A cube of ice melts without changing its shape at the uniform rate of `4(c m^3)/(min)dot` The rate of change of the surface area of the cube, in `(c m^2)/(min),` when the volume of the cube is `125c m^3,` is (a) `-4` (b) `-(16)/5` (c) `-(16)/6` (d) `-8/(15)`

Options

A. `-4`
B. `-16//5`
C. `-16//6`
D. `-8//15`

Answer

To solve the problem, we need to find the rate of change of the surface area of a cube of ice that melts at a uniform rate. Let's go through the solution step by step.

### Step 1: Understand the given information
We know:
- The volume of the cube is decreasing at a rate of \( \frac{dV}{dt} = -4 \, \text{cm}^3/\text{min} \) (negative because the volume is decreasing).
- The volume of the cube when we want to find the rate of change of the surface area is \( V = 125 \, \text{cm}^3 \).

### Step 2: Relate volume to side length
The volume \( V \) of a cube is given by:
\[ V = x^3 \]
where \( x \) is the length of a side of the cube.

### Step 3: Find the side length when volume is 125 cm³
Set the volume equal to 125 cm³:
\[ x^3 = 125 \]
Taking the cube root of both sides:
\[ x = \sqrt[3]{125} = 5 \, \text{cm} \]

### Step 4: Differentiate the volume with respect to time
Differentiate \( V = x^3 \) with respect to time \( t \):
\[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \]
Substituting \( \frac{dV}{dt} = -4 \):
\[ -4 = 3(5^2) \frac{dx}{dt} \]
Calculating \( 5^2 \):
\[ -4 = 3(25) \frac{dx}{dt} \]
\[ -4 = 75 \frac{dx}{dt} \]
Now, solve for \( \frac{dx}{dt} \):
\[ \frac{dx}{dt} = \frac{-4}{75} \, \text{cm/min} \]

### Step 5: Find the surface area of the cube
The surface area \( S \) of a cube is given by:
\[ S = 6x^2 \]

### Step 6: Differentiate the surface area with respect to time
Differentiate \( S = 6x^2 \) with respect to time \( t \):
\[ \frac{dS}{dt} = 12x \frac{dx}{dt} \]

### Step 7: Substitute \( x \) and \( \frac{dx}{dt} \)
Substituting \( x = 5 \) and \( \frac{dx}{dt} = \frac{-4}{75} \):
\[ \frac{dS}{dt} = 12(5) \left(\frac{-4}{75}\right) \]
Calculating:
\[ \frac{dS}{dt} = 60 \left(\frac{-4}{75}\right) = \frac{-240}{75} = \frac{-16}{5} \, \text{cm}^2/\text{min} \]

### Conclusion
The rate of change of the surface area of the cube when the volume is 125 cm³ is:
\[ \frac{dS}{dt} = -\frac{16}{5} \, \text{cm}^2/\text{min} \]

### Answer
The correct option is (b) \(-\frac{16}{5}\).

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