Given, radius of the sphere is R.
Let r and h be the radius and the height of the inscribed cylinder respectively
We have:
h = 2√R2−r2
Let Volume of cylinder = V
V = πr2h

= πr2×2√R2−r2
= 2πr2√R2−r2
Differentiating the above function w.r.t. r, we have,
V = 2πr2√R2−r2

dV

dr

= 4πr√R2−r2 -

4πr2

2√R2−r2

=

4πr(R2−r2−4πr3)

2√R2−r2

dV

dr

=

4πrR2−4πr3−2πr3

2√R2−r2

=

4πrR2−6πr3

2√R2−r2

For maxima or minima,

dV

dr

= 0 ⇒ 4πrR2−6πr3 = 0
⇒ 6πr3 = 4πrR2
⇒ r2 =

2R2

3

dV

dr

=

4πrR2−6πr3

2√R2−r2

Now,

d2V

dr2

=
=
=
Now, when r2 =

2R2

3

,

d2V

dr2

< 0
∴ Volume is the maximum when r2 =

2R2

3

when r2 =

2R2

3

, h = 2√R2−

2R2

3

= 2 √

R2

3

=

2R

√3

Hence, the volume of the cylinder is the maximum when the height of the cylinder is

2R

√3

OR
The equation of the given curve is x2 = 4y.
Differentiating w.r.t. x, we get

dy

dx

=

x

2

Let (h, k) be the co- ordinates of the point of contact of the normal to the curve x2 = 4y.
Now, slope of the tangent at (h, k) is given by

dy

dx

|(h,k) =

h

2

Hence, slope of the normal at (h , k) =

−2

h

Therefore, the equation of normal at (h, k) is
y - k =

−2

h

(x−h) ... (1)
Since, it passes through the point (1, 2) we have
2 - k =

−2

h

(1−h) or k = 2 +

2

h

(1 - h) ... (2)
Now, (h, k) lies on the curve x2 = 4y, so, we have:
h2 = 4k ... (3)
Solving (2) and (3), we get,
h = 2 and k = 1.
From (1), the required equation of the normal is:
y - 1 =

−2

2

(x - 2) or x + y = 3
Also, slope of the tangent = 1
∴ Equation of tangent at (1, 2) is:
y – 2 = 1(x – 1) or y = x + 1