If the marginal cost function a product is given by `MC=10-4x+3x^(2)` and fixed cost is Rs 500, then the cost function is

Options

A. `10x-2x^(2)+x^(3)`
B. `500+10x-2x^(2)+x^(3)`
C. `-4+6x`
D. `500+10-8x^(2)+9x^(3)`

Answer

To find the cost function from the given marginal cost function, we will follow these steps:

### Step 1: Understand the relationship between Marginal Cost and Cost Function
The marginal cost (MC) function represents the derivative of the cost function (C) with respect to the quantity produced (x). Therefore, we can express this relationship as:
\[ MC = \frac{dC}{dx} \]

### Step 2: Integrate the Marginal Cost Function
Given the marginal cost function:
\[ MC = 10 - 4x + 3x^2 \]
To find the cost function (C), we need to integrate the marginal cost function:
\[ C(x) = \int (10 - 4x + 3x^2) \, dx \]

### Step 3: Perform the Integration
Now, we will integrate each term:
\[ C(x) = \int 10 \, dx - \int 4x \, dx + \int 3x^2 \, dx \]
Calculating each integral:
1. \(\int 10 \, dx = 10x\)
2. \(\int 4x \, dx = 2x^2\)
3. \(\int 3x^2 \, dx = x^3\)

Putting it all together, we have:
\[ C(x) = 10x - 2x^2 + x^3 + C_1 \]
where \(C_1\) is the constant of integration.

### Step 4: Use the Fixed Cost to Find the Constant of Integration
We know that the fixed cost is Rs 500, which means when no units are produced (x = 0), the cost function should equal the fixed cost:
\[ C(0) = 500 \]
Substituting \(x = 0\) into the cost function:
\[ C(0) = 10(0) - 2(0)^2 + (0)^3 + C_1 = C_1 \]
Thus, we have:
\[ C_1 = 500 \]

### Step 5: Write the Final Cost Function
Now substituting \(C_1\) back into the cost function:
\[ C(x) = 10x - 2x^2 + x^3 + 500 \]
This can be rewritten as:
\[ C(x) = 500 + 10x - 2x^2 + x^3 \]

### Conclusion
The cost function is:
\[ C(x) = 500 + 10x - 2x^2 + x^3 \]