The total daily cost (in dollars) of producing $$ x $$ park benches is given by $$ C(x)=3125+15 x+0.2 x^{2} $$ A. Sketch the graphs of the average cost function and the marginal cost function on the same set of coordinale axes. B. Find the minimum average cost

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**A. Sketching the Graphs**

1. The total daily cost is given by  
   $$
   C(x)=3125+15x+0.2x^2.
   $$

2. The **average cost function** is defined as the total cost divided by $$ x $$:
   $$
   \overline{C}(x)=\frac{C(x)}{x}=\frac{3125}{x}+15+0.2x.
   $$
   - As $$ x $$ gets very small, the term $$\frac{3125}{x}$$ dominates, making $$\overline{C}(x)$$ very high.
   - As $$ x $$ increases, the $$\frac{3125}{x}$$ term decreases, but the $$ 0.2x $$ term increases linearly.

3. The **marginal cost function** is the derivative of the total cost function with respect to $$ x $$:
   $$
   C'(x)=\frac{d}{dx}\left(3125+15x+0.2x^2\right)=15+0.4x.
   $$
   - This is a linear function with a $$ y $$-intercept of 15 and a slope of 0.4.

4. **Graphing Together:**
   - On a set of coordinate axes, plot both functions.
   - The average cost function $$\overline{C}(x)=\frac{3125}{x}+15+0.2x$$ is a curve that decreases steeply when $$ x $$ is small, attains a minimum, and then increases roughly linearly for larger $$ x $$.
   - The marginal cost function $$ C'(x)=15+0.4x $$ is a straight line.
   - They intersect where $$\overline{C}(x)=C'(x)$$. As will be seen in part B, this occurs at $$ x=125 $$ and the common value is 65.

**B. Finding the Minimum Average Cost**

1. The average cost function is  
   $$
   \overline{C}(x)=\frac{3125}{x}+15+0.2x.
   $$

2. To find the minimum, differentiate $$\overline{C}(x)$$ with respect to $$ x $$:
   $$
   \overline{C}'(x)=-\frac{3125}{x^2}+0.2.
   $$

3. Set the derivative equal to zero to find the critical value:
   $$
   -\frac{3125}{x^2}+0.2=0.
   $$

4. Solve for $$ x $$:
   $$
   0.2=\frac{3125}{x^2} \quad \Longrightarrow \quad x^2=\frac{3125}{0.2}.
   $$
   Calculate the right-hand side:
   $$
   x^2=\frac{3125}{0.2}=15625.
   $$
   Thus,
   $$
   x=125,
   $$
   where we take the positive square root since $$ x $$ represents the number of benches.

5. Substitute $$ x=125 $$ into $$\overline{C}(x)$$ to obtain the minimum average cost:
   $$
   \overline{C}(125)=\frac{3125}{125}+15+0.2(125).
   $$
   Compute each term:
   $$
   \frac{3125}{125}=25,\quad 0.2(125)=25.
   $$
   Hence,
   $$
   \overline{C}(125)=25+15+25=65.
   $$

6. **Conclusion:**  
   The minimum average cost is $$\boxed{65}$$ dollars, which occurs when $$ x=125 $$ park benches are produced.  
   Note that at this production level, the marginal cost is  
   $$
   C'(125)=15+0.4(125)=15+50=65,
   $$
   which confirms that the graphs of $$\overline{C}(x)$$ and $$ C'(x) $$ intersect at the point $$ (125, 65) $$.
**A. Sketching the Graphs** - **Average Cost Function ($$\overline{C}(x)$$)**: Starts high when $$ x $$ is small, decreases to a minimum, then increases. - **Marginal Cost Function ($$ C'(x) $$)**: Straight line starting at 15 and increasing by 0.4 for each additional bench. **B. Minimum Average Cost** - **Minimum occurs at $$ x = 125 $$ benches**. - **Minimum Average Cost**: \$65 per bench. So, producing 125 park benches minimizes the average cost to \$65.

The total daily cost (in dollars) of producing \( x \) park benches is given by \( C(x)=3125+15 x+0.2 x^{2} \) A. Sketch the graphs of the average cost function and the marginal cost function on the same set of coordinale axes. B. Find the minimum average cost

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