---
# System prepended metadata

title: In-Class Activity 2.7
tags: [in-class-activity]

---

# In-Class Activity 2.7

## Question 1

Combine:

### a. $\dfrac{2^3 \cdot 2^4}{2^5}$


::: spoiler

<summary> Example: </summary>

\begin{align}
\dfrac{3^4 \cdot 3^6}{3^7}&= \dfrac{3^{4+6}}{3^7} \\
&= \dfrac{3^{10}}{3^7} \\
&= 3^{10-7}\\
&=3^3 \\
&=27
\end{align}

:::


::: spoiler

<summary> Solution: </summary>

\begin{align}
\dfrac{2^3 \cdot 2^4}{2^5}&=\dfrac{2^{3+4}}{2^5} \\
&=\dfrac{2^7}{2^5} \\
&=2^{7-5} \\
&=2^2 \\
&=4
\end{align}

:::

### b. $\dfrac{(e^4 \cdot e^{-2})^2}{e^3}$

::: spoiler

<summary> Example: </summary>

\begin{align}
\dfrac{(\pi^3 \cdot \pi^{-4})^2 }{\pi^5}&= \dfrac{(\pi^{3-4} )^2 }{\pi^5} \\
&= \dfrac{(\pi^{-1} )^2 }{\pi^5} \\
&= \dfrac{\pi^{-1 \cdot 2} }{\pi^5} \\
&= \dfrac{\pi^{-2} }{\pi^5} \\
&= \pi^{-2-5} \\
&=\pi^{-7} \\
&=\dfrac{1}{\pi^7}
\end{align}

:::


::: spoiler

<summary> Solution: </summary>

\begin{align}
\dfrac{(e^4 \cdot e^{-2})^2}{e^3}&=\dfrac{(e^{4-2})^2}{e^3} \\
&=\dfrac{(e^2)^2}{e^3} \\
&=\dfrac{e^4}{e^3} \\
&=e^{4-3} \\
&=e^1 \\
&=e
\end{align}

:::

## Question 2

Expand as a sum of logarithms of primes.



### a. $\ln\left(\dfrac{28}{75}\right)$


::: spoiler

<summary> Example: </summary>

Log Laws
$\ln(AB)=\ln A+\ln B$
$\ln(A/B)=\ln A-\ln B$
$\ln(A^n)=n\ln A$

\begin{align}
\ln\left(\dfrac{36}{60}\right) &=\ln 36-\ln 60 \\
&=\ln (6^2) - \ln(6 \cdot 10) \\
&=2\ln(6) - (\ln 6+\ln 10) \\
&=2\ln(6) - \ln 6-\ln 10 \\
&=\ln 6 - \ln 10 \\
&=\ln(2 \cdot 3) - \ln(2 \cdot 5) \\
&=\ln 2 + \ln 3 - (\ln 2 + \ln 5) \\
&=\ln 2 + \ln 3 - \ln 2-\ln 5 \\
&=\ln 3 - \ln 5
\end{align}

:::


::: spoiler

<summary> Solution: </summary>

\begin{align}
\ln\left(\dfrac{28}{75}\right)&=\ln (28)-\ln (75) \\
&=\ln(4 \cdot 7) - \ln (3 \cdot 25) \\
&=\ln 4+ \ln 7 - (\ln 3 + \ln 25) \\
&=\ln 4+ \ln 7 - \ln 3 - \ln 25 \\
&=\ln (2^2)+\ln 7-\ln 3 - \ln 25 \\
&=2\ln 2 +\ln 7 - \ln 3 - \ln (5^2) \\
&=2 \ln 2 +\ln 7 - \ln 3 - 2\ln 5
\end{align}

:::

### b. $\ln\left(\dfrac{105}{88}\right)$




::: spoiler

<summary> Example: </summary>

Log Laws
$\ln(AB)=\ln A+\ln B$
$\ln(A/B)=\ln A-\ln B$
$\ln(A^n)=n\ln A$

\begin{align}
\ln\left(\dfrac{108}{72}\right) &=\ln 108-\ln 72 \\
&=\ln (108) - \ln(72) \\
&=\ln (9 \cdot 12) - \ln (8 \cdot 9) \\
&=\ln 9 +\ln 12-(\ln 8 +\ln 9) \\
&=\ln 9 +\ln 12 - \ln 8 - \ln 9 \\
&=\ln 12 - \ln 8 \\
&=\ln(4 \cdot 3) - \ln(2^3) \\
&=\ln 4+\ln 3 - 3\ln 2 \\
&=\ln(2^2)+\ln 3-3 \ln 2 \\
&=2\ln 2+\ln 3 - 3\ln 2 \\
&=\ln 3 - \ln 2
\end{align}

:::


::: spoiler

<summary> Solution: </summary>

\begin{align}
\ln\left(\dfrac{105}{88}\right)&= \ln(105) - \ln (88) \\
&=\ln (5 \cdot 21) - \ln (11 \cdot 8) \\
&=\ln 5 +\ln 21 - (\ln 11 +\ln 8) \\
&=\ln 5+\ln 21 - \ln 11 - \ln 8 \\
&=\ln 5 +\ln(3 \cdot 7) - \ln 11 - \ln (2^3) \\
&=\ln 5 +\ln 3 + \ln 7-\ln 11 - 3\ln 2
\end{align}

:::




## Question 3

The total cost (in dollars) of producing $x$ food processors is $$C(x)=2000+50x-0.5x^2$$

### a. Find the exact cost to produce the 21st food processor.


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<summary> Example: </summary>

Find the exact cost to produce the 25th food processor.

The cost to produce 25 food processors is
\begin{align}
C(25)&=2000+50(25)-0.5(25)^2 \\
&=2937.50
\end{align}

The cost to produce 24 food processors is
\begin{align}
C(24)&=2000+50(24)-0.5(24)^2 \\
&=2912.00
\end{align}


Thus the cost to produce the 25th food processor is

$C(25)-C(24)=2937.50-2912=25.50$ dollars.

:::

::: spoiler

<summary> Solution: </summary>

Find the exact cost to produce the 21st food processor.

The cost to produce 21 food processors is
\begin{align}
C(21)&=2000+50(21)-0.5(21)^2 \\
&=2829.50
\end{align}

The cost to produce 20 food processors is
\begin{align}
C(20)&=2000+50(20)-0.5(20)^2 \\
&=2800.00
\end{align}


Thus the cost to produce the 21st food processor is

$C(21)-C(20)=2829.50-2800.00=29.50$ dollars.

:::

### b. Use the marginal cost to approximate the cost of producing the 21st food processor.


::: spoiler

<summary> Example: </summary>

Use the marginal cost to approximate the cost of producing the 25th food processor.

$$C(x)=2000+50x-0.5x^2$$

Taking the derivative:

\begin{align}
C'(x)&=(2000+50x-0.5x^2)' \\
&=(2000)' +50(x)'-0.5(x^2)' \\
&=0 +50(1)-0.5(2x) \\
&=50-x
\end{align}

Then plugging in $x=25$:

$C'(25)=50-25=25$ dollars.

:::

::: spoiler

<summary> Solution: </summary>

Use the marginal cost to approximate the cost of producing the 21st food processor.

$$C(x)=2000+50x-0.5x^2$$

Taking the derivative:

\begin{align}
C'(x)&=(2000+50x-0.5x^2)' \\
&=(2000)' +50(x)'-0.5(x^2)' \\
&=0 +50(1)-0.5(2x) \\
&=50-x
\end{align}

Then plugging in $x=21$:

$C'(21)=50-21=29$ dollars.

:::

## Question 4

The price $p$ and demand $x$ for a brand of running shoes are related by the equation $$x=4000-40p$$.

### a. Express the price $p$ in terms of the demand $x$, and find the domain.


::: spoiler

<summary> Example: </summary>

Express the price $p$ in terms of the demand $x$ if $x=3000-50p$.

\begin{align}
x&=3000-50p \\
x-3000&=-50p \\
-50p&=x-3000 \\
\dfrac{-50p}{-50}&=\dfrac{x-3000}{-50} \\
p&=\dfrac{x}{-50} + \dfrac{-3000}{-50} \\
p&=-\dfrac{1}{50}x+60
\end{align}

The domain is found by setting $p \geq 0$:

\begin{align}
p&\geq 0 \\
-\dfrac{1}{50}x+60 &\geq 0 \\
-\dfrac{1}{50}x &\geq -60 \\
(-50) \left(-\dfrac{1}{50}x\right) &\leq (-60)(-50) \\
x &\leq 3000
\end{align}

Also since $x \geq 0$, we have the domain as $0 \leq x \leq 3000$.

:::

::: spoiler

<summary> Solution: </summary>

Express the price $p$ in terms of the demand $x$ if $x=4000-40p$.

\begin{align}
x&=4000-40p \\
x-4000&=-40p \\
-40p&=x-4000 \\
\dfrac{-40p}{-40}&=\dfrac{x-4000}{-40} \\
p&=\dfrac{x}{-40} + \dfrac{-4000}{-40} \\
p&=-\dfrac{1}{40}x+100
\end{align}

The domain is found by setting $p \geq 0$:

\begin{align}
p&\geq 0 \\
-\dfrac{1}{40}x+100 &\geq 0 \\
-\dfrac{1}{40}x &\geq -100 \\
(-40) \left(-\dfrac{1}{40}x\right) &\leq (-100)(-40) \\
x &\leq 4000
\end{align}

Also since $x \geq 0$, we have the domain as $0 \leq x \leq 4000$.

:::

### b. Find the revenue $R(x)$ for the sale of $x$ pairs of running shoes. What is the domain of $R(x)$?

::: spoiler

<summary> Example: </summary>

Since $p=-\dfrac{1}{50}x+60$, the revenue is 
\begin{align}
R(x)&=xp \\
&=x\left(-\dfrac{1}{50}x+60\right) \\
R(x)&=-\dfrac{1}{50}x^2+60x
\end{align}

Domain is the same as before, $0 \leq x \leq 3000$.

:::

::: spoiler

<summary> Solution: </summary>

Since $p=-\dfrac{1}{40}x+100$, the revenue is 
\begin{align}
R(x)&=xp \\
&=x\left(-\dfrac{1}{40}x+100\right) \\
R(x)&=-\dfrac{1}{40}x^2+100x
\end{align}

Domain is the same as before, $0 \leq x \leq 4000$.

:::

### c. Find the marginal revenue at a production level of 1600 pairs and interpret the result.


::: spoiler

<summary> Example: </summary>

Say 500 pairs instead
Take the derivative:

\begin{align}
R(x)&=-\dfrac{1}{50}x^2+60x \\
R'(x)&=-\dfrac{1}{50} (x^2)' +60(x)' \\
R'(x)&=-\dfrac{1}{50} (2x^{2-1}) +60(1) \\
R'(x)&=-\dfrac{1}{50} (2x) +60 \\
R'(x)&=-\dfrac{2}{50} x +60 \\
R'(x)&=-\dfrac{1}{25} x +60 \\
\end{align}

Then plug in $x=500$:

\begin{align}
R'(500)&=-\dfrac{1}{25} (500) +60 \\
&=40
\end{align}

At a production level of 500 pairs of shoes, the revenue increases at a rate of $40 per shoe.

:::

::: spoiler

<summary> Solution: </summary>

Take the derivative:

\begin{align}
R(x)&=-\dfrac{1}{40}x^2+100x \\
R'(x)&=-\dfrac{1}{40} (x^2)' +100(x)' \\
R'(x)&=-\dfrac{1}{40} (2x^{2-1}) +100(1) \\
R'(x)&=-\dfrac{1}{40} (2x) +100 \\
R'(x)&=-\dfrac{2}{40} x +100 \\
R'(x)&=-\dfrac{1}{20} x +100 \\
\end{align}

Then plug in $x=1600$:

\begin{align}
R'(1600)&=-\dfrac{1}{20} (1600) +100 \\
&=-80+100 \\
&=20
\end{align}

At a production level of 1600 pairs of shoes, the revenue increases at a rate of $20 per shoe.

:::

### d. Find the marginal revenue at a production level of 2500 pairs and interpret the result.


::: spoiler

<summary> Example: </summary>

Say 2000 pairs instead


Then plug in $x=2000$:

\begin{align}
R'(x)&=-\dfrac{1}{25} x +60 \\
R'(2000)&=-\dfrac{1}{25} (2000) +60 \\
&=-20
\end{align}

At a production level of 2000 pairs of shoes, the revenue decreases at a rate of $20 per shoe.

:::

::: spoiler

<summary> Solution: </summary>

Using the marginal revenue from the previous part:

\begin{align}
R'(x)&=-\dfrac{1}{20} x +100 \\
\end{align}

Then plug in $x=2500$:

\begin{align}
R'(1600)&=-\dfrac{1}{20} (2500) +100 \\
&=-125+100 \\
&=-25
\end{align}

At a production level of 2500 pairs of shoes, the revenue decreases at a rate of $25 per shoe.

:::