---
# System prepended metadata

title: In-Class Activity 6.2
tags: [in-class-activity, '6.2']

---

# Question 1:
Find the positive solution of the following equations.

## a. 
$$50 = \frac{x^2}{600} + \frac{1}{5}x + 32$$



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

Find the positive solution of the following equation

$$30 = \frac{x^2}{2000} + \frac{11}{200}x + 26$$

Find the positive solution of the following equation

$$
30 = \dfrac{x^2}{2000} + \dfrac{11}{200}x + 26
$$

### Step-by-step Solution

1. **Subtract 30 from both sides:**

   $$ 
   0 = \dfrac{x^2}{2000} + \dfrac{11}{200}x + 26 - 30 
   $$
   Simplifying this gives:

   $$
   0 = \dfrac{x^2}{2000} + \dfrac{11}{200}x - 4
   $$

2. **Multiply the whole equation by 2000 to eliminate the denominator:**

   $$
   2000 \cdot 0 = 2000 \left( \dfrac{x^2}{2000} + \dfrac{11}{200}x - 4 \right)
   $$
   Simplifying the terms gives:

   $$
   0 = x^2 + 110x - 8000
   $$

3. **Now solve the quadratic equation:**

   The quadratic equation is:

   $$
   x^2 + 110x - 8000 = 0
   $$

4. **Use the quadratic formula:**

   The quadratic formula is given by:

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

   For this equation, $a = 1$ , $b = 110$, and $c = -8000$.

5. **Substitute the values into the quadratic formula:**

   $$
   x = \dfrac{-110 \pm \sqrt{110^2 - 4(1)(-8000)}}{2(1)}
   $$
   Simplifying inside the square root:

   $$
   x = \dfrac{-110 \pm \sqrt{12100 + 32000}}{2}
   $$
   $$
   x = \dfrac{-110 \pm \sqrt{44100}}{2}
   $$

6. **Calculate the square root:**

   $$
   \sqrt{44100} = 210
   $$

7. **Substitute back:**

   $$
   x = \dfrac{-110 \pm 210}{2}
   $$

8. **Find the two possible values for \(x\):**

   - For the positive root:

     $$
     x = \dfrac{-110 + 210}{2} = \dfrac{100}{2} = 50
     $$

   - For the negative root:

     $$ 
     x = \dfrac{-110 - 210}{2} = \dfrac{-320}{2} = -160
     $$

9. **Since we are only interested in the positive solution, the final answer is:**

   $$
   x = 50
   $$

Thus, the positive solution is:

$$
x = 50
$$


</details>

## b. 
$$30 = \frac{x^2}{64} + \frac{1}{4}x + 6$$


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

$$\dfrac{x^2}{375}+\dfrac{2x}{75}+39=40$$


Find the positive solution of the following equation

$$
\dfrac{x^2}{375} + \dfrac{2x}{75} + 39 = 40
$$

### Step-by-step Solution

1. **Subtract 40 from both sides:**

   $$
   0 = \dfrac{x^2}{375} + \dfrac{2x}{75} + 39 - 40
   $$

   Simplifying this gives:

   $$
   0 = \dfrac{x^2}{375} + \dfrac{2x}{75} - 1
   $$

2. **Multiply the whole equation by 375 to eliminate the denominators:**

   $$
   375 \cdot 0 = 375 \left( \dfrac{x^2}{375} + \dfrac{2x}{75} - 1 \right)
   $$

   This simplifies to:

   $$
   0 = x^2 + 10x - 375
   $$

3. **Now solve the quadratic equation:**

   The quadratic equation is:

   $$
   x^2 + 10x - 375 = 0
   $$

4. **Use the quadratic formula:**

   The quadratic formula is given by:

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

   For this equation, $a = 1$, $b = 10$, and $c = -375$.

5. **Substitute the values into the quadratic formula:**

   $$
   x = \dfrac{-10 \pm \sqrt{10^2 - 4(1)(-375)}}{2(1)}
   $$

   Simplifying inside the square root:

   $$
   x = \dfrac{-10 \pm \sqrt{100 + 1500}}{2}
   $$

   $$
   x = \dfrac{-10 \pm \sqrt{1600}}{2}
   $$

6. **Calculate the square root:**

   $$
   \sqrt{1600} = 40
   $$

7. **Substitute back:**

   $$
   x = \dfrac{-10 \pm 40}{2}
   $$

8. **Find the two possible values for $x$:**

   - For the positive root:

     $$
     x = \dfrac{-10 + 40}{2} = \dfrac{30}{2} = 15
     $$

   - For the negative root:

     $$ 
     x = \dfrac{-10 - 40}{2} = \dfrac{-50}{2} = -25
     $$

9. **Since we are only interested in the positive solution, the final answer is:**

   $$
   x = 15
   $$

Thus, the positive solution is:

$$
x = 15
$$

</details>

# Question 2:
Find the following derivatives

## a. For the function $$f(x) = \ln(3x^9)$$ find: $$f'(x) = $$

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

To solve for the derivative of the function:

$$
f(x) = \ln(4x^8)
$$

we will apply the rules of logarithmic differentiation and the chain rule.

### Step-by-step Solution

1. **Rewrite the logarithmic expression using logarithmic properties:**

   Using the property of logarithms that $\ln(ab) = \ln(a) + \ln(b)$, we can rewrite the function as:

   $$
   f(x) = \ln(4) + \ln(x^8)
   $$

2. **Simplify the logarithmic expression:**

   Using the property $\ln(x^n) = n \ln(x)$, we simplify $\ln(x^8)$ as:

   $$
   f(x) = \ln(4) + 8\ln(x)
   $$

3. **Differentiate both terms:**

   - The derivative of $\ln(4)$ is $0$, since $\ln(4)$ is a constant.
   - The derivative of $8\ln(x)$ is $\dfrac{8}{x}$, using the basic derivative rule for $\ln(x)$.

4. **Final derivative:**

   Therefore, the derivative of $f(x)$ is:

   $$
   f'(x) = \dfrac{8}{x}
   $$

Thus, the derivative of $f(x)$ is:

$$
f'(x) = \dfrac{8}{x}
$$


</details>

## b. For the function $$g(x) = \frac{x + 8}{x - 5}$$ find: $$g'(x) = $$

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

To find the derivative of the function:

$$
g(x) = \dfrac{x + 9}{x - 4}
$$

we will use the quotient rule. The quotient rule is given by:

$$
\left( \dfrac{f(x)}{h(x)} \right)' = \dfrac{f'(x)h(x) - f(x)h'(x)}{[h(x)]^2}
$$

For our function, we identify:

- Top function: $f(x) = x + 9$
- Bottom function: $h(x) = x - 4$

### Step-by-step Differentiation

| Original / Derivative  | Top Function $f(x)$  | Bottom Function $h(x)$ |
|------------------------|----------------------|------------------------|
| Original               | $x + 9$              | $x - 4$                |
| Derivative             | $f'(x) = 1$          | $h'(x) = 1$            |

Now, applying the quotient rule:

$$
g'(x) = \dfrac{(1)(x - 4) - (x + 9)(1)}{(x - 4)^2}
$$

Simplifying the numerator:

$$
g'(x) = \dfrac{x - 4 - x - 9}{(x - 4)^2}
$$

This simplifies further:

$$
g'(x) = \dfrac{-13}{(x - 4)^2}
$$

### Final Answer

Thus, the derivative of the function is:

$$
g'(x) = \dfrac{-13}{(x - 4)^2}
$$



</details>

## c. For the function $$h(x) = -2x^5 + 4x^4 - 2x^2 + 1$$ find: $$h'(x) = $$

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


To find the derivative of the function:

$$
h(x) = -3x^6 + 7x^5 - 2x^3 + 8
$$

we will apply the basic rules of differentiation, specifically the power rule, which states that for any term $ax^n$, the derivative is $a \cdot n \cdot x^{n-1}$.

### Step-by-step Differentiation

1. **Differentiate each term individually:**

   - The derivative of $-3x^6$ is:

     $$
     \dfrac{d}{dx}[-3x^6] = -3 \cdot 6x^{6-1} = -18x^5
     $$

   - The derivative of $7x^5$ is:

     $$
     \dfrac{d}{dx}[7x^5] = 7 \cdot 5x^{5-1} = 35x^4
     $$

   - The derivative of $-2x^3$ is:

     $$
     \dfrac{d}{dx}[-2x^3] = -2 \cdot 3x^{3-1} = -6x^2
     $$

   - The derivative of the constant $8$ is $0$ since the derivative of a constant is always zero.

2. **Combine all the derivatives:**

   The derivative of the entire function $h(x)$ is:

   $$
   h'(x) = -18x^5 + 35x^4 - 6x^2
   $$

### Final Answer

Thus, the derivative of the function is:

$$
h'(x) = -18x^5 + 35x^4 - 6x^2
$$



</details>


# Question 3:
Find the consumers' and producers' surplus at the equilibrium price level for the given price-demand and price-supply equations.

$$p = D(x) = 50 - \frac{x}{10}, \quad p = S(x) = 11 + \frac{x}{20}$$

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


Find the consumers' and producers' surplus at the equilibrium price level for the given price-demand and price-supply equations.

$$p = D(x) = -0.5x+500, \quad p = S(x) = 0.25x+260$$

### Given:
- Demand equation: 
  $$ 
  p = D(x) = -0.5x + 500 
  $$
- Supply equation: 
  $$ 
  p = S(x) = 0.25x + 260 
  $$

We need to find the consumers' and producers' surplus at the equilibrium price level using the formula for consumers' and producers' surplus. The formula for consumers' surplus is:

$$
\text{Consumers' Surplus} = \int_0^{\bar{x}} \left(D(x) - \bar{p}\right) \, dx
$$

The formula for producers' surplus is:

$$
\text{Producers' Surplus} = \int_0^{\bar{x}} \left(\bar{p} - S(x)\right) \, dx
$$

Where:
- $\bar{x}$ is the equilibrium quantity.
- $\bar{p}$ is the equilibrium price.

### Step 1: Find the equilibrium quantity and price

At equilibrium, the demand equals the supply:
$$
D(x) = S(x)
$$

Substitute the equations for $D(x)$ and $S(x)$:

$$
-0.5x + 500 = 0.25x + 260
$$

Solve for $x$:

\begin{align}
500 - 260 &= 0.25x + 0.5x \\
240 &= 0.75x \\
x &= \dfrac{240}{0.75} \\
x &= 320
\end{align}

Thus, the equilibrium quantity is $\bar{x} = 320$.

Now, substitute $x = 320$ into either the demand or supply equation to find the equilibrium price:

$$
p = D(320) = -0.5(320) + 500
$$

\begin{align}
p &= -160 + 500 \\
p &= 340
\end{align}

Thus, the equilibrium price is $\bar{p} = 340$.

### Step 2: Consumers' Surplus

The formula for consumers' surplus is:

$$
\text{Consumers' Surplus} = \int_0^{320} \left(D(x) - 340\right) \, dx
$$

Substitute the demand function $D(x) = -0.5x + 500$:

$$
\text{Consumers' Surplus} = \int_0^{320} \left(-0.5x + 500 - 340\right) \, dx
$$

Simplify the integrand:

$$
\text{Consumers' Surplus} = \int_0^{320} \left(-0.5x + 160\right) \, dx
$$

Now calculate the integral:

$$
\int (-0.5x + 160) \, dx = -0.25x^2 + 160x
$$

Evaluate the integral from $0$ to $320$:

\begin{align}
\left[-0.25(320)^2 + 160(320)\right] - [0] &= -0.25(102400) + 160(320) \\
&= -25600 + 51200 \\
&= 25600
\end{align}

Thus, the consumers' surplus is:

$$
\text{Consumers' Surplus} = 25600
$$

### Step 3: Producers' Surplus

The formula for producers' surplus is:

$$
\text{Producers' Surplus} = \int_0^{320} \left(340 - S(x)\right) \, dx
$$

Substitute the supply function $S(x) = 0.25x + 260$:

$$
\text{Producers' Surplus} = \int_0^{320} \left(340 - (0.25x + 260)\right) \, dx
$$

Simplify the integrand:

$$
\text{Producers' Surplus} = \int_0^{320} \left(340 - 0.25x - 260\right) \, dx
$$

$$
\text{Producers' Surplus} = \int_0^{320} \left(80 - 0.25x\right) \, dx
$$

Now calculate the integral:

$$
\int (80 - 0.25x) \, dx = 80x - 0.125x^2
$$

Evaluate the integral from $0$ to $320$:

\begin{align}
\left[80(320) - 0.125(320)^2\right] - [0] &= 80(320) - 0.125(102400) \\
&= 25600 - 12800 \\
&= 12800
\end{align}

Thus, the producers' surplus is:

$$
\text{Producers' Surplus} = 12800
$$

### Final Answer:
- **Consumers' Surplus:** $25600$
- **Producers' Surplus:** $12800$


</details>

---

# Question 4:
Find the producers' surplus at a price level of $$\bar{p} = 67$$ for the price supply equation

$$p = S(x) = 10 + \frac{x}{10} + \frac{3}{10000}x^2$$

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

Let the price supply equation be $S(x)=15 + \frac{x}{10} + \frac{3}{10000}x^2$ with an equilibrium price of $103.


### Given:
- Supply equation: 
  $$
  S(x) = 15 + \frac{x}{10} + \frac{3}{10000}x^2
  $$
- Equilibrium price: 
  $$
  \bar{p} = 103
  $$

We need to find the producers' surplus at the equilibrium price. The formula for producers' surplus is:

$$
\text{Producers' Surplus} = \int_0^{\bar{x}} \left(\bar{p} - S(x)\right) \, dx
$$

Where $\bar{x}$ is the equilibrium quantity. To find $\bar{x}$, we solve for $x$ when $S(x) = 103$.

### Step 1: Solve for equilibrium quantity

Set the supply function equal to 103:

$$
15 + \frac{x}{10} + \frac{3}{10000}x^2 = 103
$$

Subtract 15 from both sides:

$$
\frac{x}{10} + \frac{3}{10000}x^2 = 88
$$

Multiply through by 10000 to eliminate fractions:

$$
1000x + 3x^2 = 880000
$$

Rearrange into standard quadratic form:

$$
3x^2 + 1000x - 880000 = 0
$$

Solve this quadratic equation using the quadratic formula:

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

Substitute $a = 3$, $b = 1000$, and $c = -880000$:

$$
x = \frac{-1000 \pm \sqrt{1000^2 - 4(3)(-880000)}}{2(3)}
$$

Simplify:

$$
x = \frac{-1000 \pm \sqrt{1000000 + 10560000}}{6}
$$

$$
x = \frac{-1000 \pm \sqrt{11560000}}{6}
$$

$$
x = \frac{-1000 \pm 3400}{6}
$$

Thus, the two possible values for $x$ are:

$$
x = \frac{-1000 + 3400}{6} = 400
$$

or

$$
x = \frac{-1000 - 3400}{6} = -733.33
$$

Since quantity $x$ cannot be negative, we take $x = 400$.

Thus, the equilibrium quantity is $\bar{x} = 400$.

### Step 2: Find Producers' Surplus

The formula for producers' surplus is:

$$
\text{Producers' Surplus} = \int_0^{400} \left(103 - \left(15 + \frac{x}{10} + \frac{3}{10000}x^2\right)\right) \, dx
$$

Simplify the integrand:

$$
\text{Producers' Surplus} = \int_0^{400} \left(88 - \frac{x}{10} - \frac{3}{10000}x^2\right) \, dx
$$


To solve the integral for the producer's surplus:

$$
\int_0^{400} \left(88 - \frac{x}{10} - \frac{3}{10000}x^2\right) \, dx
$$

We calculate the integral using the following steps:


\begin{align*}
\int_0^{400} \left(88 - \frac{x}{10} - \frac{3}{10000}x^2\right) \, dx &= \left[ 88x - \frac{x^2}{20} - \frac{x^3}{10000} \right]_0^{400} \\
&= \left[ 88(400) - \frac{400^2}{20} - \frac{400^3}{10000} \right] \\
&= \left[ 35200 - \frac{160000}{20} - \frac{64000000}{10000} \right] \\
&= \left[ 35200 - 8000 - 6400 \right] \\
&= 20800
\end{align*}


Thus, the producer's surplus is:

$$
20800
$$

</details>