Question

The slant height of a right circular cone is $\sqrt{3}\text{ cm}$. The height of the cone for maximum volume is

• A. $\sqrt{5}\text{ cm}$
• B. $5\text{ cm}$
• C. $\sqrt{3}\text{ cm}$
• D. $1\text{ cm}$

Hint:

For any right circular cone with a fixed slant height $l$, the volume is always maximized at a unique fixed ratio: $h = \frac{l}{\sqrt{3}}$. Substituting our given value $l = \sqrt{3}$ into this ratio gives $h = \frac{\sqrt{3}}{\sqrt{3}} = 1\text{ cm}$ instantly!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a right circular cone with a fixed slant height $l = \sqrt{3}\text{ cm}$. We need to find the specific vertical height $h$ that maximizes the volume of the cone.

Step 2: Key Formula or Approach:
The formula for the volume of a cone is: $$V = \frac{1}{3}\pi r^2 h$$ Using the geometric relation between slant height, radius, and vertical height ($l^2 = r^2 + h^2$), we substitute $r^2 = l^2 - h^2$ to express the volume solely in terms of a single variable, $h$. We then find the maximum value by setting $\frac{dV}{dh} = 0$.

Step 3: Detailed Explanation:
Substitute the given slant height $l = \sqrt{3}$ into the Pythagorean relation: $$(\sqrt{3})^2 = r^2 + h^2 \implies r^2 = 3 - h^2$$ Now, substitute this expression for $r^2$ into the volume equation: $$V = \frac{1}{3}\pi (3 - h^2)h = \frac{1}{3}\pi (3h - h^3)$$ Differentiate $V$ with respect to $h$: $$\frac{dV}{dh} = \frac{1}{3}\pi \frac{d}{dh}(3h - h^3) = \frac{1}{3}\pi (3 - 3h^2) = \pi(1 - h^2)$$ To maximize the volume, set the first derivative equal to zero: $$\pi(1 - h^2) = 0 \implies 1 - h^2 = 0 \implies h^2 = 1 \implies h = 1\text{ cm}$$ (Since height must be a positive length, we discard $h = -1$). To verify that this point is a maximum, let's check the sign of the second derivative: $$\frac{d^2V}{dh^2} = \frac{d}{dh}(\pi(1 - h^2)) = -2\pi h$$ Evaluating at $h = 1$: $$\frac{d^2V}{dh^2} = -2\pi(1) = -2\pi \lt 0$$ Since the second derivative is negative, the volume reaches its maximum value when $h = 1\text{ cm}$.

Step 4: Final Answer:
The height of the cone for maximum volume is $1\text{ cm}$, which corresponds to option (D).