3. [-/3 Points $$ ] $$ DETAILS MY NOTES HARMATHAP12 2.3.009.EP. Suppose that in a monopoly market, the demand function for a product is $$ p=130-0.50 x $$ and the revenue function is $$ R=p x $$, where $$ x $$ is the number of units sold and $$ p $$ is the price per unit (in dollars). Form the revenue function $$ R(x) $$. Find the number of units that must be soldito maximize revenue. $$ x= $$ Find the price of the product (in dollars) that will maximize revenue. $$ \$ \square $$

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Sure, let's work through each part of the problem step by step.

### **1. Form the Revenue Function $$ R(x) $$**

**Given:**
- Demand function: $$ p = 130 - 0.50x $$
- Revenue function: $$ R = p \times x $$

**Step-by-Step:**
1. **Substitute the demand function into the revenue function:**
   $$
   R(x) = p \times x = (130 - 0.50x) \times x
   $$
2. **Expand the equation:**
   $$
   R(x) = 130x - 0.50x^2
   $$

**Revenue Function:**
$$
R(x) = 130x - 0.50x^2
$$

---

### **2. Find the Number of Units $$ x $$ That Must Be Sold to Maximize Revenue**

To find the value of $$ x $$ that maximizes the revenue, we can use the vertex formula for a quadratic function. The revenue function $$ R(x) = -0.50x^2 + 130x $$ is a downward-opening parabola, and its maximum point (vertex) can be found using:

$$
x = -\frac{b}{2a}
$$

**Where:**
- $$ a = -0.50 $$
- $$ b = 130 $$

**Calculations:**
$$
x = -\frac{130}{2 \times (-0.50)} = -\frac{130}{-1} = 130
$$

**Number of Units to Maximize Revenue:**
$$
x = 130 \text{ units}
$$

---

### **3. Find the Price $$ p $$ That Will Maximize Revenue**

Now that we know $$ x = 130 $$ units maximizes revenue, we can find the corresponding price using the demand function.

**Given Demand Function:**
$$
p = 130 - 0.50x
$$

**Substitute $$ x = 130 $$:**
$$
p = 130 - 0.50 \times 130 = 130 - 65 = 65
$$

**Price to Maximize Revenue:**
$$
\$65
$$

---

### **Summary of Answers**

1. **Revenue Function:**
   $$
   R(x) = 130x - 0.50x^2
   $$

2. **Number of Units to Maximize Revenue:**
   $$
   x = 130 \text{ units}
   $$

3. **Price to Maximize Revenue:**
   $$
   \$65
   $$
To maximize revenue, sell **130 units** at a price of **\$65** each.

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