---
# System prepended metadata

title: Math 132 Practice Exam 2
tags: [exam 2, practice exam]

---

[toc] 

# Math 132, Preparation for Exam 2

## Question 1: 
A manufacturer of a clock reports that when there is a demand for $x$ clocks, his retailer sets the price as:

$$p(x)=100-\frac{x}{40}$$

where $p(x)$ is in dollars.

### (a) Find the equation for $R(x)$, the retailer's revenue from the sale of $x$ clocks.

<details>

<summary> Solution: </summary>

Revenue $R(x)$ is the product of the price per clock $p(x)$ and the number of clocks sold $x$. So,

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

Substitute $p(x)=100-\frac{x}{40}$:

$$R(x) = x \left(100 - \frac{x}{40}\right)$$
$$R(x) = 100x - \frac{x^2}{40}$$

Thus, the revenue function is:

$$R(x)=100x-\frac{x^2}{40}$$

</details>

### (b) Find the marginal revenue function $R'(x)$.

<details>
<summary> Solution: </summary>


To find the marginal revenue function, take the derivative of $R(x)$ with respect to $x$:

$$R'(x) = \frac{d}{dx} \left(100x - \frac{x^2}{40}\right)$$
$$R'(x) = 100 - \frac{2x}{40}$$
$$R'(x) = 100 - \frac{x}{20}$$

Thus, the marginal revenue function is:

$$R'(x)=100-\frac{x}{20}$$

</details>

### `(c)` Find the marginal revenue when 2300 clocks are sold.

<details>
<summary> Solution: </summary>

Substitute $x=2300$ into the marginal revenue function $R'(x)=100-\dfrac{x}{20}$ :

$$R'(2300) = 100 - \frac{2300}{20}$$
$$R'(2300) = 100 - 115$$
$$R'(2300) = -15$$

Thus, the marginal revenue when 2300 clocks are sold is **-15 dollars**.
</details>

### (d) Explain what your answer to `1(c)` means in this situation.

<details>
<summary> Solution: </summary>

The marginal revenue of -15 dollars means that for each additional clock sold beyond 2300, the retailer will lose $15 in revenue. This indicates that selling more than 2300 clocks results in decreasing returns.
</details>


## 2. Find the derivative for each of the following functions. Do not use the 4-step process, use the rules for derivatives.

### (a) $g(x)=\ln(3x)+4e^x$

<details>
<summary> Solution: </summary>

Using the chain rule for $\ln(3x)$ and the basic derivative of $e^x$:

$$g'(x) = \frac{1}{3x} \cdot 3 + 4e^x$$
$$g'(x) = \frac{1}{x} + 4e^x$$

Thus, the derivative is:

$$g'(x)=\frac{1}{x} + 4e^x$$
</details>

### (b) $h(x)=\frac{x^5+6}{x^5}$

<details>
<summary> Solution: </summary>

First, simplify $h(x)$:

$$h(x) = 1 + \frac{6}{x^5}=1+6x^{-5}$$

Now, take the derivative:

$$h'(x) = 0 + 6 \cdot (-5x^{-6})=-30x^{-6}$$
$$h'(x) = -\frac{30}{x^6}$$

Thus, the derivative is:

$$h'(x)=-\frac{30}{x^6}$$

</details>

### `(c)` $K(x)=(3x+5)(x+2)$

<details>
<summary> Solution: </summary>

Use the product rule or expand first:

$$K(x) = 3x^2 + 6x + 5x + 10 = 3x^2 + 11x + 10$$

Now, take the derivative:

$$K'(x) = 6x + 11$$

Thus, the derivative is:

$$K'(x)=6x+11$$
</details>

### (d) $f(x)=\frac{8\sqrt{x}}{x^4+3x}$

<details>
<summary> Solution: </summary>

We will use the quotient rule to differentiate $f(x) = \frac{8\sqrt{x}}{x^4 + 3x}$. The quotient rule states:

$$f'(x) = \frac{(h(x) \cdot g'(x)) - (g(x) \cdot h'(x))}{[h(x)]^2}$$

Where:
- $g(x)$ is the **Top Function** (numerator)
- $h(x)$ is the **Bottom Function** (denominator)

#### Step 1: Identify the components for the quotient rule

- The top function is $g(x)=8x^{1/2}$
- The bottom function is $h(x)=x^4+3x$

#### Step 2: Differentiate the Top and Bottom Functions

The derivative of the top function $g(x)=8x^{1/2}$ is:

$$g'(x) = 8 \cdot \frac{1}{2}x^{1/2 - 1} = 4x^{-1/2}$$

The derivative of the bottom function $h(x)=x^4+3x$ is:

$$h'(x)=4x^3 + 3$$

|               | Top Function                              | Bottom Function |               
|---------------|-------------------------------------------|-----------------|
| Original      | $8x^{1/2}$                                | $x^4 + 3x$      |
| Derivative    | $4x^{-1/2}$                               | $4x^3 + 3$      |

#### Step 3: Apply the quotient rule

Now apply the quotient rule:

$$f'(x) = \frac{(x^4 + 3x) \cdot 4x^{-1/2} - 8x^{1/2} \cdot (4x^3 + 3)}{(x^4 + 3x)^2}$$

#### Step 4: Simplify the expression

Now simplify each term:

1. The first part of the numerator is $(x^4 + 3x) \cdot 4x^{-1/2}$:

   $$ (x^4 + 3x) \cdot 4x^{-1/2} = 4x^{7/2} + 12x^{1/2} $$

2. The second part of the numerator is $8x^{1/2} \cdot (4x^3 + 3)$:

   $$ 8x^{1/2} \cdot (4x^3 + 3) = 32x^{7/2} + 24x^{1/2} $$

Now, substitute these into the quotient rule:

$$f'(x) = \frac{(4x^{7/2} + 12x^{1/2}) - (32x^{7/2} + 24x^{1/2})}{(x^4 + 3x)^2}$$

#### Step 5: Combine like terms

Combine the like terms in the numerator:

$$f'(x) = \frac{(4x^{7/2} - 32x^{7/2}) + (12x^{1/2} - 24x^{1/2})}{(x^4 + 3x)^2}$$
$$f'(x) = \frac{-28x^{7/2} - 12x^{1/2}}{(x^4 + 3x)^2}$$

#### Final Answer:

Thus, the derivative is:

$$f'(x) = \frac{-28x^{7/2} - 12x^{1/2}}{(x^4 + 3x)^2}$$

</details>



### (e) $f(x)=e^x(x-4)^5$

<details>
<summary> Solution: </summary>

We will use the **product rule** and **chain rule** to differentiate $f(x) = e^x(x-4)^5$. The product rule states:

$$f'(x) = g'(x)h(x) + g(x)h'(x)$$

Where:
- $g(x) = e^x$ is the first function.
- $h(x) = (x - 4)^5$ is the second function.

#### Step 1: Identify the components of the product rule and chain rule

| Function       | First Function $g(x)$  | Second Function $h(x)$        | Chain Rule: Outside Function | Chain Rule: Inside Function |
|----------------|------------------------|-------------------------------|------------------------------|-----------------------------|
| Original       | $e^x$                  | $(x - 4)^5$                   | $u^5$                        | $x - 4$                     |
| Derivative     | $e^x$                  | $5(x - 4)^4$                  | $5u^4$                       | $1$                         |

Here, $g(x) = e^x$ is the first function, and $h(x) = (x-4)^5$ is the second function. To differentiate $h(x)$, we use the **chain rule**:
- The outside function is $u^5$ (where $u = x - 4$), and its derivative is $5u^4 = 5(x - 4)^4$.
- The inside function is $x - 4$, with its derivative being $1$.

#### Step 2: Apply the product rule

Now apply the product rule:

$$f'(x) = e^x \cdot (x-4)^5 + e^x \cdot 5(x-4)^4$$

#### Step 3: Simplify the expression

Factor out $e^x(x-4)^4$ from both terms:

$$f'(x) = e^x(x-4)^4 \left[(x - 4) + 5\right]$$

Simplify further:

$$f'(x) = e^x(x-4)^4(x + 1)$$

#### Final Answer:

Thus, the derivative is:

$$f'(x) = e^x(x-4)^4(x + 1)$$

</details>




### (f) $f(x)=(8-5x^2)^4$

<details>
<summary> Solution: </summary>

We will use the **chain rule** to differentiate $f(x) = (8 - 5x^2)^4$. The chain rule states:

$$f'(x) = \frac{d}{dx}\left[\text{outside function}\right] \cdot \frac{d}{dx}\left[\text{inside function}\right]$$

#### Step 1: Identify the components of the chain rule

| Function       | Outside               | Inside         |
|----------------|-----------------------|----------------|
| Original       | $(u)^4$               | $8 - 5x^2$     |
| Derivative     | $4u^3$                | $-10x$         |

Here, $u = 8 - 5x^2$ is the inside function.

#### Step 2: Apply the chain rule

Apply the chain rule by first differentiating the outside function and then multiplying by the derivative of the inside function:

$$f'(x) = 4(8 - 5x^2)^3 \cdot (-10x)$$

#### Step 3: Simplify the expression

Now simplify the derivative:

$$f'(x) = -40x(8 - 5x^2)^3$$

#### Final Answer:

Thus, the derivative is:

$$f'(x) = -40x(8 - 5x^2)^3$$

</details>


### (g) $f(x)=3\ln(2x+3)$

<details>
<summary> Solution: </summary>

We will use the **chain rule** to differentiate $f(x) = 3\ln(2x+3)$. The chain rule states:

$$f'(x) = \frac{d}{dx}\left[\text{outside function}\right] \cdot \frac{d}{dx}\left[\text{inside function}\right]$$

#### Step 1: Identify the components of the chain rule

| Function       | Outside               | Inside          |
|----------------|-----------------------|-----------------|
| Original       | $3\ln(u)$             | $2x + 3$        |
| Derivative     | $\frac{3}{u}$         | $2$             |

Here, $u = 2x + 3$ is the inside function.

#### Step 2: Apply the chain rule

Apply the chain rule by first differentiating the outside function and then multiplying by the derivative of the inside function:

$$f'(x) = 3 \cdot \frac{1}{2x+3} \cdot 2$$

#### Step 3: Simplify the expression

Now simplify the derivative:

$$f'(x) = \frac{6}{2x+3}$$

#### Final Answer:

Thus, the derivative is:

$$f'(x) = \frac{6}{2x+3}$$

</details>



## 3. Suppose $20,000$ is invested at an interest rate of 7% compounded continuously. How long will it take:

### (a) For the amount in the account to grow to $28,000$?

<details>
<summary> Solution: </summary>

For continuously compounded interest, the formula is:

$$A = P e^{rt}$$

Where:
- $A$ is the amount of money after time $t$
- $P$ is the principal amount ($20,000$)
- $r$ is the interest rate ($0.07$)
- $t$ is the time in years

We need to solve for $t$ when $A = 28,000$:

$$28,000 = 20,000 e^{0.07t}$$

Divide both sides by 20,000:

$$1.4 = e^{0.07t}$$

Take the natural logarithm of both sides:

$$\ln(1.4) = 0.07t$$

Solve for $t$:

$$t = \frac{\ln(1.4)}{0.07}$$

Using a calculator:

$$t \approx \frac{0.3365}{0.07} \approx 4.81 \ \text{years}$$

Thus, it will take approximately **4.81 years** for the amount to grow to $28,000$.
</details>

### (b) For the money to double?

<details>
<summary> Solution: </summary>

For the money to double, $A = 2P$, so the equation becomes:

$$2(20,000) = 20,000 e^{0.07t}$$

Simplify:

$$2 = e^{0.07t}$$

Take the natural logarithm of both sides:

$$\ln(2) = 0.07t$$

Solve for $t$:

$$t = \frac{\ln(2)}{0.07}$$

Using a calculator:

$$t \approx \frac{0.6931}{0.07} \approx 9.90 \ \text{years}$$

Thus, it will take approximately **9.90 years** for the money to double.
</details>


## 4. The price-demand equation of an office-supply company selling $x$ paper shredders per year is given by:

$$p=300-\frac{x}{30}$$

### (a) How many paper shredders should the company sell to maximize their revenue? What would the maximum revenue be?

<details>
<summary> Solution: </summary>

Revenue is calculated as:

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

Substitute $p = 300 - \frac{x}{30}$ into the revenue equation:

$$R(x) = \left(300 - \frac{x}{30}\right) \cdot x$$
$$R(x) = 300x - \frac{x^2}{30}$$

To maximize revenue, take the derivative of $R(x)$ and set it equal to zero:

$$R'(x) = 300 - \frac{2x}{30}$$
$$R'(x) = 300 - \frac{x}{15}$$

Set $R'(x) = 0$:

$$300 - \frac{x}{15} = 0$$

Solve for $x$:

$$\frac{x}{15} = 300$$
$$x = 300 \times 15 = 4500$$

Thus, the company should sell **4500 paper shredders** to maximize revenue.

Now, substitute $x = 4500$ into the revenue function to find the maximum revenue:

$$R(4500) = 300(4500) - \frac{4500^2}{30}$$
$$R(4500) = 135000 - \frac{20250000}{30}$$
$$R(4500) = 135000 - 675000 = 675000$$

Thus, the maximum revenue is **$675,000**.

</details>

### (b) Suppose the cost function for producing and selling $x$ paper shredders per year is $C(x)=1000+100x$. How many paper shredders should the company sell to maximize the profit? What price should they charge for the paper shredders?

<details>
<summary> Solution: </summary>

The profit function is the revenue minus the cost:

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

Substitute the revenue function $R(x) = 300x - \frac{x^2}{30}$ and the cost function $C(x) = 1000 + 100x$:

$$P(x) = \left(300x - \frac{x^2}{30}\right) - (1000 + 100x)$$
$$P(x) = 300x - \frac{x^2}{30} - 1000 - 100x$$
$$P(x) = 200x - \frac{x^2}{30} - 1000$$

To maximize profit, take the derivative of $P(x)$ and set it equal to zero:

$$P'(x) = 200 - \frac{2x}{30}$$
$$P'(x) = 200 - \frac{x}{15}$$

Set $P'(x) = 0$:

$$200 - \frac{x}{15} = 0$$

Solve for $x$:

$$\frac{x}{15} = 200$$
$$x = 200 \times 15 = 3000$$

Thus, the company should sell **3000 paper shredders** to maximize profit.

Now, substitute $x = 3000$ into the price-demand equation to find the price:

$$p = 300 - \frac{3000}{30}$$
$$p = 300 - 100 = 200$$

Thus, the company should charge **$200** per paper shredder.

Finally, calculate the maximum profit by substituting $x = 3000$ into the profit function:

$$P(3000) = 200(3000) - \frac{3000^2}{30} - 1000$$
$$P(3000) = 600000 - \frac{9000000}{30} - 1000$$
$$P(3000) = 600000 - 300000 - 1000 = 299000$$

Thus, the maximum profit is **$299,000**.
</details>

## NO 4.5 on Exam 2. Skip problem 5.

## 5. (SKIP THIS PROBLEM) Find the absolute maximum and absolute minimum (if they exist) of the function $f(x)=x^3-6x^2$:

### (a) on the interval $[-1,7]$

<details>
<summary> Solution: </summary>

To find the absolute maximum and minimum on the interval $[-1,7]$, we first take the derivative of the function and find the critical points.

#### Step 1: Find the derivative of $f(x)$

$$f'(x) = 3x^2 - 12x$$
$$f'(x) = 3x(x - 4)$$

Set $f'(x) = 0$ to find the critical points:

$$3x(x - 4) = 0$$

Solve for $x$:

$$x = 0 \quad \text{or} \quad x = 4$$

Thus, the critical points are $x = 0$ and $x = 4$.

#### Step 2: Evaluate $f(x)$ at the critical points and endpoints

Now, evaluate $f(x)$ at the critical points and the endpoints of the interval $[-1,7]$.

- At $x = -1$:

$$f(-1) = (-1)^3 - 6(-1)^2 = -1 - 6 = -7$$

- At $x = 0$:

$$f(0) = 0^3 - 6(0)^2 = 0$$

- At $x = 4$:

$$f(4) = 4^3 - 6(4)^2 = 64 - 96 = -32$$

- At $x = 7$:

$$f(7) = 7^3 - 6(7)^2 = 343 - 294 = 49$$

#### Step 3: Determine the absolute maximum and minimum

From the evaluations:

- $f(-1) = -7$
- $f(0) = 0$
- $f(4) = -32$
- $f(7) = 49$

Thus, the **absolute maximum** is **$49$** at $x = 7$, and the **absolute minimum** is **$-32$** at $x = 4$.


</details>

### (b) on the interval $(0,\infty)$

<details>
<summary> Solution: </summary>

We will follow a similar process, but this time the interval is $(0,\infty)$, so we only need to consider the critical points within this interval.

#### Step 1: Find the critical points

We already know the critical points from part (a):

$$x = 0 \quad \text{and} \quad x = 4$$


Since the interval is $(0,\infty)$, we discard $x = 0$ and only consider $x = 4$.

#### Step 2: Analyze the behavior as $x \to \infty$

As $x \to \infty$, the dominant term in the function $f(x) = x^3 - 6x^2$ is $x^3$, which grows without bound. Therefore, $f(x) \to \infty$ as $x \to \infty$.

#### Step 3: Evaluate $f(x)$ at $x = 4$

We already know from part (a) that:

$$f(4) = -32$$

#### Step 4: Conclusion

On the interval $(0,\infty)$:

- The **absolute minimum** is **$-32$** at $x = 4$.
- There is no absolute maximum because $f(x) \to \infty$ as $x \to \infty$.
</details>


## 6. For the function $f(x)=(x-3)(x^2-6x-3)$:

### (a) Find $f'(x)$, the critical values of $f(x)$, all local extrema, and the intervals on which $f(x)$ is increasing and decreasing.

<details>
<summary> Solution: </summary>

#### Step 1: Expand the function $f(x)$

First, expand $f(x)$:

$$f(x) = (x-3)(x^2-6x-3)$$
$$f(x) = x(x^2 - 6x - 3) - 3(x^2 - 6x - 3)$$
$$f(x) = x^3 - 6x^2 - 3x - 3x^2 + 18x + 9$$
$$f(x) = x^3 - 9x^2 + 15x + 9$$

#### Step 2: Find $f'(x)$

Now, take the derivative of $f(x)$:

$$f'(x) = 3x^2 - 18x + 15$$

#### Step 3: Find the critical points

Set $f'(x) = 0$ to find the critical points:

$$3x^2 - 18x + 15 = 0$$

Divide the equation by 3:

$$x^2 - 6x + 5 = 0$$

Factor the quadratic:

$$(x - 5)(x - 1) = 0$$

Thus, the critical points are $x = 1$ and $x = 5$.

#### Step 4: Determine intervals of increasing and decreasing

To find where $f(x)$ is increasing or decreasing, analyze the sign of $f'(x)$ around the critical points:

| Interval    | Test point $x$ | $f'(x)$ evaluation        | Sign        |
|-------------|-----------------|---------------------------|-------------|
| $(-\infty, 1)$ | 0               | $f'(0) = 3(0)^2 - 18(0) + 15 = 15$ | Positive    |
| $(1, 5)$    | 3               | $f'(3) = 3(3)^2 - 18(3) + 15 = -12$ | Negative    |
| $(5, \infty)$ | 6               | $f'(6) = 3(6)^2 - 18(6) + 15 = 15$ | Positive    |

Thus:
- $f(x)$ is increasing on the intervals $(-\infty,1) \cup (5,\infty)$.
- $f(x)$ is decreasing on the interval $(1,5)$.

#### Step 5: Find local extrema

- At $x = 1$, $f(x)$ changes from increasing to decreasing, so there is a **local maximum** at $x = 1$.
- At $x = 5$, $f(x)$ changes from decreasing to increasing, so there is a **local minimum** at $x = 5$.

The $y$ values are found by plugging the $x$ values into the original function:
- At $x=1$, $y=f(1)=(1-3)(1^2-6(1)-3)=(-2)(-8)=16$
- At $x=5$, $y=f(5)=(5-3)(5^2-6(5)-3)=(2)(25-30-3)=(2)(-8)=-16$

Thus, the local extrema are:
- Local maximum at $x = 1$
- Local minimum at $x = 5$

</details>

### (b) Find $f''(x)$, the inflection points, and the intervals on which $f(x)$ is concave up and concave down.

<details>
<summary> Solution: </summary>

#### Step 1: Find $f''(x)$

Take the derivative of $f'(x)$ to find $f''(x)$:

$$f''(x) = 6x - 18$$

#### Step 2: Find inflection points

Set $f''(x) = 0$ to find the inflection points:

$$6x - 18 = 0$$
$$x = 3$$

The $y$-value is found by plugging the $x=3$ into the original function:
- - At $x=3$, $y=f(3)=(3-3)(3^2-6(3)-3)=(0)(-12)=0$


To conclude that $x=3$ is an inflection point, we have to determine if the concavity changes around it.

#### Step 3: Determine concavity

To find where $f(x)$ is concave up or concave down, analyze the sign of $f''(x)$ around the inflection point:

| Interval    | Test point $x$ | $f''(x)$ evaluation        | Sign        |
|-------------|-----------------|---------------------------|-------------|
| $(-\infty, 3)$ | 2               | $f''(2) = 6(2) - 18 = -6$ | Negative    |
| $(3, \infty)$ | 4               | $f''(4) = 6(4) - 18 = 6$ | Positive    |

Thus:
- $f(x)$ is concave down on $(-\infty, 3)$.
- $f(x)$ is concave up on $(3, \infty)$.
- $f(x)$ has an inflection point at $x=3$, the point $(3,0)$.
- 
</details>

### `(c)` Sketch a graph of the function $f(x)$.

<details>
<summary> Solution: </summary>

Based on the critical points, local extrema, and concavity:

- There is a local maximum at the point $(1,16)$
- There is a local minimum at the point $(5,-16)$
- There is an inflection point at the point $(3,0)$, where the concavity changes from concave down to concave up.

The $y$-intercept is when $x=0$, so $y=f(0)=(0-3)(0^2-6(0)-3)=(-3)(-3)=9$. 
- The $y$-intercept is $(0,9)$.

The graph will increase until $x = 1$, then decrease until $x = 5$, and increase again after $x = 5$. The concavity changes at $x = 3$.

A rough sketch would reflect these features, showing a turning point at $x = 1$ (local max), an inflection point at $x = 3$, and a local min at $x = 5$.


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



</details>




## 7. An art gallery has determined that the demand function for a limited edition print they are preparing is:

$$p = 475e^{-0.02q}$$

### (a) What is the elasticity of demand for a price of $300$?

<details>
<summary> Solution: </summary>

#### Step 1: Find the demand function $q(p)$

We are given the price-demand equation $p = 475e^{-0.02q}$. First, solve for $q$ in terms of $p$.

$$p = 475e^{-0.02q}$$

Divide both sides by 475:

$$\frac{p}{475} = e^{-0.02q}$$

Take the natural logarithm of both sides:

$$\ln\left(\frac{p}{475}\right) = -0.02q$$

Solve for $q$:

$$q = -\frac{1}{0.02}\ln\left(\frac{p}{475}\right)$$
$$q = -50\ln\left(\frac{p}{475}\right)$$

#### Step 2: Find the derivative of $q(p)$

Differentiate $q(p)$ with respect to $p$:

$$\frac{dq}{dp} = -50 \cdot \frac{1}{p} = -\frac{50}{p}$$

#### Step 3: Use the formula for elasticity of demand

The formula for the elasticity of demand is:

$$E(p) = -\left(\frac{p}{q}\right) \cdot \frac{dq}{dp}$$

Substitute $q = -50\ln\left(\frac{p}{475}\right)$ and $\frac{dq}{dp} = -\frac{50}{p}$:

$$E(p) = -\left(\frac{p}{-50\ln\left(\frac{p}{475}\right)}\right) \cdot \left(-\frac{50}{p}\right)$$

Simplify:

$$E(p) = \frac{50}{-50\ln\left(\frac{p}{475}\right)}$$
$$E(p) = -\frac{1}{\ln\left(\frac{p}{475}\right)}$$

#### Step 4: Calculate the elasticity at $p = 300$

Substitute $p = 300$ into the elasticity formula:

$$E(300) =- \frac{1}{\ln\left(\frac{300}{475}\right)}$$

Calculate the logarithm:

$$E(300) = -\frac{1}{\ln(0.6316)} \approx -\frac{1}{-0.4596} \approx 2.175$$

Thus, the elasticity of demand for a price of $300$ is approximately **2.175**.

</details>

### (b) Use elasticity to find the price that will maximize revenue.

<details>
<summary> Solution: </summary>

Revenue is maximized when the elasticity of demand $E(p) = 1$.

#### Step 1: Set $E(p) = 1$

We know that:

$$E(p) = -\frac{1}{\ln\left(\frac{p}{475}\right)}$$

Set $E(p) = 1$:

$$-\frac{1}{\ln\left(\frac{p}{475}\right)} = 1$$

Solve for $\ln\left(\frac{p}{475}\right)$:

$$\ln\left(\frac{p}{475}\right) = -1$$

#### Step 2: Solve for $p$

Exponentiate both sides to remove the logarithm:

$$\frac{p}{475} = e^{-1}$$

Multiply both sides by 475:

$$p = 475e^{-1}$$

Since $e^{-1} \approx 0.3679$:

$$p \approx 475 \times 0.3679 \approx 174.76$$

Thus, the price that will maximize revenue is approximately **$174.76**.

</details>


## Alternate 7. Given $p + 0.004x = 32$ and $0 \leq p \leq 32$.

### (a) Find $E(p)$ when $p = 12$.

<details>
<summary> Solution: </summary>

#### Step 1: Find the demand function $x(p)$

The price-demand equation is given as:

$$p + 0.004x = 32$$

Solve for $x$ in terms of $p$:

$$0.004x = 32 - p$$
$$x = \frac{32 - p}{0.004}$$
$$x = 8000 - 250p$$

Thus, the demand function is $x(p) = 8000 - 250p$.

#### Step 2: Find the derivative of $x(p)$

Differentiate $x(p)$ with respect to $p$:

$$\frac{dx}{dp} = -250$$

#### Step 3: Use the formula for elasticity of demand

The formula for elasticity of demand is:

$$E(p) = -\left(\frac{p}{x}\right) \cdot \frac{dx}{dp}$$

Substitute $x = 8000 - 250p$ and $\frac{dx}{dp} = -250$:

$$E(p) = -\left(\frac{p}{8000 - 250p}\right) \cdot (-250)$$

Simplify:

$$E(p) = \frac{250p}{8000 - 250p}$$

#### Step 4: Calculate the elasticity at $p = 12$

Substitute $p = 12$ into the elasticity formula:

$$E(12) = \frac{250(12)}{8000 - 250(12)}$$
$$E(12) = \frac{3000}{8000 - 3000}$$
$$E(12) = \frac{3000}{5000} = 0.6$$

Thus, the elasticity of demand when $p = 12$ is **0.6**.
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### (b) Should you raise or lower prices to maximize revenue?

<details>
<summary> Solution: </summary>

Revenue is maximized when the elasticity of demand $E(p) = 1$. Since $E(12) = 0.6$, and $0.6 < 1$, the demand is **inelastic** at this price.

When demand is inelastic ($E(p) < 1$), raising prices will increase revenue. Therefore, you should **raise prices** to maximize revenue.
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### `(c)` What price will maximize the revenue?

<details>
<summary> Solution: </summary>

Revenue is maximized when $E(p) = 1$. Set the elasticity formula equal to $1$:

$$\frac{250p}{8000 - 250p} = 1$$

Solve for $p$:

$$250p = 8000 - 250p$$
$$500p = 8000$$
$$p = \frac{8000}{500} = 16$$

Thus, the price that will maximize revenue is **$16**.

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