# Math 132, Activity for Section 4.6

Show your work, explain your solution.

## Question 1:
For each price demand equation, find the revenue function $R(x)$ and its domain.

1. $x + 100p = 6000$

<details> <summary> Example: </summary>

For the price-demand equation:
$$x + 200p = 8000$$

Find the revenue function $R(x)$ and its domain.

### Step 1: Rewrite the price-demand equation in slope-intercept form ($mx + b$ form)

Solve for $p$ in terms of $x$:
$$x + 200p = 8000$$
$$200p = 8000 - x$$
$$p = -\frac{1}{200}x + \frac{8000}{200}$$
$$p = -0.005x + 40$$

Now, we have the price-demand equation in $mx + b$ form:
$$p = -0.005x + 40$$

### Step 2: Find the revenue function $R(x)$

The revenue function is defined as:
$$R(x) = x \cdot p$$

Substitute $p = -0.005x + 40$ into the revenue function:
$$R(x) = x \cdot (-0.005x + 40)$$

Simplify the expression:
$$R(x) = -0.005x^2 + 40x$$

### Step 3: Domain of $R(x)$

For the revenue function to be valid, $x$ must satisfy the original price-demand equation. Since $p \geq 0$ (the price cannot be negative):
$$-0.005x + 40 \geq 0$$

Solve for $x$:
$$40 \geq 0.005x$$
$$x \leq \frac{40}{0.005}$$
$$x \leq 8000$$

Additionally, $x \geq 0$ because the quantity demanded cannot be negative.

Thus, the domain of $R(x)$ is:
$$0 \leq x \leq 8000$$

### Final Answer:

The revenue function is:
$$R(x) = -0.005x^2 + 40x$$

The domain of the revenue function is:
$$0 \leq x \leq 8000$$

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2. $5x + 300p = 9000$

<details> <summary> Example: </summary>

Similar to part 1.

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3. $2x + 600p = 12000$

<details> <summary> Example: </summary>

Similar to part 1.

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4. $10x + e^p = 3000$

<details> <summary> Example: </summary>

For the price-demand equation:
$$20x + e^p = 4000$$

Find the revenue function $R(x)$ and its domain.

### Step 1: Rewrite the price-demand equation:

Solve for $p$ in terms of $x$:
$$20x + e^p = 4000$$
$$e^p = 4000 - 20x$$

Take the natural logarithm on both sides to solve for $p$:
$$p = \ln(4000 - 20x)$$

Now, we have the price-demand equation in $p = \ln(4000 - 20x)$ form.

### Step 2: Find the revenue function $R(x)$

The revenue function is defined as:
$$R(x) = x \cdot p$$

Substitute $p = \ln(4000 - 20x)$ into the revenue function:
$$R(x) = x \cdot \ln(4000 - 20x)$$

### Step 3: Domain of $R(x)$

For the revenue function to be valid, $x$ must satisfy the original price-demand equation. Since $e^p > 0$, this implies:
$$4000 - 20x > 0$$

Solve for $x$:
$$4000 > 20x$$
$$x < \frac{4000}{20}$$
$$x < 200$$

Additionally, $x \geq 0$ because the quantity demanded cannot be negative.

Thus, the domain of $R(x)$ is:
$$0 \leq x < 200$$

### Final Answer:

The revenue function is:
$$R(x) = x \cdot \ln(4000 - 20x)$$

The domain of the revenue function is:
$$0 \leq x < 200$$

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## Question 2:
A company manufactures and sells $x$ smartphones per week. The weekly price–demand and cost equation are, respectively:

$$
p = 500 - 0.4x, \quad C(x) = 20,000 + 20x
$$

What is the maximum weekly profit? How much should the company charge for the phone, and how many phones should be produced to realize the maximum weekly profit?

<details> <summary> Example: </summary>

# Problem:
A company manufactures and sells $x$ smartphones per week. The weekly price–demand and cost equations are, respectively:

$$
p = 400 - 0.2x, \quad C(x) = 5000 + 10x
$$

Find the number of smartphones that should be produced to maximize profit, the price at which they should be sold, and the maximum weekly profit.

### Step 1: Revenue Function

The revenue function is given by:

$$
R(x) = x \cdot p
$$

Substitute the price equation $p = 400 - 0.2x$ into the revenue equation:

$$
R(x) = x(400 - 0.2x)
$$

Simplify:

$$
R(x) = 400x - 0.2x^2
$$

### Step 2: Profit Function

The profit function $P(x)$ is the revenue function minus the cost function:

$$
P(x) = R(x) - C(x)
$$

Substitute the revenue function $R(x) = 400x - 0.2x^2$ and the cost function $C(x) = 5000 + 10x$:

$$
P(x) = (400x - 0.2x^2) - (5000 + 10x)
$$

Simplify:

$$
P(x) = 400x - 0.2x^2 - 5000 - 10x
$$
$$
P(x) = -0.2x^2 + 390x - 5000
$$

### Step 3: Find the Critical Points

To maximize the profit, we take the derivative of the profit function and set it equal to zero:

$$
P'(x) = -0.4x + 390
$$

Set $P'(x) = 0$:

$$
-0.4x + 390 = 0
$$

Solve for $x$:

$$
0.4x = 390
$$
$$
x = \frac{390}{0.4} = 975
$$

Thus, the company should produce **975 smartphones** to maximize profit.

### Step 4: Find the Price

To find the price at which the company should sell the smartphones, substitute $x = 975$ into the price-demand equation $p = 400 - 0.2x$:

$$
p = 400 - 0.2(975)
$$
$$
p = 400 - 195 = 205
$$

Thus, the company should charge **$205** for each smartphone.

### Step 5: Find the Maximum Profit

To find the maximum weekly profit, substitute $x = 975$ into the profit function $P(x) = -0.2x^2 + 390x - 5000$:

$$
P(975) = -0.2(975)^2 + 390(975) - 5000
$$

Calculate:

$$
P(975) = -0.2(950625) + 390(975) - 5000
$$
$$
P(975) = -190125 + 380250 - 5000
$$
$$
P(975) = 185125
$$

Thus, the maximum weekly profit is **$185,125**.

### Final Answer:

- The company should produce **975 smartphones** to maximize profit.
- The price should be set at **$205** per smartphone.
- The maximum weekly profit is **$185,125**.

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## Question 3:
A company manufactures and sells $x$ digital cameras per week. The weekly price–demand and cost equations are, respectively:

$$
p = 400 - 0.4x, \quad C(x) = 2000 + 160x
$$

What is the maximum weekly profit? How much should the company charge for the cameras, and how many cameras should be produced to realize the maximum weekly profit?

<details> <summary> Example: </summary>

A company manufactures and sells $x$ digital cameras per week. The weekly price–demand and cost equations are, respectively:

$$
p = 600 - 0.6x, \quad C(x) = 7000 + 200x
$$

Find the number of cameras that should be produced to maximize profit, the price at which they should be sold, and the maximum weekly profit.

### Step 1: Revenue Function

The revenue function is given by:

$$
R(x) = x \cdot p
$$

Substitute the price equation $p = 600 - 0.6x$ into the revenue equation:

$$
R(x) = x(600 - 0.6x)
$$

Simplify:

$$
R(x) = 600x - 0.6x^2
$$

### Step 2: Profit Function

The profit function $P(x)$ is the revenue function minus the cost function:

$$
P(x) = R(x) - C(x)
$$

Substitute the revenue function $R(x) = 600x - 0.6x^2$ and the cost function $C(x) = 7000 + 200x$:

$$
P(x) = (600x - 0.6x^2) - (7000 + 200x)
$$

Simplify:

$$
P(x) = 600x - 0.6x^2 - 7000 - 200x
$$
$$
P(x) = -0.6x^2 + 400x - 7000
$$

### Step 3: Find the Critical Points

To maximize the profit, we take the derivative of the profit function and set it equal to zero:

$$
P'(x) = -1.2x + 400
$$

Set $P'(x) = 0$:

$$
-1.2x + 400 = 0
$$

Solve for $x$:

$$
1.2x = 400
$$
$$
x = \frac{400}{1.2} = 333.33
$$

Thus, the company should produce **333 digital cameras** to maximize profit.

### Step 4: Find the Price

To find the price at which the company should sell the cameras, substitute $x = 333$ into the price-demand equation $p = 600 - 0.6x$:

$$
p = 600 - 0.6(333)
$$
$$
p = 600 - 199.8 = 400.2
$$

Thus, the company should charge approximately **$400.20** for each camera.

### Step 5: Find the Maximum Profit

To find the maximum weekly profit, substitute $x = 333$ into the profit function $P(x) = -0.6x^2 + 400x - 7000$:

$$
P(333) = -0.6(333)^2 + 400(333) - 7000
$$

Calculate:

$$
P(333) = -0.6(110889) + 133200 - 7000
$$
$$
P(333) = -66533.4 + 133200 - 7000
$$
$$
P(333) = 59666.6
$$

Thus, the maximum weekly profit is **$59,666.60**.

### Final Answer:

- The company should produce **333 digital cameras** to maximize profit.
- The price should be set at approximately **$400.20** per camera.
- The maximum weekly profit is **$59,666.60**.

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