Calculus
Applications of Derivatives - Maxima and Minima
GRB_1000_SCQ
Grade Class 12

Question:

From a given solid cone of height $H$, another inverted cone is carved whose height is $h$, such that its volume is maximum, then the ratio $\dfrac{H}{h}$ is equal to:
2
3
4
6

Step-by-Step Solution

Key Concept: Optimization of volume of inscribed cone
Step 1: Set up the geometry of the original cone. Let the original cone have height $H$ and base radius $R$. The apex is at the top and the base is at the bottom. We need to understand how the radius of the original cone varies with height. Step 2: Establish the relationship between radius and height in the original cone. By similar triangles, at any height $y$ measured from the base of the original cone, the radius of the original cone is: $$r_{\text{original}}(y) = R\left(1 - \frac{y}{H}\right) = R\frac{H-y}{H}$$ Step 3: Describe the configuration of the inverted cone. The inverted cone is carved from the original cone with its base coinciding with the base of the original cone and its apex pointing upward into the original cone. Let the inverted cone have height $h$ and base radius $r$. Step 4: Determine the constraint for the inverted cone to fit inside the original cone. For the inverted cone to fit inside the original cone without protruding, its lateral surface must not extend beyond the lateral surface of the original cone. At height $h$ from the base, the inverted cone's radius must equal the original cone's radius at that height. This gives us the constraint: $$r = R\frac{H-h}{H}$$ Step 5: Express the volume of the inverted cone. The volume of the inverted cone with base radius $r$ and height $h$ is: $$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(R\frac{H-h}{H}\right)^2 h$$ $$V = \frac{\pi R^2}{3H^2}(H-h)^2 h$$ Step 6: Maximize the volume by finding the critical points. To maximize $V$, we need to maximize the function $f(h) = (H-h)^2 h$. Taking the derivative with respect to $h$: $$\frac{df}{dh} = (H-h)^2 + h \cdot 2(H-h) \cdot (-1)$$ $$= (H-h)^2 - 2h(H-h)$$ $$= (H-h)[(H-h) - 2h]$$ $$= (H-h)(H-3h)$$ Step 7: Solve for the critical point. Setting $\frac{df}{dh} = 0$: $$(H-h)(H-3h) = 0$$ This gives $h = H$ or $h = \frac{H}{3}$. Since $h = H$ is trivial (the entire original cone), the maximum occurs at: $$h = \frac{H}{3}$$ Step 8: Calculate the ratio $\frac{H}{h}$. $$\frac{H}{h} = \frac{H}{H/3} = 3$$ **Final Answer:** The ratio $\dfrac{H}{h} = 3$ The correct option is **Option 2: 3**
Correct Answer: 4

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