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 `H/h` is

Explanation

To solve the problem of finding the ratio \( \frac{H}{h} \) where the volume of the inverted cone carved from a solid cone is maximized, we can follow these steps:

### Step 1: Understand the Geometry of the Problem
We have a solid cone with height \( H \) and radius \( R \). An inverted cone with height \( h \) and radius \( r \) is carved out from the top of the solid cone. The goal is to maximize the volume of the inverted cone.

### Step 2: Establish the Relationship Between the Radii and Heights
The radius of the inverted cone at height \( h \) can be expressed in terms of \( H \) and \( h \). Since the radius decreases linearly, we can write:
\[ r = \frac{R}{H} (H - h) \]
This equation shows that as \( h \) increases, the radius \( r \) decreases.

### Step 3: Write the Volume of the Inverted Cone
The volume \( V \) of a cone is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting the expression for \( r \):
\[ V = \frac{1}{3} \pi \left(\frac{R}{H} (H - h)\right)^2 h \]
This simplifies to:
\[ V = \frac{1}{3} \pi \frac{R^2}{H^2} (H - h)^2 h \]

### Step 4: Differentiate the Volume with Respect to \( h \)
To find the height \( h \) that maximizes the volume, we need to differentiate \( V \) with respect to \( h \) and set the derivative to zero:
\[ \frac{dV}{dh} = \frac{1}{3} \pi \frac{R^2}{H^2} \left[2(H - h)(-1)h + (H - h)^2\right] \]
Setting \( \frac{dV}{dh} = 0 \) gives:
\[ 2(H - h)(-h) + (H - h)^2 = 0 \]

### Step 5: Solve the Equation
Expanding and rearranging the equation:
\[ (H - h)^2 = 2h(H - h) \]
This can be simplified to:
\[ H^2 - 2Hh + h^2 = 2hH - 2h^2 \]
Combining like terms leads to:
\[ H^2 - 4Hh + 3h^2 = 0 \]
This is a quadratic equation in terms of \( h \).

### Step 6: Use the Quadratic Formula
Using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = -4H, c = H^2 \):
\[ h = \frac{4H \pm \sqrt{(-4H)^2 - 4 \cdot 3 \cdot H^2}}{2 \cdot 3} \]
\[ h = \frac{4H \pm \sqrt{16H^2 - 12H^2}}{6} \]
\[ h = \frac{4H \pm 2H}{6} \]
This gives two potential solutions:
1. \( h = H \) (not valid since the inverted cone cannot have the same height as the solid cone)
2. \( h = \frac{H}{3} \)

### Step 7: Find the Ratio \( \frac{H}{h} \)
Substituting \( h = \frac{H}{3} \) into the ratio:
\[ \frac{H}{h} = \frac{H}{\frac{H}{3}} = 3 \]

Thus, the ratio \( \frac{H}{h} \) is \( 3 \).

### Final Answer
The ratio \( \frac{H}{h} \) is \( 3 \).

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