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.
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}
If $ \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p $, then $ 96 \log_e p $ is equal to _______