# Question 1. Solve for $x$ as a function of $p$. $x=f(p)$. Determine the domain of $f(p)$.

## a. $p=42-0.3x$ ; $\qquad 0 \leq x \leq 140$

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<summary> Example: </summary>

Question: Solve for $x$ as a function of $p$. $x=f(p)$. Determine the range of $x$, where $p=40 - 0.5x$ and $0 \leq x \leq 80$.

Given the equation:

$$
p = 40 - 0.5x
$$

### Step 1: Solve for $x$

To solve for $x$, rearrange the equation to isolate $x$:

$$
p = 40 - 0.5x
$$

Subtract $40$ from both sides:

$$
p - 40 = -0.5x
$$

Now, divide by $-0.5$:

$$
x = \frac{p - 40}{-0.5}
$$

Simplify the expression:

$$
x = -2(p - 40) = 80 - 2p
$$

Thus, the function $x = f(p)$ is:

$$
x = f(p) = 80 - 2p
$$

### Step 2: Determine the range of $p$

We are given the constraint $0 \leq x \leq 80$. Now we will calculate the corresponding values of $p$.

1. **When $x = 0$:**

$$
p = 40 - 0.5(0) = 40
$$

2. **When $x = 80$:**

$$
p = 40 - 0.5(80) = 40 - 40 = 0
$$

### Conclusion:

The function $x = f(p)$ is:

$$
x = 80 - 2p
$$

The range of $p$ is:

$$
0 \leq p \leq 40
$$

Thus, for $0 \leq x \leq 80$, the corresponding range for $p$ is $0 \leq p \leq 40$.

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## b. $p=50-0.5x^2$ ; $\qquad 0 \leq x \leq 10$

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<summary> Example: </summary>

To solve for $x$ as a function of $p$ from the equation $p = 32 - 0.5x^2$ where $0\leq x \leq 8$, we start by isolating $x$:

1. Rearranging the equation, we have:
$$0.5x^2 = 32 - p$$

2. Multiplying both sides by 2 gives:
$$x^2 = 64 - 2p$$

3. Taking the square root of both sides yields:
$$x = \sqrt{64 - 2p}$$

Given that $x$ must also satisfy the constraints $0 \leq x \leq 8$, we need to determine the range of $p$ for which $x$ is defined and non-negative.

### Finding the Domain of $f(p)$

1. Since $x$ must be non-negative:
$$64 - 2p \geq 0$$
This simplifies to:
$$2p \leq 64 \implies p \leq 32$$

2. Now we need to check when $x$ reaches its upper bound:
$$x = 8 \implies 8 = \sqrt{64 - 2p}$$
Squaring both sides gives:
$$64 = 64 - 2p \implies 2p = 0 \implies p = 0$$

Thus, the function $x = f(p)$ is defined for $p$ in the interval where $x$ is non-negative:
- The lower bound for $p$ is $p = 0$ (when $x = 8$).
- The upper bound is $p = 32$ (when $x = 0$).

### Final Results

The function can be expressed as:
$$x=f(p) = \sqrt{64 - 2p}$$
with the domain:
$$0 \leq p \leq 32$$

In summary, the function $f(p)$ is given by:
$$f(p) = \sqrt{64 - 2p}, \quad \text{for } 0 \leq p \leq 32.$$

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## c. $p=25e^{-x/20}$ ; $\qquad 0 \leq x \leq 20$

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<summary> Example: </summary>

$p=35e^{-x/40}$ and $0 \leq x \leq 40$.

To solve for $x$ as a function of $p$ from the equation $p = 35e^{-x/40}$, we start by isolating $x$:

1. Dividing both sides by 35 gives:
$$\frac{p}{35} = e^{-x/40}$$

2. Taking the natural logarithm of both sides yields:
$$\ln\left(\frac{p}{35}\right) = -\frac{x}{40}$$

3. Multiplying both sides by -40 gives:
$$x = -40 \ln\left(\frac{p}{35}\right)$$

Next, we need to determine the domain of $f(p)$ based on the constraints $0 \leq x \leq 40$.

To solve for $x$ as a function of $p$ from the equation $p = 35e^{-x/40}$, we will isolate $x$ step by step:

1. **Divide by 35**: Start by dividing both sides of the equation by 35:
$$
\frac{p}{35} = e^{-x/40}
$$

2. **Take the natural logarithm**: Next, take the natural logarithm of both sides:
$$
\ln\left(\frac{p}{35}\right) = \ln\left(e^{-x/40}\right)
$$
Using the property of logarithms that $\ln(e^y) = y$, we can simplify the right side:
$$
\ln\left(\frac{p}{35}\right) = -\frac{x}{40}
$$

3. **Isolate $x$**: To isolate $x$, multiply both sides by -40:
$$
x = -40 \ln\left(\frac{p}{35}\right)
$$

Next, we need to determine the domain of $f(p)$ based on the constraints $0 \leq x \leq 40$.

### Finding the Domain of $f(p)$

1. **Non-negativity of $x$**: Since $x$ must be non-negative:
$$
-40 \ln\left(\frac{p}{35}\right) \geq 0
$$
This inequality can be simplified:
- Dividing by -40 (which reverses the inequality):
$$
\ln\left(\frac{p}{35}\right) \leq 0
$$
- Exponentiating both sides (since the exponential function is increasing):

$$e^{\ln\left(\frac{p}{35}\right)} \leq e^0$$
$$
\frac{p}{35} \leq 1 \implies p \leq 35
$$

2. **Upper bound of $x$**: Now we check the upper bound of $x$:
- Set $x = 40$:
$$
40 = -40 \ln\left(\frac{p}{35}\right)
$$
- Dividing by -40 gives:
$$
-1 = \ln\left(\frac{p}{35}\right)
$$
- Exponentiating both sides yields:
$$
\frac{p}{35} = e^{-1} \implies p = 35e^{-1}$$

### Final Results

The function can be expressed as:
$$f(p) = -40 \ln\left(\frac{p}{35}\right)$$
with the domain:
$$35e^{-1} \leq p \leq 35$$

In summary, the function $f(p)$ is given by:
$$f(p) = -40 \ln\left(\frac{p}{35}\right), \quad \text{for } 35e^{-1} \leq p \leq 35.$$

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# Question 2. Find the percent rate of change of $f(x)$ at the indicated value of $x$. Round to the nearest percent.

$f(x)=5100-x^2$; $\hspace{1cm}$ $x=41$

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<summary> Example: </summary>

Given the function:

$$
f(x) = 5300 - x^3
$$

and $x = 19$, we need to find the percent rate of change at this value.

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

The rate of change of a function is given by its derivative. So we first differentiate $f(x)$:

$$
f'(x) = \frac{d}{dx} \left( 5300 - x^3 \right) = -3x^2
$$

### Step 2: Evaluate $f(x)$ and $f'(x)$ at $x = 19$

First, calculate $f(19)$:

$$
f(19) = 5300 - (19)^3 = 5300 - 6859 = -1559
$$

Now, calculate $f'(19)$:

$$
f'(19) = -3(19)^2 = -3(361) = -1083
$$

### Step 3: Calculate the percent rate of change

The percent rate of change is given by:

$$
\text{Percent Rate of Change} = \frac{f'(x)}{f(x)} \times 100
$$

Substitute the values of $f(19)$ and $f'(19)$:

$$
\text{Percent Rate of Change} = \frac{-1083}{-1559} \times 100 \approx 69.46\%
$$

### Step 4: Round to the nearest percent

Rounding 69.46% to the nearest percent gives:

$$
\text{Percent Rate of Change} \approx 69\%
$$

### Conclusion:

The percent rate of change of $f(x)$ at $x = 19$ is approximately **69%**.

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# Question 3. Use the price-demand equation $p+0.004x=32$; $\hspace{1cm}$ $0 \leq p \leq 32$

a. Find the elasticity of demand when $p=$$22. If the $22 price is decreased by 6%, what is the approximate percent change in demand?

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<summary> Example: </summary>

Use the price-demand equation $p + 0.005x = 30$; $\hspace{1cm} 0 \leq p \leq 30$

### a. Find the elasticity of demand when $p = 21$. If the $21 price is decreased by 5%, what is the approximate percent change in demand?

### Step 1: Solve for $x$ as a function of $p$

Given the price-demand equation:

$$
p + 0.005x = 30
$$

Rearrange to solve for $x$:

$$
0.005x = 30 - p
$$

Now, divide both sides by $0.005$:

$$
x = \frac{30 - p}{0.005}
$$

Simplify:

$$
x = 200(30 - p)
$$

Thus, the demand function is:

$$
x = 6000 - 200p
$$

### Step 2: Find the derivative of $x$ with respect to $p$

To calculate the elasticity of demand, we need to differentiate $x$ with respect to $p$:

$$
\frac{dx}{dp} = \frac{d}{dp} \left( 6000 - 200p \right) = -200
$$

### Step 3: Find the elasticity of demand

The formula for the elasticity of demand is:

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

Substitute the values of $x$ and $\frac{dx}{dp}$ into the formula. At $p = 21$:

1. First, calculate $x$ when $p = 21$:

$$
x(21) = 6000 - 200(21) = 6000 - 4200 = 1800
$$

2. Now, substitute into the elasticity formula:

$$
E(21) = -\frac{21}{1800} \cdot (-200) = -\frac{21 \times (-200)}{1800} =- \frac{-4200}{1800} = 2.33
$$

Thus, the elasticity of demand when $p = 21$ is approximately $2.33$.

### Step 4: Determine the percent change in demand when price is decreased by 5%

If the price is decreased by 5%, the approximate percent change in demand can be found using the elasticity of demand. The formula is:

$$
\text{Percent change in demand} = -E(p) \cdot \text{Percent change in price}
$$

The percent change in price is $-5\%$, and $E(21) = 2.33$:

$$
\text{Percent change in demand} = -(2.33) \cdot (-5) = 11.65\%
$$

### Conclusion:

- The elasticity of demand when $p = 21$ is approximately $2.33$.
- If the price is decreased by 5%, the approximate percent change in demand is increased by **11.65%**.

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b. Find the elasticity of demand when $p=$16. If the $16 price is increased by 9%, what is the approximate percent change in demand?

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<summary> Example: </summary>

Use the price-demand equation $p + 0.005x = 30$; $\hspace{1cm} 0 \leq p \leq 30$

### Step 1: Express the demand function
We are given the price-demand equation:

$$
p + 0.005x = 30
$$

First, solve for $x$ in terms of $p$. Rearranging the equation:

$$
0.005x = 30 - p
$$

Now, solve for $x$:

$$
x = \frac{30 - p}{0.005} = 200(30 - p)
$$

Thus, the demand function is:

$$
x(p) = 200(30 - p)
$$

### Step 2: Find the derivative of the demand function
To find the elasticity of demand, we need the derivative of the demand function with respect to $p$. The demand function is:

$$
x(p) = 200(30 - p)
$$

Taking the derivative with respect to $p$:

$$
\frac{dx}{dp} = -200
$$

### Step 3: Calculate the elasticity of demand
The elasticity of demand $E$ is given by the formula:

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

We are asked to find the elasticity at $p = 11$. First, find $x(11)$:

$$
x(11) = 200(30 - 11) = 200 \times 19 = 3800
$$

Now, substitute $p = 11$, $x(11) = 3800$, and $\frac{dx}{dp} = -200$ into the elasticity formula:

$$
E = -\frac{11}{3800} \cdot (-200) = -\frac{11 \times (-200)}{3800} = 0.579
$$

Thus, the elasticity of demand at $p = 11$ is approximately $0.579$.

### Step 4: Determine the effect of an 8% price increase

The percentage change in demand can be approximated using the elasticity formula:

$$
\% \text{change in demand} \approx -E \times \% \text{change in price}
$$

Substitute $E = 0.579$ and $\% \text{change in price} = 8\%$:

$$
\% \text{change in demand} \approx -0.579 \times 8\% = -4.63\%
$$

### Conclusion:
The elasticity of demand at $p = 11$ is approximately $0.579$, and if the price is increased by 8%, the demand is expected to decrease by approximately 4.63%.

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# Question 4 The price-demand equation for hamburgers at a fast food restaurant is $$x+400p=3000$$ Currently the price of a hamburger is $3. If the price is increased by 10%, will the revenue increase or decrease?

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<summary> Example: </summary>

The price-demand equation for hamburgers at a fast food restaurant is $$x+300p=4000$$ Currently the price of a hamburger is $3.50. If the price is increased by 10%, will the revenue increase or decrease?

### Step 1: Express the demand function
We are given the price-demand equation:

$$
x + 300p = 4000
$$

First, solve for $x$ in terms of $p$. Rearranging the equation:

$$
x = 4000 - 300p
$$

Thus, the demand function is:

$$
x(p) = 4000 - 300p
$$

### Step 2: Find the derivative of the demand function
To analyze the impact on revenue, we first calculate the derivative of the demand function with respect to $p$. The demand function is:

$$
x(p) = 4000 - 300p
$$

Taking the derivative with respect to $p$:

$$
\frac{dx}{dp} = -300
$$

### Step 3: Express the revenue function
Revenue $R$ is the product of price and demand:

$$
R(p) = p \cdot x(p) = p \cdot (4000 - 300p)
$$

This simplifies to:

$$
R(p) = 4000p - 300p^2
$$

### Step 4: Find the elasticity of demand
The elasticity of demand $E$ is given by the formula:

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

At $p = 3.50$, first find $x(3.50)$:

$$
x(3.50) = 4000 - 300 \times 3.50 = 4000 - 1050 = 2950
$$

Now, substitute $p = 3.50$, $x(3.50) = 2950$, and $\frac{dx}{dp} = -300$ into the elasticity formula:

$$
E = -\frac{3.50}{2950} \cdot (-300) = -\frac{3.50 \times (-300)}{2950} = 0.356
$$

### Step 5: Analyze the impact on revenue
To understand whether revenue will increase or decrease, consider the relationship between elasticity and revenue:

- If $|E| > 1$ (elastic demand), increasing the price will decrease revenue.
- If $|E| < 1$ (inelastic demand), increasing the price will increase revenue.

Since the elasticity at $p = 3.50$ is $0.356$, which is less than 1 in absolute value, the demand is inelastic. Therefore, increasing the price will increase revenue.

### Conclusion:
If the price of the hamburger is increased by 10%, the revenue will increase because the demand is inelastic at the current price of $3.50$.

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