Problem: Maximize the volume of an open cylinder (closed at the base, open at the top) for a fixed surface area \( S \).
Let:
Surface Area (S) of the open cylinder:
\[ S = \pi r^2 + 2\pi r h \]
Solve for height \( h \):
\[ 2\pi r h = S - \pi r^2 \Rightarrow h = \frac{S - \pi r^2}{2\pi r} = \frac{S}{2\pi r} - \frac{r}{2} \]
Volume (V) of the cylinder:
\[ V = \pi r^2 h = \pi r^2 \left( \frac{S}{2\pi r} - \frac{r}{2} \right) = \frac{S r}{2} - \frac{\pi r^3}{2} \]
Differentiate V with respect to \( r \) to find the maximum:
\[ \frac{dV}{dr} = \frac{S}{2} - \frac{3\pi r^2}{2} \]
Set \( \frac{dV}{dr} = 0 \) for critical point:
\[ \frac{S}{2} - \frac{3\pi r^2}{2} = 0 \Rightarrow S = 3\pi r^2 \]
Substitute back to find \( h \):
\[ h = \frac{S}{2\pi r} - \frac{r}{2} = \frac{3\pi r^2}{2\pi r} - \frac{r}{2} = \frac{3r}{2} - \frac{r}{2} = r \]
So, at maximum volume:
\[ h = r \Rightarrow \text{Radius} = \text{Height} \]
Second derivative test:
\[ \frac{d^2V}{dr^2} = -3\pi r < 0 \quad \text{(for } r > 0 \text{)} \]
This confirms a maximum.
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |
If $f(x) = \frac{1 - x + \sqrt{9x^2 + 10x + 1}}{2x}$, then $\lim_{x \to -1^-} f(x) =$
Given \( f(x) = \begin{cases} \frac{1}{2}(b^2 - a^2), & 0 \le x \le a \\[6pt] \frac{1}{2}b^2 - \frac{x^2}{6} - \frac{a^3}{3x}, & a<x \le b \\[6pt] \frac{1}{3} \cdot \frac{b^3 - a^3}{x}, & x>b \end{cases} \). Then: