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Prove that the radius of the right circular cylinder of greate

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Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.

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Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.

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Text solutionVerified

Solution: OE=x
OC=r
AO=h

∠ A O C = \sim ∠ Q E C

\frac{A O}{O C} = \frac{Q E}{E C}

(Q E) = \frac{A O}{O C} (E C)

\frac{h}{r} (r - x)

δ x = 2 π x (\frac{h}{r} (r - x))

\frac{2 π h}{r} (r x - x^{2})

δ^{'} (x) = 0

x = \frac{r}{2}

derivating again
\displaystyle\delta{''}{\left(\frac{{r}}{{2}}\right)}=\frac{{{2}\pi{h}}}{{r}}{\left(-{2}\right)}

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Question Text	Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.
Answer Type	Text solution:1
Upvotes	150

Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.

Text solutionVerified

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