The total cost associated with provision of free mid-day meals to x students of a school in primary classes is given by `C(x)= 0.005x^(3)- 0.02x^(2) + 30x + 50`. If the marginal cost is given by rate of change `(dC)/(dx)` total cost, write the marginal cost of food for 300 students.

Answer

To find the marginal cost of food for 300 students, we will follow these steps:

### Step 1: Write down the cost function
The total cost function is given by:
\[ C(x) = 0.005x^3 - 0.02x^2 + 30x + 50 \]

### Step 2: Differentiate the cost function
To find the marginal cost, we need to differentiate the cost function \( C(x) \) with respect to \( x \):
\[ \frac{dC}{dx} = \frac{d}{dx}(0.005x^3 - 0.02x^2 + 30x + 50) \]

### Step 3: Apply the power rule of differentiation
Using the power rule, we differentiate each term:
- The derivative of \( 0.005x^3 \) is \( 0.005 \cdot 3x^{3-1} = 0.015x^2 \)
- The derivative of \( -0.02x^2 \) is \( -0.02 \cdot 2x^{2-1} = -0.04x \)
- The derivative of \( 30x \) is \( 30 \)
- The derivative of the constant \( 50 \) is \( 0 \)

Putting it all together, we have:
\[ \frac{dC}{dx} = 0.015x^2 - 0.04x + 30 \]

### Step 4: Substitute \( x = 300 \)
Next, we will substitute \( x = 300 \) into the marginal cost function:
\[ \frac{dC}{dx} \bigg|_{x=300} = 0.015(300^2) - 0.04(300) + 30 \]

### Step 5: Calculate \( 300^2 \)
Calculating \( 300^2 \):
\[ 300^2 = 90000 \]

### Step 6: Substitute and simplify
Now substituting \( 300^2 \) back into the equation:
\[ \frac{dC}{dx} \bigg|_{x=300} = 0.015(90000) - 0.04(300) + 30 \]
Calculating each term:
- \( 0.015 \times 90000 = 1350 \)
- \( 0.04 \times 300 = 12 \)

Putting it all together:
\[ \frac{dC}{dx} \bigg|_{x=300} = 1350 - 12 + 30 \]
\[ = 1350 - 12 + 30 = 1368 \]

### Final Answer
The marginal cost of food for 300 students is:
\[ \boxed{1368} \]