Question 1198377
### **1. Cost Function \( C(x) \):**

The total cost \( C(x) \) includes three components:
1. **Material and manufacturing cost**:
   - The total number of widgets demanded annually is \( N \).
   - Each widget costs \( b \) dollars to make.
   - The **total manufacturing cost** is:
     \[
     \text{Manufacturing Cost} = N \cdot b
     \]

2. **Setup cost**:
   - A production run is done every \( x \) widgets.
   - The total number of production runs per year is \( \frac{N}{x} \) (since \( N \) widgets are needed annually, and \( x \) widgets are produced per run).
   - Each production run incurs a setup cost \( P \).
   - The **total setup cost** is:
     \[
     \text{Setup Cost} = \frac{N}{x} \cdot P
     \]

3. **Storage cost**:
   - Widgets are produced in batches of \( x \) and consumed uniformly over the year.
   - The average number of widgets stored at any time is \( \frac{x}{2} \) (half of the production run, assuming constant consumption).
   - The storage cost per widget per year is \( c \).
   - The **total storage cost** is:
     \[
     \text{Storage Cost} = \frac{x}{2} \cdot c
     \]

Combining these components, the total cost function is:
\[
C(x) = N \cdot b + \frac{N}{x} \cdot P + \frac{x}{2} \cdot c
\]

---

### **2. Constraints on \( x \):**

1. \( x > 0 \): The production run size must be positive.
2. \( x \leq N \): The production run size cannot exceed the total annual demand.

Thus, \( x \) must satisfy:
\[
0 < x \leq N
\]

---

### **3. Minimizing \( C(x) \):**

To minimize \( C(x) \), take the derivative of \( C(x) \) with respect to \( x \) and set it to zero.

\[
C(x) = N \cdot b + \frac{N}{x} \cdot P + \frac{x}{2} \cdot c
\]
\[
\frac{dC}{dx} = -\frac{N \cdot P}{x^2} + \frac{c}{2}
\]

Set \( \frac{dC}{dx} = 0 \):
\[
-\frac{N \cdot P}{x^2} + \frac{c}{2} = 0
\]
\[
\frac{N \cdot P}{x^2} = \frac{c}{2}
\]
\[
x^2 = \frac{2 \cdot N \cdot P}{c}
\]
\[
x = \sqrt{\frac{2 \cdot N \cdot P}{c}}
\]

---

### **4. Verifying Minimum:**

To confirm that this value of \( x \) gives a global minimum, examine the second derivative:
\[
\frac{d^2C}{dx^2} = \frac{2N \cdot P}{x^3}
\]

For \( x > 0 \), \( \frac{d^2C}{dx^2} > 0 \), indicating that \( C(x) \) is convex and has a global minimum at \( x = \sqrt{\frac{2 \cdot N \cdot P}{c}} \).

---

### **5. Number of Production Runs per Year:**

The number of production runs per year is:
\[
\text{Production Runs} = \frac{N}{x}
\]

Substitute \( x = \sqrt{\frac{2 \cdot N \cdot P}{c}} \):
\[
\text{Production Runs} = \frac{N}{\sqrt{\frac{2 \cdot N \cdot P}{c}}} = \sqrt{\frac{N \cdot c}{2 \cdot P}}
\]

---

### **Summary of Results:**

1. **Cost Function**:
   \[
   C(x) = N \cdot b + \frac{N \cdot P}{x} + \frac{x \cdot c}{2}
   \]

2. **Optimal Production Run Size**:
   \[
   x = \sqrt{\frac{2 \cdot N \cdot P}{c}}
   \]

3. **Number of Production Runs per Year**:
   \[
   \text{Production Runs} = \sqrt{\frac{N \cdot c}{2 \cdot P}}
   \]