To find the number of units $x$ that minimizes the average cost per unit, we start by calculating the average cost function.

Step 1: Define the Average Cost Function

Given the total cost function: $C(x) = 1280 + 39x + 0.8x^2$

The average cost per unit, $\overline{C}(x)$, is: $\overline{C}(x) = \frac{C(x)}{x} = \frac{1280 + 39x + 0.8x^2}{x}$

Simplify the expression: $\overline{C}(x) = \frac{1280}{x} + 39 + 0.8x$

Step 2: Find the Minimum of the Average Cost Function

To find the minimum of $\overline{C}(x)$, we take the derivative of $\overline{C}(x)$ with respect to $x$, set it to zero, and solve for $x$.

1. Differentiate $\overline{C}(x)$ with respect to $x$: $\overline{C}'(x) = -\frac{1280}{x^2} + 0.8$

2. Set $\overline{C}'(x) = 0$ to find the critical points: $-\frac{1280}{x^2} + 0.8 = 0$

3. Solve for $x$: $\frac{1280}{x^2} = 0.8$ $x^2 = \frac{1280}{0.8}$ $x^2 = 1600$ $x = \sqrt{1600} = 40$

Thus, the number of units $x$ that minimizes the average cost per unit is $x = 40$.

Step 3: Calculate the Minimum Average Cost

Now, substitute $x = 40$ back into the average cost function $\overline{C}(x)$ to find the minimum average cost.

$\overline{C}(40) = \frac{1280}{40} + 39 + 0.8 \times 40$

Calculate each term: $\overline{C}(40) = 32 + 39 + 32 = 103$

Answer

• The number of units that minimizes the average cost per unit is 40 units.
• The minimum average cost per unit is $103.

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Related Questions

1. How would the result change if the cost function included a cubic term?
2. What does the second derivative tell us about the behavior of the average cost function?
3. Can we minimize the total cost instead of the average cost with similar techniques?
4. How would a change in fixed costs (e.g., from 1280 to 1500) impact the minimum average cost?
5. What role does marginal cost play in finding the optimal production level?

Tip:

When optimizing functions with respect to costs, always differentiate carefully, as small changes can impact the outcome.