\sqrt{3} \sqrt{3}}=\dfrac{\sqrt{6}l}{3}[/tex]
regards.
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Understanding maximum and minimum in SOS is essential for analyzing and interpreting data accurately. It helps identify outliers and extreme values, which can greatly impact the results of a study or experiment.
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