The total cost function of a firm is given by `C(x)=1/(3)x^(3)-5x^(2)+30x-15` where the selling price per unit is given as Rs 6. Find for what value of x will the profit be maximum.

Answer

To find the value of \( x \) that maximizes the profit for the given total cost function \( C(x) = \frac{1}{3}x^3 - 5x^2 + 30x - 15 \) and a selling price per unit of Rs 6, we will follow these steps:

### Step 1: Define the Profit Function
The profit function \( P(x) \) can be defined as the difference between the total revenue and the total cost. The total revenue \( R(x) \) when selling \( x \) units at Rs 6 per unit is given by:
\[ R(x) = 6x \]
Thus, the profit function can be expressed as:
\[ P(x) = R(x) - C(x) = 6x - C(x) \]
Substituting the cost function:
\[ P(x) = 6x - \left(\frac{1}{3}x^3 - 5x^2 + 30x - 15\right) \]
Simplifying this, we get:
\[ P(x) = 6x - \frac{1}{3}x^3 + 5x^2 - 30x + 15 \]
Combining like terms:
\[ P(x) = -\frac{1}{3}x^3 + 5x^2 - 24x + 15 \]

### Step 2: Differentiate the Profit Function
To find the maximum profit, we need to differentiate \( P(x) \) with respect to \( x \):
\[ P'(x) = -x^2 + 10x - 24 \]

### Step 3: Set the Derivative to Zero
To find the critical points, we set the derivative equal to zero:
\[ -x^2 + 10x - 24 = 0 \]
Multiplying through by -1:
\[ x^2 - 10x + 24 = 0 \]

### Step 4: Solve the Quadratic Equation
We can solve this quadratic equation using the factorization method:
\[ (x - 6)(x - 4) = 0 \]
Thus, the solutions are:
\[ x = 6 \quad \text{and} \quad x = 4 \]

### Step 5: Determine the Maximum Profit
To determine which of these values gives a maximum profit, we can use the second derivative test. We differentiate \( P'(x) \):
\[ P''(x) = -2 \]
Since \( P''(x) < 0 \), this indicates that the profit function is concave down, meaning that both critical points are local maxima. However, we need to evaluate the profit at both points to find the maximum.

### Step 6: Calculate Profit at Critical Points
Calculating \( P(4) \):
\[ P(4) = -\frac{1}{3}(4^3) + 5(4^2) - 24(4) + 15 \]
\[ = -\frac{64}{3} + 80 - 96 + 15 \]
\[ = -\frac{64}{3} + \frac{240}{3} - \frac{288}{3} + \frac{45}{3} \]
\[ = -\frac{64 + 288 - 240 - 45}{3} = -\frac{64 + 288 - 240 - 45}{3} = -\frac{64 + 48}{3} = -\frac{16}{3} \]

Calculating \( P(6) \):
\[ P(6) = -\frac{1}{3}(6^3) + 5(6^2) - 24(6) + 15 \]
\[ = -\frac{216}{3} + 180 - 144 + 15 \]
\[ = -72 + 180 - 144 + 15 = -72 + 51 = -21 \]

### Conclusion
The maximum profit occurs at \( x = 6 \).