(1 point) The cost of producing $$ x $$ units of a product is given by $$ C(x)=600+110 x-110 \ln (x), \quad x \geq 1 $$ Find the minimum average cost. Minimum Average Cost = $$ \square $$

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We begin with the cost function

$$
C(x)=600+110x-110\ln(x),\quad x\ge1.
$$

The average cost is

$$
A(x)=\frac{C(x)}{x}=\frac{600}{x}+110-\frac{110\ln(x)}{x}.
$$

To find the minimum average cost we first compute the derivative $$ A'(x) $$.

1. Write the average cost as

$$
A(x)=\frac{600}{x}+110-\frac{110\ln(x)}{x}.
$$

2. The derivative of $$\frac{600}{x}$$ is

$$
\frac{d}{dx}\left(\frac{600}{x}\right)=-\frac{600}{x^2}.
$$

3. For the term $$-\frac{110\ln(x)}{x}$$, write it as

$$
-110\ln(x)\cdot x^{-1}.
$$

Using the product rule, its derivative is

$$
-110\left[\frac{1}{x}\cdot x^{-1} + \ln(x)\cdot \left(-x^{-2}\right)\right] = -110\left[\frac{1}{x^2}-\frac{\ln(x)}{x^2}\right] = -\frac{110\left(1-\ln(x)\right)}{x^2}.
$$

4. Thus, the derivative of the average cost is

$$
A'(x)=-\frac{600}{x^2}-\frac{110\left(1-\ln(x)\right)}{x^2}=-\frac{600+110\left(1-\ln(x)\right)}{x^2}.
$$

To find a critical point, set $$ A'(x)=0 $$:

$$
-\frac{600+110\left(1-\ln(x)\right)}{x^2}=0.
$$

Since $$ x^2>0 $$ for $$ x\ge 1 $$, the numerator must be zero:

$$
600+110\left(1-\ln(x)\right)=0.
$$

Solve for $$\ln(x)$$:

$$
600+110-110\ln(x)=0 \quad\Longrightarrow\quad 710-110\ln(x)=0,
$$
$$
110\ln(x)=710 \quad\Longrightarrow\quad \ln(x)=\frac{710}{110}=\frac{71}{11}.
$$

Thus,

$$
x=e^{\frac{71}{11}}.
$$

Next, we substitute $$ x=e^{\frac{71}{11}} $$ into the cost function $$ C(x) $$:

$$
C\left(e^{\frac{71}{11}}\right)=600+110e^{\frac{71}{11}}-110\ln\left(e^{\frac{71}{11}}\right).
$$

Since $$\ln\left(e^{\frac{71}{11}}\right)=\frac{71}{11}$$, we have

$$
C\left(e^{\frac{71}{11}}\right)=600+110e^{\frac{71}{11}}-110\left(\frac{71}{11}\right)=600+110e^{\frac{71}{11}}-710.
$$

This simplifies to

$$
C\left(e^{\frac{71}{11}}\right)=110e^{\frac{71}{11}}-110.
$$

The minimum average cost is then

$$
A_{\min}=\frac{C\left(e^{\frac{71}{11}}\right)}{e^{\frac{71}{11}}}=\frac{110e^{\frac{71}{11}}-110}{e^{\frac{71}{11}}}=110-\frac{110}{e^{\frac{71}{11}}}.
$$

Thus, the minimum average cost is

$$
\boxed{110\left(1-\frac{1}{e^{\frac{71}{11}}}\right)}.
$$
The minimum average cost is $$ 110\left(1 - \frac{1}{e^{\frac{71}{11}}}\right) $$.

(1 point) The cost of producing units of a product is given by

Find the minimum average cost.
Minimum Average Cost =

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(1 point) The cost of producing units of a product is given by Find the minimum average cost. Minimum Average Cost =