Given r + h = 6
⇒ h = 6 - r ...(1)
Let f(r) = V = πr²h
= πr²(6-r) = π(6r² -r³)
f'(r) = π (12r - 3r²)
∴f'(r) = 0
⇒π (12r- 3r²) = 0
⇒ 12r-3r² = 0
⇒ 3r(4-r) = 0
⇒ r = 0 or r = 4
ஃThe critical numbers are 0, 4
f"(r) = π (12 - 6r)
When r = 4, f"(r) = π(12 - 24) < 0
ஃ f(r) maximum when r = 4
When r = 4, h = 6 - 4 = 2
When r = 0, h = 6 - 0 = 6
ஃ Volume of the cylinder V = πr² h = π (4)² (2)
= 32 πCu. units
or volume of the cylinder V = π (0²) (6) = 0
Cu. Units.

\(\underset { x\rightarrow \frac { { \pi }^{ - } }{ 2 } }{ lim } \frac { secx }{ tanx } =\underset { x\rightarrow \frac { { \pi }^{ - } }{ 2 } }{ lim } \frac { \frac { 1 }{ \frac { cosx }{ sinx } } }{ \frac { sinx }{ cosx } } =\underset { x\rightarrow \frac { { \pi }^{ - } }{ 2 } }{ lim } \frac { 1 }{ sinx } \)
\(=\frac { 1 }{ sin\frac { \pi }{ 2 } } =\frac { 1 }{ 1 } =1\)

\(\underset { x\rightarrow \infty }{ lim } \frac { x }{ logx } =\frac { \infty }{ \infty } \)
Which is in indeterminate form
∴ By L' Hopital rule we get,
\(\underset { x\rightarrow \infty }{ lim } \frac { \frac { 1 }{ 1 } }{ x } =\underset { x\rightarrow \infty }{ lim } x=\infty \)

Since f(x) = x² - 2x - 3 , \(f'(x)=2x-2>0 \forall x\in (2, \infty)\). Hence f (x) is strictly increasing in \((2, \infty)\)