The cost function of a firm per-day for x units is given by `C(x)= 3000+ 271 x +( x^(3))/(6), ` whereas the revenue function is given by `R(x) = 3300 + 1000x – (x^(3))/(6),0 lt xlt 30 `. Calculate:
(i) the number of units that maximize the profit
(ii) the profit per unit when the maximum profit level been achieved.

Answer

To solve the problem, we need to find the number of units that maximize the profit and the profit per unit when the maximum profit level is achieved. We will follow these steps:

### Step 1: Define the Profit Function
The profit function \( P(x) \) is defined as the difference between the revenue function \( R(x) \) and the cost function \( C(x) \).

Given:
- Cost function:
\[ C(x) = 3000 + 271x + \frac{x^3}{6} \]
- Revenue function:
\[ R(x) = 3300 + 1000x - \frac{x^3}{6} \]

Thus, the profit function is:
\[ P(x) = R(x) - C(x) \]

### Step 2: Substitute the Functions
Substituting the given functions into the profit function:
\[ P(x) = \left(3300 + 1000x - \frac{x^3}{6}\right) - \left(3000 + 271x + \frac{x^3}{6}\right) \]

### Step 3: Simplify the Profit Function
Now, simplify the profit function:
\[ P(x) = 3300 + 1000x - \frac{x^3}{6} - 3000 - 271x - \frac{x^3}{6} \]
\[ P(x) = (3300 - 3000) + (1000x - 271x) - \left(\frac{x^3}{6} + \frac{x^3}{6}\right) \]
\[ P(x) = 300 + 729x - \frac{2x^3}{6} \]
\[ P(x) = 300 + 729x - \frac{x^3}{3} \]

### Step 4: Find the Derivative of the Profit Function
To maximize the profit, we need to find the derivative of \( P(x) \) and set it to zero:
\[ P'(x) = 729 - \frac{d}{dx}\left(\frac{x^3}{3}\right) \]
\[ P'(x) = 729 - x^2 \]

### Step 5: Set the Derivative to Zero
Set the derivative equal to zero to find the critical points:
\[ 729 - x^2 = 0 \]
\[ x^2 = 729 \]
\[ x = \sqrt{729} = 27 \]

### Step 6: Verify the Maximum
Since \( x \) must be non-negative, we take \( x = 27 \).

### Step 7: Calculate Maximum Profit
Now, we substitute \( x = 27 \) back into the profit function to find the maximum profit:
\[ P(27) = 300 + 729(27) - \frac{27^3}{3} \]
Calculating each term:
- \( 729 \times 27 = 19683 \)
- \( 27^3 = 19683 \)
- Thus,
\[ P(27) = 300 + 19683 - \frac{19683}{3} \]
Calculating \( \frac{19683}{3} = 6561 \):
\[ P(27) = 300 + 19683 - 6561 = 13422 \]

### Step 8: Calculate Profit Per Unit
Finally, to find the profit per unit when the maximum profit is achieved:
\[ \text{Profit per unit} = \frac{P(27)}{27} = \frac{13422}{27} \approx 497.11 \]

### Final Answers
(i) The number of units that maximize the profit is \( x = 27 \).

(ii) The profit per unit when the maximum profit level is achieved is approximately \( 497.11 \).