The radius of the base of a cone is increasing at the rate of 3 cm/min and the altitude is decreasing at the rate of 4 cm/min. The rate of change of lateral surface when the radius is 7 cm and altitude is 24cm is (a) `108pic m^2//min` (b) `7pic m^2//min` (c) `27pic m^2//min` (d) none of these

Options

A. `108picm^(2)//min`
B. `7picm^(2)//min`
C. `27picm^(2)//min`
D. none of these

Answer

To solve the problem step by step, we will follow the given information and apply the necessary formulas.

### Step 1: Understand the given information
- The radius of the base of the cone (r) is increasing at the rate of \( \frac{dr}{dt} = 3 \) cm/min.
- The altitude (height, h) is decreasing at the rate of \( \frac{dh}{dt} = -4 \) cm/min (negative because it is decreasing).
- We need to find the rate of change of the lateral surface area (S) when \( r = 7 \) cm and \( h = 24 \) cm.

### Step 2: Formula for lateral surface area
The lateral surface area \( S \) of a cone is given by the formula:
\[ S = \pi r l \]
where \( l \) is the slant height of the cone.

### Step 3: Find the slant height (l)
The slant height \( l \) can be calculated using the Pythagorean theorem:
\[ l = \sqrt{h^2 + r^2} \]
Substituting the values of \( h \) and \( r \):
\[ l = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ cm} \]

### Step 4: Differentiate the lateral surface area with respect to time
To find \( \frac{dS}{dt} \), we differentiate \( S = \pi r l \) with respect to time \( t \):
\[ \frac{dS}{dt} = \pi \left( r \frac{dl}{dt} + l \frac{dr}{dt} \right) \]

### Step 5: Find \( \frac{dl}{dt} \)
To find \( \frac{dl}{dt} \), we differentiate the equation \( l^2 = h^2 + r^2 \):
\[ 2l \frac{dl}{dt} = 2h \frac{dh}{dt} + 2r \frac{dr}{dt} \]
Substituting the known values:
- \( l = 25 \) cm
- \( h = 24 \) cm
- \( \frac{dh}{dt} = -4 \) cm/min
- \( \frac{dr}{dt} = 3 \) cm/min

This gives us:
\[ 2(25) \frac{dl}{dt} = 2(24)(-4) + 2(7)(3) \]
Calculating the right side:
\[ 50 \frac{dl}{dt} = -192 + 42 \]
\[ 50 \frac{dl}{dt} = -150 \]
\[ \frac{dl}{dt} = -3 \text{ cm/min} \]

### Step 6: Substitute values into the derivative of surface area
Now we can substitute \( r = 7 \) cm, \( l = 25 \) cm, \( \frac{dr}{dt} = 3 \) cm/min, and \( \frac{dl}{dt} = -3 \) cm/min into the equation for \( \frac{dS}{dt} \):
\[ \frac{dS}{dt} = \pi \left( 7(-3) + 25(3) \right) \]
Calculating:
\[ \frac{dS}{dt} = \pi \left( -21 + 75 \right) = \pi (54) \]
Thus,
\[ \frac{dS}{dt} = 54\pi \text{ cm}^2/\text{min} \]

### Final Answer
The rate of change of the lateral surface area when the radius is 7 cm and the altitude is 24 cm is:
\[ \frac{dS}{dt} = 54\pi \text{ cm}^2/\text{min} \]