---
# System prepended metadata

title: Project 3 with Examples
tags: [projects]

---

# Question 1. A company produces smart watches. At a production level of 500 watches per week they noticed that each additional watch increased weekly revenue by about $100. When the company reached a production level of 700 watches per week they noticed that each additional watch increased the revenue by about $80. The weekly cost to produce x smart watches is $$C(x)=32375+40x$$

## a. Assuming that the marginal revenue is a linear function, find the equation for the marginal revenue $R'(x)$.

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

A company produces smart watches. At a production level of 400 watches per week they noticed that each additional watch increased weekly revenue by about $150. When the company reached a production level of 800 watches per week they noticed that each additional watch increased the revenue by about $70. The weekly cost to produce x smart watches is $$C(x)=27350+50x$$ 

a. Assuming that the marginal revenue is a linear function, find the equation for the marginal revenue $R'(x)$.
    
### Step 1: Marginal Revenue as a Linear Function

We are given that the marginal revenue, $R'(x)$, is a linear function, so we can express it in the form:

$$
R'(x) = mx + b
$$

Where:
- $m$ is the slope of the marginal revenue function,
- $b$ is the y-intercept.

### Step 2: Use the Information Provided

We know two points that describe how the marginal revenue changes with the production level:

1. At $x = 400$, the marginal revenue is $R'(400) = 150$.
2. At $x = 800$, the marginal revenue is $R'(800) = 70$.

We will use these two points to find the slope and intercept.

### Step 3: Find the Slope $m$

The slope $m$ is the rate of change of marginal revenue with respect to the production level. We can calculate it using the formula for the slope between two points $(x_1, R'_1)$ and $(x_2, R'_2)$:

$$
m = \frac{R'_2 - R'_1}{x_2 - x_1}
$$

Substitute the values:

$$
m = \frac{70 - 150}{800 - 400}
$$

$$
m = \frac{-80}{400} = -0.2
$$

### Step 4: Find the Y-Intercept $b$

Now that we know the slope $m = -0.2$, we can use one of the points to find the y-intercept $b$. Using the point $(400, 150)$:

$$
R'(400) = -0.2(400) + b
$$

Substitute $R'(400) = 150$:

$$
150 = -80 + b
$$

$$
b = 230
$$

### Step 5: Write the Equation for $R'(x)$

Now that we have both the slope $m = -0.2$ and the intercept $b = 230$, the equation for the marginal revenue function is:

$$
R'(x) = -0.2x + 230
$$

### Final Answer:

The equation for the marginal revenue function is:

$$
R'(x) = -0.2x + 230
$$


</details>

## b. Find the revenue function $R(x)$ using the fact that the revenue is $0 if no watches are produced.


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

### Step 1: Relationship Between Revenue and Marginal Revenue

The marginal revenue function $R'(x)$ represents the rate of change of the revenue function $R(x)$. To find the revenue function, we need to integrate the marginal revenue function $R'(x)$.

We are given the marginal revenue function:

$$
R'(x) = -0.2x + 230
$$

### Step 2: Integrate the Marginal Revenue Function

To find the revenue function $R(x)$, we integrate the marginal revenue function $R'(x)$:

$$
R(x) = \int (-0.2x + 230) \, dx
$$

### Step 3: Perform the Integration

Now, we integrate each term:

$$
R(x) = -0.2 \int x \, dx + 230 \int 1 \, dx
$$

The integrals are:

$$
R(x) = -0.2 \cdot \frac{x^2}{2} + 230x + C
$$

Simplify:

$$
R(x) = -0.1x^2 + 230x + C
$$

### Step 4: Find the Constant $C$

We are told that the revenue is $0$ if no watches are produced. This means that when $x = 0$, $R(0) = 0$. Substitute $x = 0$ and $R(0) = 0$ into the equation:

$$
0 = -0.1(0)^2 + 230(0) + C
$$

$$
0 = C
$$

Thus, $C = 0$.

### Step 5: Write the Revenue Function $R(x)$

Now that we know $C = 0$, the revenue function is:

$$
R(x) = -0.1x^2 + 230x
$$

### Final Answer:

The revenue function is:

$$
R(x) = -0.1x^2 + 230x
$$


</details>

## c. Find the price–demand equation and its domain.

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

    
    
### Step 1: Relationship Between Revenue and Price

The revenue function is related to the price and quantity of watches sold. The revenue function is given by:

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

Where:
- $R(x)$ is the revenue,
- $p(x)$ is the price per watch,
- $x$ is the quantity of watches sold.

From the previous result, we have the revenue function:

$$
R(x) = -0.1x^2 + 230x
$$

Thus, we can write:

$$
p(x) \cdot x = -0.1x^2 + 230x
$$

### Step 2: Solve for the Price $p(x)$

To find the price per watch $p(x)$, divide both sides of the equation by $x$ (assuming $x \neq 0$):

$$
p(x) = \frac{R(x)}{x} = \frac{-0.1x^2 + 230x}{x}
$$

Simplify:

$$
p(x) = -0.1x + 230
$$

### Step 3: Determine the Domain

The domain of the price-demand equation is the range of values for $x$ such that the price $p(x)$ is non-negative $(p(x) \geq 0)$.

From the equation $p(x) = -0.1x + 230$, we require:

$$
-0.1x + 230 \geq 0
$$

Solve for $x$:

$$
-0.1x \geq -230
$$

$$
x \leq 2300
$$

Since the company cannot produce a negative number of watches, we also have $x \geq 0$.

Thus, the domain of the price-demand function is:

$$
0 \leq x \leq 2300
$$

### Final Answer:

The price-demand equation is:

$$
p(x) = -0.1x + 230
$$

The domain of the price-demand equation is:

$$
0 \leq x \leq 2300
$$


</details>

## d. Plot $R(x)$ and $C(x)$, find the break–even points and verify your result in the plot.


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

    
![image](https://hackmd.io/_uploads/BkPFdmskJx.png)


### Step 1: Revenue and Cost Functions

We already have the revenue function $R(x)$:

$$
R(x) = -0.1x^2 + 230x
$$

The cost function is given as:

$$
C(x) = 27350 + 50x
$$

### Step 2: Find the Break-Even Points

The break-even points occur when the revenue equals the cost, i.e., when:

$$
R(x) = C(x)
$$

Substitute the functions for $R(x)$ and $C(x)$:

$$
-0.1x^2 + 230x = 27350 + 50x
$$

Rearrange the equation:

$$
-0.1x^2 + 230x - 50x = 27350
$$

Simplify:

$$
-0.1x^2 + 180x = 27350
$$

Multiply the entire equation by $10$ to eliminate the decimal:

$$
-x^2 + 1800x = 273500
$$

Rearrange the equation:

$$
x^2 - 1800x + 273500 = 0
$$

### Step 3: Solve the Quadratic Equation

We can solve this quadratic equation using the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

For the equation $x^2 - 1800x + 273500 = 0$, we have:
- $a = 1$
- $b = -1800$
- $c = 273500$

Substitute these values into the quadratic formula:

$$
x = \frac{-(-1800) \pm \sqrt{(-1800)^2 - 4(1)(273500)}}{2(1)}
$$

Simplify:

$$
x = \frac{1800 \pm \sqrt{3240000 - 1094000}}{2}
$$

$$
x = \frac{1800 \pm \sqrt{2146000}}{2}
$$

$$
x = \frac{1800 \pm 1465.65}{2}
$$

Thus, the two solutions are:

$$
x_1 = \frac{1800 + 1465.65}{2} = 1632.83
$$

$$
x_2 = \frac{1800 - 1465.65}{2} = 167.17
$$

### Step 4: Verify with a Plot

The break even points agree with the graph of $R(x)$ and $C(x)$ above.


</details>

## e. Find the profit function $P(x)$ and its domain. Plot $P(x)$.

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

### Step 1: Profit Function Definition

The profit function $P(x)$ is the difference between the revenue function $R(x)$ and the cost function $C(x)$:

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

From the previous results, we know:
- The revenue function is:

$$
R(x) = -0.1x^2 + 230x
$$

- The cost function is:

$$
C(x) = 27350 + 50x
$$

### Step 2: Write the Profit Function

Substitute the revenue and cost functions into the profit function formula:

$$
P(x) = \left(-0.1x^2 + 230x\right) - \left(27350 + 50x\right)
$$

Simplify the equation:

$$
P(x) = -0.1x^2 + 230x - 27350 - 50x
$$

$$
P(x) = -0.1x^2 + 180x - 27350
$$

### Step 3: Determine the Domain

The domain of the profit function is determined by the price-demand equation, which has a domain of $0 \leq x \leq 2300$. This is because beyond 2300 units, the price becomes negative, which is not feasible.

Thus, the domain of the profit function is:

$$
0 \leq x \leq 2300
$$

### Final Answer:

The profit function is:

$$
P(x) = -0.1x^2 + 180x - 27350
$$

The domain of the profit function is:

$$
0 \leq x \leq 2300
$$

### Step 4: Plot $P(x)$

We will now plot the profit function $P(x)$ over its domain.

![image](https://hackmd.io/_uploads/HJxRuXskye.png)


</details>

## f. How many watches are produced per week to maximize weekly profit? What is the price per watch and what is the maximum weekly profit?

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

### Step 1: Profit Function

The profit function is:

$$
P(x) = -0.1x^2 + 180x - 27350
$$

### Step 2: Find the Marginal Profit (Derivative of $P(x)$)

The marginal profit is the derivative of the profit function $P(x)$ with respect to $x$. Differentiate $P(x)$:

$$
P'(x) = \frac{d}{dx} \left(-0.1x^2 + 180x - 27350\right)
$$

The derivative is:

$$
P'(x) = -0.2x + 180
$$

### Step 3: Set the Marginal Profit Equal to Zero

To find the critical points, set $P'(x) = 0$:

$$
-0.2x + 180 = 0
$$

Solve for $x$:

$$
-0.2x = -180
$$

$$
x = \frac{180}{0.2} = 900
$$

Thus, 900 watches are produced per week to maximize profit.

### Step 4: Verify Maximum with Sign Chart of $P'(x)$

We verify that $x = 900$ is a maximum by checking the sign of $P'(x)$ on either side of $x = 900$.

| Interval         | Test $x$ value | Work                          | Sign of $P'(x)$ |
|------------------|----------------|-------------------------------|-----------------|
| $(0, 900)$       | $x = 500$      | $P'(500) = -0.2(500) + 180 = 80$ | Positive        |
| $(900, \infty)$  | $x = 1000$     | $P'(1000) = -0.2(1000) + 180 = -20$ | Negative        |

Since $P'(x)$ changes from positive to negative at $x = 900$, this confirms that $P(x)$ has a maximum at $x = 900$.

### Step 5: Find the Price per Watch

The price-demand equation is:

$$
p(x) = -0.1x + 230
$$

Substitute $x = 900$ into the price-demand equation:

$$
p(900) = -0.1(900) + 230
$$

$$
p(900) = -90 + 230 = 140
$$

Thus, the price per watch to maximize profit is $140.

### Step 6: Find the Maximum Weekly Profit

Substitute $x = 900$ into the profit function $P(x)$ to find the maximum weekly profit:

$$
P(900) = -0.1(900)^2 + 180(900) - 27350= 53750
$$

Thus, the maximum weekly profit is $53,750.

### Final Answer:

- The company should produce 900 watches per week to maximize profit.
- The price per watch should be $140.
- The maximum weekly profit is $53,750.


</details>