# Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base

**Solution:**

Let r and h be the radius and height of the cylinder respectively.

Then, the surface area (S) of the cylinder is given by,

S = 2πr^{2} + 2πrh

Therefore,

h = S - (2π r^{2})/2π r

= S / 2π (1/r) - r

Let V be the volume of the cylinder.

Then,

V = π r^{2}h

= = πr^{2} [S / 2π (1/r) - r]

= Sr/2 - (πr^{3})

dV/dr = S/2 - 3π r^{2}

d^{2}V/dr^{2} = - 6π r

Now,

dV/dr = 0

⇒ S/2 - 3πr^{2} = 0

S/2 = 3πr^{2}

⇒ r^{2} = S/6π

When, r^{2} = S/6π

Then,

d^{2}V/dr^{2}

= - 6π (√S / 6π) < 0

By second derivative test,

the volume is the maximum when r^{2} = S/6π.

Now, when r^{2} = S/6π

Then,

h = (6π r^{2})/2π (1/r) - r

= 3r - r

= 2r

Hence, the volume is the maximum when the height is twice the radius i.e. when the height is equal to the diameter

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 20

## Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base

**Summary:**

Hence we have shown that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base

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