Question

A solid \(S\) is made from a cylinder surmounted by a hemisphere on top with both its circular faces sharing a common centre. The radius of cylinder and radius of hemisphere are \(x\) cm. The height of the cylinder is \((20-4x)\) cm and the volume of \(S\) is \(V=\frac{1}{3}\pi y\). Find the maximum value of \(y\).

• A. 480
• B. 360
• C. 320
• D. 160

Hint:

For maximum volume problems, first express the required quantity as a function of one variable, then differentiate and use the second derivative test.

Answer 1

The Correct Option is C

Step 1: Write the volume of the solid.
The solid consists of a cylinder and a hemisphere. 
\[ V = \text{Volume of cylinder}+\text{Volume of hemisphere}. \]

Step 2: Use volume formulas. 
Radius of both cylinder and hemisphere is \(x\). 
Height of cylinder is: 
\[ 20-4x. \]
So, 
\[ V=\pi x^2(20-4x)+\frac{2}{3}\pi x^3. \]

Step 3: Simplify the volume. 
\[ V=20\pi x^2-4\pi x^3+\frac{2}{3}\pi x^3. \]
\[ V=20\pi x^2-\frac{10}{3}\pi x^3. \]

Step 4: Compare with given form. 
Given: 
\[ V=\frac{1}{3}\pi y. \]
Therefore: 
\[ \frac{1}{3}\pi y=20\pi x^2-\frac{10}{3}\pi x^3. \]
Multiplying both sides by \(\frac{3}{\pi}\), we get: 
\[ y=60x^2-10x^3. \]

Step 5: Differentiate \(y\) with respect to \(x\). 
\[ \frac{dy}{dx}=120x-30x^2. \]
\[ \frac{dy}{dx}=30x(4-x). \]

Step 6: Find critical points. 
For maximum or minimum: 
\[ \frac{dy}{dx}=0. \]
\[ 30x(4-x)=0. \]
So, 
\[ x=0 \quad \text{or} \quad x=4. \]
Since \(x=0\) gives zero volume, we take: 
\[ x=4. \]

Step 7: Verify maximum using second derivative. 
\[ \frac{d^2y}{dx^2}=120-60x. \]
At \(x=4\): 
\[ \frac{d^2y}{dx^2}=120-240=-120<0. \]
Hence, \(y\) is maximum at \(x=4\). 

Step 8: Find maximum value of \(y\). 
\[ y=60x^2-10x^3. \]
Substitute \(x=4\): 
\[ y=60(4)^2-10(4)^3. \]
\[ y=60(16)-10(64). \]
\[ y=960-640=320. \]
Final Answer: 
\[ \boxed{320} \]