# Written Homework 4

## 2.7.46. The price-demand equation and the cost function for the production of HDTVs are given, respectively, by $x=9000-30p$, and $C(x)=150000+30x$, where $x$ is the number of HDTVs that can be sold at a price of $p per TV and $C(x)$ is the total cost (in dollars) of producing $x$ TVs.

### a. Express the price $p$ as a function of the demand $x$, and find the domain of this function.

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<summary> Example: </summary>

If $x=6000-20p$, then solving for $p$:

\begin{align}
x&=6000-20p \\
x-6000&=-20p \\
-20p&=x-6000 \\
\dfrac{-20p}{-20}&=\dfrac{x-6000}{-20} \\
p&=\dfrac{x}{-20}+\dfrac{-6000}{-20} \\
p&=-\dfrac{1}{20}x+300
\end{align}

The domain is found by setting $p \geq 0$ and knowing that $x \geq 0$:

\begin{align}
p&\geq 0 \\
-\dfrac{1}{20}x+300 &\geq 0 \\
-\dfrac{1}{20}x&\geq -300 \\
(-20) \left(-\dfrac{1}{20}x\right)&\leq (-300)(-20) \\
x &\leq 6000
\end{align}

This combined with $x \geq 0$ gives the domain $0 \leq x \leq 6000$.

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<summary> Solution: </summary>

### Given:
The price-demand equation is $x = 9000 - 30p$, where $x$ is the number of HDTVs sold at price $p$ per TV.

### Step 1: Express the price $p$ as a function of the demand $x$.

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

\begin{align}
x &= 9000 - 30p \\
30p &= 9000 - x \\
p &= \dfrac{9000 - x}{30} \\
p&=\dfrac{9000}{30}-\dfrac{1}{30}x \\
p&=-\dfrac{1}{30}x+300
\end{align}

Thus, the price $p$ as a function of demand $x$ is:

\begin{align}
p(x) &= -\dfrac{1}{30}x+300
\end{align}

### Step 2: Find the domain of the function $p(x)$.

The domain of $p(x)$ corresponds to the values of $x$ for which the price is non-negative, i.e., $p(x) \geq 0$:

\begin{align}
p(x) &\geq 0 \\
-\dfrac{1}{30}x+300&\geq 0 \\
\left(-\dfrac{1}{30}x+300\right)(-30)&\leq 0(-30) \\
x- 9000 &\leq 0 \\
x &\leq 9000
\end{align}

Since the demand $x$ cannot be negative, we also have $x \geq 0$.

Thus, the domain of the function $p(x)$ is:

\begin{align}
0 \leq x \leq 9000
\end{align}

### Final Answer:

The price as a function of demand is:

\begin{align}
p(x) &=-\dfrac{1}{30}x+300
\end{align}

The domain of this function is $0 \leq x \leq 9000$.

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### b. Find the marginal cost $C'(x)$.

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<summary> Example: </summary>

If $C(x)=23000+25x$, then the marginal cost is the derivative of the cost function by using the power rule $(x^n)'=nx^{n-1}$.

Can rewrite the cost function as $C(x)=23000x^0+25x^1$.

\begin{align}
C'(x)&=23000(x^0)'+25(x^1)' \\
&=23000(0x^{0-1})+25(1x^{1-1}) \\
&=23000(0)+25(1) \\
&=25
\end{align}

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<summary> Solution: </summary>

If $C(x)=150000+30x$, then the marginal cost is the derivative of the cost function by using the power rule $(x^n)'=nx^{n-1}$.

Can rewrite the cost function as $C(x)=150000x^0+30x^1$.

\begin{align}
C'(x)&=150000(x^0)'+30(x^1)' \\
&=150000(0x^{0-1})+30(1x^{1-1}) \\
&=150000(0)+30(1) \\
&=30
\end{align}

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### c. Find the revenue function and state its domain.

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<summary> Example: </summary>

If the price demand equation is $p=-\dfrac{1}{20}x+300$, then the revenue is $xp$:

\begin{align}
R(x)&=xp \\
&=x\left(-\dfrac{1}{20}x+300\right) \\
&=-\dfrac{1}{20}x^2+300x
\end{align}

The domain is the same as in part a. $0 \leq x \leq 6000$.

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<summary> Solution: </summary>

If the price demand equation is $p=-\dfrac{1}{30}x+300$, then the revenue is $xp$:

\begin{align}
R(x)&=xp \\
&=x\left(-\dfrac{1}{30}x+300\right) \\
&=-\dfrac{1}{30}x^2+300x
\end{align}

The domain is the same as in part a. $0 \leq x \leq 9000$.

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

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<summary> Example: </summary>

The marginal revenue is the derivative of the revenue function.

Rewriting the revenue as a sum of power functions:

$R(x)=-\dfrac{1}{20}x^2+300x^1$.

Then taking the derivative by the power rule $(x^n)'=nx^{n-1}$:

\begin{align}
R'(x)&=-\dfrac{1}{20}(x^2)' +300(x^1)' \\
&=-\dfrac{1}{20} (2x^{2-1})+300(1x^{1-1}) \\
&=-\dfrac{1}{20}(2x^1) +300(1x^0) \\
&=-\dfrac{1}{20}(2x)+300(1) \\
&=-\dfrac{2}{20}x +300 \\
&=-\dfrac{1}{10}x+300
\end{align}

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<summary> Solution: </summary>

The marginal revenue is the derivative of the revenue function.

Rewriting the revenue as a sum of power functions:

$R(x)=-\dfrac{1}{30}x^2+300x^1$.

Then taking the derivative by the power rule $(x^n)'=nx^{n-1}$:

\begin{align}
R'(x)&=-\dfrac{1}{30}(x^2)' +300(x^1)' \\
&=-\dfrac{1}{30} (2x^{2-1})+300(1x^{1-1}) \\
&=-\dfrac{1}{30}(2x^1) +300(1x^0) \\
&=-\dfrac{1}{30}(2x)+300(1) \\
&=-\dfrac{2}{30}x +300 \\
&=-\dfrac{1}{15}x+300
\end{align}

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### e. Find $R'(3000)$ and $R'(6000)$ and interpret these quantities.

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<summary> Example: </summary>

If the marginal revenue is $R'(x)=-\dfrac{1}{10}x+300$ and if we want to find $R'(2000)$, we simply plug in $x=2000$:

\begin{align}
R'(2000)&=-\dfrac{1}{10}(2000)+300 \\
&=-200+300 \\
&=100
\end{align}

This means at a production level of 2000, the revenue increases at a rate of $100 per TV.

If we want to find $R'(4000)$, we simply plug in $x=4000$:

\begin{align}
R'(4000)&=-\dfrac{1}{10}(4000)+300 \\
&=-400+300 \\
&=-100
\end{align}

This means at a production level of 4000, the revenue decreases at a rate of $100 per TV.

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::: spoiler

<summary> Solution: </summary>

If the marginal revenue is $R'(x)=-\dfrac{1}{15}x+300$ and if we want to find $R'(3000)$, we simply plug in $x=3000$:

\begin{align}
R'(3000)&=-\dfrac{1}{15}(3000)+300 \\
&=-200+300 \\
&=100
\end{align}

This means at a production level of 3000, the revenue increases at a rate of $100 per TV.

If we want to find $R'(6000)$, we simply plug in $x=6000$:

\begin{align}
R'(6000)&=-\dfrac{1}{15}(6000)+300 \\
&=-400+300 \\
&=-100
\end{align}

This means at a production level of 6000, the revenue decreases at a rate of $100 per TV.

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### f. Graph the cost function and the revenue function on the same coordinate system for $0 \leq x \leq 9000$. Find the break-even points and indicate regions of loss and profit.

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<summary> Example: </summary>

Here we can graph $R(x)=-\dfrac{1}{20}x^2+300x$ and $C(x)=C(x)=23000+25x$ on the interval $0 \leq x \leq 6000$:


This means the break even points are when $x=84.95$ and $x=5415.05$.

The revenue exceeds cost (profit) when $84.95 < x < 5415.05$, and the cost exceeds revenue (loss) when $0\leq x<84.95$ and when $5415.05<x<6000$.

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<summary> Solution: </summary>

Here we can graph $R(x)=-\dfrac{1}{15}x^2+300x$ and $C(x)=C(x)=150000+30x$ on the interval $0 \leq x \leq 9000$:


This means the break even points are when $x=600$ and $x=7500$.

The revenue exceeds cost (profit) when $600 < x < 7500$, and the cost exceeds revenue (loss) when $0\leq x<600$ and when $7500<x\leq 9000$.

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### g. Find the profit function in terms of $x$.

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<summary> Example: </summary>

If $R(x)=-\dfrac{1}{20}x^2+300x$ and $C(x)=23000+25x$, then the profit is $P(x)=R(x)-C(x)$.

\begin{align}
P(x)&=-\dfrac{1}{20}x^2+300x- (23000+25x) \\
&=-\dfrac{1}{20}x^2+300x-23000-25x \\
&=-\dfrac{1}{20}x^2+275x-23000
\end{align}

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<summary> Solution: </summary>

If $R(x)=-\dfrac{1}{30}x^2+300x$ and $C(x)=150000+30x$, then the profit is $P(x)=R(x)-C(x)$.

\begin{align}
P(x)&=-\dfrac{1}{30}x^2+300x- (150000+30x) \\
&=-\dfrac{1}{30}x^2+300x-150000-30x \\
&=-\dfrac{1}{30}x^2+270x-150000
\end{align}

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### h. Find the marginal profit $P'(x)$.

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<summary> Example: </summary>

If the profit function is $P(x)=-\dfrac{1}{20}x^2+275x-23000$, then the marginal profit is the derivative.

We can first rewrite as sum of power functions:

$P(x)=-\dfrac{1}{20}x^2+275x^1-23000x^0$.

Then taking the derivative:

\begin{align}
P'(x)&=(-\dfrac{1}{20}x^2+275x^1-23000x^0)' \\
&=-\dfrac{1}{20}(x^2)'+275(x^1)' - 23000(x^0)' \\
&=-\dfrac{1}{20}(2x^{2-1})+275(1x^{1-1})-23000(0x^{0-1}) \\
&=-\dfrac{1}{20}(2x^1)+275(1x^0)-23000(0x^{-1}) \\
&=-\dfrac{1}{20}(2x)+275(1)-0 \\
&=-\dfrac{2}{20}x+275 \\
&=-\dfrac{1}{10}x+275
\end{align}

Thus the marginal profit is $P'(x)=-\dfrac{1}{10}x+275$.

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<summary> Solution: </summary>

If the profit function is $P(x) = -\dfrac{1}{30}x^2 + 270x - 150000$, then the marginal profit is the derivative.

### Step 1: Rewrite as a sum of power functions

$$
P(x) = -\dfrac{1}{30}x^2 + 270x^1 - 150000x^0
$$

### Step 2: Take the derivative

\begin{align}
P'(x) &= \left( -\dfrac{1}{30}x^2 + 270x^1 - 150000x^0 \right)' \\
&= -\dfrac{1}{30}(x^2)' + 270(x^1)' - 150000(x^0)' \\
&= -\dfrac{1}{30}(2x^{2-1}) + 270(1x^{1-1}) - 150000(0x^{0-1}) \\
&= -\dfrac{1}{30}(2x^1) + 270(1x^0) - 150000(0x^{-1}) \\
&= -\dfrac{2}{30}x + 270(1) - 0 \\
&= -\dfrac{1}{15}x + 270
\end{align}

Thus, the marginal profit is $P'(x) = -\dfrac{1}{15}x + 270$.

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### i. Find $P'(1500)$ and $P'(4500)$ and interpret these quantities.

::: spoiler

<summary> Example: </summary>

If the marginal profit is $P'(x)=-\dfrac{1}{10}x+275$,

Then we can find $P'(2000)$ by plugging in $x=2000$ into the derivative.

\begin{align}
P'(2000)&=-\dfrac{1}{10}(2000)+275 \\
&=-200+275 \\
&=75
\end{align}

This means at a production level of 2000 TVs, the profit will increase at a rate of $75 per TV.

Then we can find $P'(3000)$ by plugging in $x=3000$ into the derivative.

\begin{align}
P'(3000)&=-\dfrac{1}{10}(3000)+275 \\
&=-300+275 \\
&=-25
\end{align}

This means at a production level of 3000 TVs, the profit will decrease at a rate of $25 per TV.

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::: spoiler

<summary> Solution: </summary>

### Step 1: Find $P'(1500)$

Substitute $x = 1500$ into the marginal profit function:

$$
P'(1500) = -\dfrac{1}{15}(1500) + 270
$$

Simplifying:

$$
P'(1500) = -100 + 270 = 170
$$

Thus, $P'(1500) = 170$.

### Step 2: Find $P'(4500)$

Substitute $x = 4500$ into the marginal profit function:

$$
P'(4500) = -\dfrac{1}{15}(4500) + 270
$$

Simplifying:

$$
P'(4500) = -300 + 270 = -30
$$

Thus, $P'(4500) = -30$.

### Interpretation of these quantities:

- **$P'(1500) = 170$:** This means that when 1,500 HDTVs are produced and sold, the profit is increasing at a rate of $170 per TV. Producing and selling one additional HDTV beyond 1,500 will increase profit by approximately $170.

- **$P'(4500) = -30$:** This means that when 4,500 HDTVs are produced and sold, the profit is decreasing at a rate of $30 per TV. Producing and selling one additional HDTV beyond 4,500 will decrease profit by approximately $30. This suggests that the company might be overproducing at this level, leading to diminishing profits.

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