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Question
Solve the following : Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is `(2"R")/sqrt(3)`. Also, find the maximum volume.
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Solution
Let R be the radius and h be the height of the cylinder which is inscribed in a sphere of radius r cm.
Then from the figure,
`"R"^2 + (h/2)^2` = r2
∴ R2 = `r^2 - h^2/(4)` ...(1)
Let V be the volume of the cylinder.
Then V = πR2h
= `pi(r^2 - h^2/(4))h` ...[By (1)]
= `pi(r^2 - h^3/(4))`
∴ `"dV"/"dh" = pid/"dh"(r^2h - h^3/(4))`
= `pi(r^2 xx 1 - 1/4 xx 3h^2)`
= `pi(r^2 - 3/4h^2)`
and
`(d^2V)/("dh"^2) = pid/"dh"(r^2 - 3/4h^2)`
= `pi(0 - 3/4 xx 2h)`
= `-(3)/(2)pih`
Now, `"dV"/"dh" = 0 "gives", pi(r^2 - 3/4h^2)` = 0
∴ `r^2 - 3/4h^2` = 0
∴ `(3)/(4)h^2` = r2
∴ h2 = `(4r^2)/(3)`
∴ h = `(2r)/sqrt(3)` ...[∵ h > 0]
and
`((d^2V)/(dh^2))_("at" h = (2r)/sqrt(3)`
= `-(3)/(2)pi xx (2r)/sqrt(3) < 0`
∴ V is maximum at h = `(2r)/sqrt(3)`
If h = `(2r)/sqrt(3)`, then from (1)
R2 = `r^2 - (1)/(4) xx (4r^2)/(3) = (2r^2)/(3)`
∴ volumeof the largest cylinder
= `pi xx (2r^2)/(3) xx (2r)/sqrt(3) = (4pir^3)/(3sqrt(3)`cu cm.
Hence, the volume of the largest cylinder inscribed in a sphere of radius 'r' cm = `(4R^3)/(3sqrt(3)`cu cm.
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