Question

suppose $c(x) = 0.02x^2 + 2x + 3800$ is the total cost of a company to produce $x$ units of a certain product. first find the derivative of the average cost $a(x) = \frac{c(x)}{x}$. then determine the production level $x$ that minimizes $a(x)$. (round the production level to two decimal places.)
$a(x) = \square$
$x = \square$ units
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Explanation:

Step1: Define average cost function

$A(x) = \frac{C(x)}{x} = \frac{0.02x^2 + 2x + 3800}{x} = 0.02x + 2 + \frac{3800}{x}$

Step2: Compute derivative $A'(x)$

Use power rule: $\frac{d}{dx}(x^n)=nx^{n-1}$, $\frac{d}{dx}(\frac{k}{x})=-\frac{k}{x^2}$
$A'(x) = 0.02 - \frac{3800}{x^2}$

Step3: Set $A'(x)=0$ to find critical points

$0.02 - \frac{3800}{x^2} = 0$
Rearrange: $0.02x^2 = 3800$
$x^2 = \frac{3800}{0.02} = 190000$
$x = \sqrt{190000} \approx 435.89$ (we discard negative $x$ as units can't be negative)

Step4: Verify minimum (second derivative test)

$A''(x) = \frac{7600}{x^3}$, which is positive for $x>0$, so $x$ is a minimum.

Answer:

$A'(x) = 0.02 - \frac{3800}{x^2}$
$x = 435.89$ units