Find the production level that minimizes the average cost per unit for the cost function $C(x) = 0.004x^2 + 40x + 16000$. Show that it is a minimum by using a graphing calculator to sketch the graph of the average cost function.

Neetesh Kumar | November 13, 2024

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Step-by-step solution:

We first need to find the average cost function to find the production level that minimizes the average cost. The average cost function $\bar{C}(x)$ is defined as:

$\bar{C}(x) = \frac{C(x)}{x} = \dfrac{0.004x^2 + 40x + 16000}{x}$

Simplify $\bar{C}(x)$ by dividing each term by $x$:

$\bar{C}(x) = 0.004x + 40 + \dfrac{16000}{x}$

To find the minimum average cost, we differentiate $\bar{C}(x)$ with respect to $x$ and set the derivative equal to zero.

Step 1: Differentiate the Average Cost Function

Calculate $\bar{C}'(x)$:

$\bar{C}'(x) = \frac{d}{dx} \left( 0.004x + 40 + \frac{16000}{x} \right)$

Using the power rule and the derivative of $\frac{1}{x}$, we get:

$\bar{C}'(x) = 0.004 - \frac{16000}{x^2}$

Step 2: Set $\bar{C}'(x) = 0$ to Find Critical Points

Set the derivative equal to zero:

$0.004 - \frac{16000}{x^2} = 0$

Solve for $x$:

$\frac{16000}{x^2} = 0.004$

Multiply both sides by $x^2$:

$16000 = 0.004x^2$

Now, divide by $0.004$:

$x^2 = \frac{16000}{0.004} = 4000000$

Take the square root of both sides:

$x = \sqrt{4000000} = 2000$

So, the production level that minimizes the average cost is $x = 2000$.

Step 3: Verify that this is a Minimum

To confirm that $x = 2000$ is a minimum.
We can use the second derivative test or observe the behavior of $\bar{C}'(x)$.
In this case, $\bar{C}'(x)$ changes from negative to positive as $x$ increases through $2000$, indicating a minimum at $x = 2000$.

Final Answer

The production level that minimizes the average cost is $x = \boxed{2000}$ units.

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