# 3.1.27 Continuous compound interest. Provident Bank offers a 10-year CD that earns 2.15% compounded continuously.
## a. If $10,000 is invested in this CD, how much will it be worth in 10 years?

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

Provident Bank offers a 5-year CD that earns 3.45% compounded continuously. If $10,000 is invested in this CD, how much will it be worth in 5 years?

$$
A = Pe^{rt}
$$

where:
- $A$ is the amount of money accumulated after $n$ years, including interest.
- $P$ is the principal amount (the initial amount of money).
- $r$ is the annual interest rate (decimal).
- $t$ is the time the money is invested for in years.
- $e$ is the base of the natural logarithm, approximately equal to 2.71828.

Given:
- $P = 10000$
- $r = 0.0345$
- $t = 5$

Let's plug these values into the formula:

$$
A = 10000 \cdot e^{0.0345 \cdot 5}
$$

Now, calculate the exponent:

$$
0.0345 \cdot 5 = 0.1725
$$

Then, compute the value of $e^{0.1725}$:

$$
e^{0.1725} \approx 1.188311
$$

Now, multiply by the principal:

$$
A = 10000 \cdot 1.188311 \approx 11883.11
$$

Therefore, the investment will be worth approximately $11,883.11 in 5 years.

</details>

## b. How long will it take for the account to be worth $18,000?

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

Provident Bank offers a 5-year CD that earns 3.45% compounded continuously. If $10,000 is invested in this CD, how long will it take for the account to be worth $19,000?

We are given the formula for continuously compounded interest:

$$
A = Pe^{rt}
$$

Where:
- $A = 19000$ (the future amount we want),
- $P = 10000$ (the initial investment),
- $r = 0.0345$ (the annual interest rate in decimal form),
- $t$ is the time we want to solve for.

### Step 1: Rearrange the formula to solve for $t$

We start with:

$$
A = Pe^{rt}
$$

To isolate $t$, first divide both sides of the equation by $P$:

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

Next, take the natural logarithm (ln) of both sides to eliminate the $e$:

$$
\ln\left(\frac{A}{P}\right) = rt
$$

Finally, solve for $t$:

$$
t = \frac{\ln\left(\frac{A}{P}\right)}{r}
$$

### Step 2: Plug in the known values

We are given:
- $A = 19000$
- $P = 10000$
- $r = 0.0345$

Substitute these into the equation:

$$
t = \frac{\ln\left(\frac{19000}{10000}\right)}{0.0345}
$$

### Step 3: Simplify and calculate

First, calculate the ratio:

$$
\frac{19000}{10000} = 1.9
$$

Then take the natural logarithm:

$$
\ln(1.9) \approx 0.6419
$$

Now, divide by the interest rate:

$$
t = \frac{0.6419}{0.0345} \approx 18.6 \text{ years}
$$

Therefore, it will take approximately 18.6 years for the investment to grow to $19,000.

</details>

# 3.2.28 Given $f(x)=3x^3-2e^x$, find $f'(x)$.

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

We are given the function:

$$
f(x) = 4x^5 - 3e^x
$$

### Step 1: Differentiate each term separately

To find $f'(x)$, we will differentiate each term of the function.

For the first term, $4x^5$:
- Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$, we get:
$$\frac{d}{dx}(4x^5) = 4 \cdot 5x^{4} = 20x^4$$

For the second term, $-3e^x$:
- The derivative of $e^x$ with respect to $x$ is just $e^x$, so:
$$\frac{d}{dx}(-3e^x) = -3e^x$$

### Step 2: Combine the results

Now, combining both terms, we get:

$$
f'(x) = 20x^4 - 3e^x
$$

Thus, the derivative of the function is:

$$
f'(x) = 20x^4 - 3e^x
$$

</details>

# 3.3.20 Given $f(x)=5x\ln x$, find $f'(x)$ and simplify.

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

We are given the function:

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

### Step 1: Apply the product rule

To differentiate $f(x)$, we need to use the **product rule**, which states:

$$
\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)
$$

In our case:
- $u(x) = 4x$
- $v(x) = \ln x$

### Step 2: Differentiate each part

First, differentiate $u(x) = 4x$:
$$
u'(x) = 4
$$

Next, differentiate $v(x) = \ln x$:
$$
v'(x) = \frac{1}{x}
$$

### Step 3: Apply the product rule

Now, apply the product rule:

$$
f'(x) = u'(x) v(x) + u(x) v'(x)
$$

Substitute the values of $u'(x)$, $u(x)$, $v(x)$, and $v'(x)$:

$$
f'(x) = 4 \cdot \ln x + 4x \cdot \frac{1}{x}
$$

### Step 4: Simplify

Simplify the second term:

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

Thus, the derivative of the function is:

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

</details>

# 3.3.30 Given $f(x)=\dfrac{x^2-4}{x^2+5}$, find $f'(x)$ and simplify.

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

We are given the function:

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

### Step 1: Apply the quotient rule

To differentiate $f(x)$, we need to use the **quotient rule**, which states:

$$
\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
$$

In our case:
- $u(x) = x^2 - 3$
- $v(x) = x^2 + 6$

### Step 2: Differentiate $u(x)$ and $v(x)$

First, differentiate $u(x) = x^2 - 3$:
$$
u'(x) = 2x
$$

Next, differentiate $v(x) = x^2 + 6$:
$$
v'(x) = 2x
$$

### Step 3: Apply the quotient rule

Now, apply the quotient rule:

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

### Step 4: Simplify the numerator

Expand both parts of the numerator:

$$
(2x)(x^2 + 6) = 2x^3 + 12x
$$

$$
(2x)(x^2 - 3) = 2x^3 - 6x
$$

Now subtract the second expression from the first:

$$
2x^3 + 12x - (2x^3 - 6x) = 2x^3 + 12x - 2x^3 + 6x = 18x
$$

### Step 5: Write the final simplified derivative

Thus, the derivative of the function is:

$$
f'(x) = \frac{18x}{(x^2 + 6)^2}
$$

</details>

# 3.4.54 Find $y'$ if $y=\left[ \ln (x^2+3) \right]^{3/2}$

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

We are given the function:

$$
y = \left[ \ln (x^3 + 4) \right]^{\frac{2}{3}}
$$

### Step 1: Apply the chain rule

To differentiate $y$, we need to use the **chain rule**, which states:

$$
\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)
$$

Here, let:
- $f(u) = u^{\frac{2}{3}}$
- $g(x) = \ln (x^3 + 4)$

### Step 2: Differentiate $f(u) = u^{2/3}$

Using the power rule, differentiate $f(u)$:

$$
f'(u) = \frac{2}{3} u^{-\frac{1}{3}}
$$

Substitute $u = \ln (x^3 + 4)$ back in:

$$
f'(u) = \frac{2}{3} \left[ \ln (x^3 + 4) \right]^{-\frac{1}{3}}
$$

### Step 3: Differentiate $g(x) = \ln (x^3 + 4)$

Now, differentiate $g(x) = \ln (x^3 + 4)$ using the chain rule again:

$$
g'(x) = \frac{1}{x^3 + 4} \cdot 3x^2
$$

So,

$$
g'(x) = \frac{3x^2}{x^3 + 4}
$$

### Step 4: Combine the results

Now, apply the chain rule:

$$
y' = f'(u) \cdot g'(x)
$$

Substitute the values:

$$
y' = \frac{2}{3} \left[ \ln (x^3 + 4) \right]^{-\frac{1}{3}} \cdot \frac{3x^2}{x^3 + 4}
$$

### Step 5: Simplify the result

The $3$ in the numerator cancels with the $\frac{2}{3}$, so the final derivative is:

$$
y' = 2x^2 \cdot \frac{\left[ \ln (x^3 + 4) \right]^{-\frac{1}{3}}}{x^3 + 4}
$$

Thus, the derivative of the function is:

$$
y' = \frac{2x^2 \left[ \ln (x^3 + 4) \right]^{-\frac{1}{3}}}{x^3 + 4}
$$

</details>

# 3.7.48 Use the price–demand equation to determine whether demand is elastic, is inelastic, or has unit elasticity at the indicated values of $p$. $$x=f(p)=12000-10p^2$$

## a. $p=10$

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

Use the price–demand equation to determine whether demand is elastic, is inelastic, or has unit elasticity at the indicated values of $p$. $$x=f(p)=13000-12p^2$$

We are given the price–demand equation:

$$
x = f(p) = 13000 - 12p^2
$$

The elasticity of demand $E(p)$ is given by the formula:

$$
E(p) = \left| \frac{p \cdot f'(p)}{f(p)} \right|
$$

Where:
- $f(p)$ is the demand function,
- $f'(p)$ is the derivative of $f(p)$ with respect to $p$,
- $p$ is the price.

### Step 1: Differentiate $f(p)$

First, differentiate the demand function $x = f(p) = 13000 - 12p^2$ with respect to $p$:

$$
f'(p) = \frac{d}{dp}(13000 - 12p^2) = -24p
$$

### Step 2: Calculate $E(p)$ for each value of $p$

#### a. $p = 15$

1. Compute $f(15)$:

$$
f(15) = 13000 - 12(15)^2 = 13000 - 12(225) = 13000 - 2700 = 10300
$$

2. Compute $f'(15)$:

$$
f'(15) = -24(15) = -360
$$

3. Now calculate $E(15)$:

$$
E(15) = \left| \frac{15 \cdot (-360)}{10300} \right| = \left| \frac{-5400}{10300} \right| \approx 0.5243
$$

Since $E(15) < 1$, the demand is **inelastic** at $p = 15$.

#### b. $p = 40$

1. Compute $f(40)$:

$$
f(40) = 13000 - 12(40)^2 = 13000 - 12(1600) = 13000 - 19200 = -6200
$$

However, since negative demand doesn't make sense, let's assume this value is unrealistic for demand.

2. Compute $f'(40)$:

$$
f'(40) = -24(40) = -960
$$

Even though $f(40)$ is negative, we can still calculate $E(40)$:

$$
E(40) = \left| \frac{40 \cdot (-960)}{-6200} \right| = \left| \frac{-38400}{-6200} \right| = 6.19
$$

Since $E(40) > 1$, the demand is **elastic** at $p = 40$.

### Conclusion:

- For $p = 15$, the demand is **inelastic**.
- For $p = 40$, the demand is **elastic**.

</details>

## b. $p=20$

## c. $p=30$

# 4.1.44 Find the intervals on which $f(x)$ is increasing, the intervals on which $f(x)$ is decreasing, and the local extrema. $$f(x)=x^4-8x^3+32$$

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

We are given the function:

$$
f(x) = x^4 - 12x^3 + 13
$$

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

To find the intervals where the function is increasing or decreasing, we first compute the first derivative of $f(x)$:

$$
f'(x) = 4x^3 - 36x^2
$$

Factoring the derivative:

$$
f'(x) = 4x^2(x - 9)
$$

### Step 2: Find critical points

Setting $f'(x) = 0$:

$$
4x^2(x - 9) = 0
$$

This gives the critical points $x = 0$ and $x = 9$.

### Step 3: Test the intervals

We now test the sign of $f'(x)$ on the intervals determined by the critical points: $(-\infty, 0)$, $(0, 9)$, and $(9, \infty)$. Below is the table for testing:

| Interval | Test Value | Calculation for $f'(x)$ | Sign |
|----------------|-------------|------------------------------------------------|-----------|
| $(-\infty, 0)$ | $x = -1$ | $f'(-1) = 4(-1)^2(-1 - 9) = 4(1)(-10) = -40$ | Negative |
| $(0, 9)$ | $x = 1$ | $f'(1) = 4(1)^2(1 - 9) = 4(1)(-8) = -32$ | Negative |
| $(9, \infty)$ | $x = 10$ | $f'(10) = 4(10)^2(10 - 9) = 4(100)(1) = 400$ | Positive |

### Step 4: Conclusion

- $f(x)$ is **decreasing** on $(-\infty, 0)$ and $(0, 9)$.
- $f(x)$ is **increasing** on $(9, \infty)$.
- There is a **local minimum** at $x = 9$.

### Step 5: Find the local minimum

To find the value of the local minimum:

$$
f(9) = (9)^4 - 12(9)^3 + 13 = 6561 - 11664 + 13 = -5090
$$

Thus, the local minimum occurs at $(9, -5090)$.

</details>

# 4.2.35 Find the intervals on which the graph of $f$ is concave upward, the intervals on which the graph of $f$ is concave downward, and the $x$,$y$ coordinates of the inflection points. $$f(x)=-x^4+12x^3-7x+10$$

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

We are given the function:

$$
f(x) = -x^4 + 8x^3 - 3x + 8
$$

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

To determine concavity, we first compute the first derivative $f'(x)$:

$$
f'(x) = \frac{d}{dx}(-x^4 + 8x^3 - 3x + 8) = -4x^3 + 24x^2 - 3
$$

Now, compute the second derivative $f''(x)$:

$$
f''(x) = \frac{d}{dx}(-4x^3 + 24x^2 - 3) = -12x^2 + 48x
$$

### Step 2: Find inflection points

To find the inflection points, set $f''(x) = 0$ and solve for $x$:

$$
-12x^2 + 48x = 0
$$

Factor the equation:

$$
-12x(x - 4) = 0
$$

This gives the critical points for concavity:

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

### Step 3: Test the intervals for concavity

We test the sign of $f''(x)$ on the intervals determined by the inflection points: $(-\infty, 0)$, $(0, 4)$, and $(4, \infty)$. Below is the table for testing:

| Interval | Test Value | Calculation for $f''(x)$ | Sign |
|----------------|-------------|--------------------------------------------------|----------------|
| $(-\infty, 0)$ | $x = -1$ | $f''(-1) = -12(-1)^2 + 48(-1) = -12 - 48 = -60$ | Negative |
| $(0, 4)$ | $x = 1$ | $f''(1) = -12(1)^2 + 48(1) = -12 + 48 = 36$ | Positive |
| $(4, \infty)$ | $x = 5$ | $f''(5) = -12(5)^2 + 48(5) = -12(25) + 48(5) = -300 + 240 = -60$ | Negative |

### Step 4: Conclusion

- The graph of $f(x)$ is **concave downward** on $(-\infty, 0)$ and $(4, \infty)$.
- The graph of $f(x)$ is **concave upward** on $(0, 4)$.
- There are **inflection points** where the concavity changes at $x = 0$ and $x = 4$.

### Step 5: Find the $y$-coordinates of the inflection points

To find the $y$-coordinates of the inflection points, evaluate $f(x)$ at $x = 0$ and $x = 4$.

1. For $x = 0$:

$$
f(0) = -(0)^4 + 8(0)^3 - 3(0) + 8 = 8
$$

So, the inflection point is at $(0, 8)$.

2. For $x = 4$:

$$
f(4) = -(4)^4 + 8(4)^3 - 3(4) + 8 = -256 + 512 - 12 + 8 = 252
$$

So, the inflection point is at $(4, 252)$.

### Final Answer

- The graph of $f(x)$ is **concave downward** on $(-\infty, 0)$ and $(4, \infty)$.
- The graph of $f(x)$ is **concave upward** on $(0, 4)$.
- The inflection points are at $(0, 8)$ and $(4, 252)$.

</details>

# 4.5.62 Find the absolute minimum value on $(0,\infty)$ for $$f(x)=4x \ln x - 7x$$

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

We are given the function:

$$
f(x) = 5x \ln x - 6x
$$

We are tasked with finding the absolute minimum value of $f(x)$ on the interval $(0, \infty)$.

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

To find critical points, we first compute the derivative of $f(x)$:

Using the product rule for $5x \ln x$:
$$
f'(x) = \frac{d}{dx}(5x \ln x - 6x)
$$

For the first term $5x \ln x$, we apply the product rule:
$$
\frac{d}{dx}[5x \ln x] = 5 \ln x + 5
$$

For the second term $-6x$, the derivative is:
$$
\frac{d}{dx}[-6x] = -6
$$

So, the first derivative is:
$$
f'(x) = 5 \ln x + 5 - 6 = 5 \ln x - 1
$$

### Step 2: Find critical points

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

$$
5 \ln x - 1 = 0
$$

Solving for $\ln x$:

$$
\ln x = \frac{1}{5}
$$

Exponentiate both sides to solve for $x$:

$$
x = e^{\frac{1}{5}} \approx 1.221
$$

### Step 3: Check if it's a minimum by using the second derivative

Now, we compute the second derivative $f''(x)$ to check if the critical point is a minimum.

The derivative of $f'(x) = 5 \ln x - 1$ is:

$$
f''(x) = \frac{d}{dx}[5 \ln x - 1] = \frac{5}{x}
$$

For $x = e^{\frac{1}{5}}$, the second derivative is:

$$
f''(e^{\frac{1}{5}}) = \frac{5}{e^{\frac{1}{5}}} > 0
$$

Since $f''(x) > 0$ at $x = e^{\frac{1}{5}}$, this indicates that the function is concave upward, and the critical point is a **local minimum**.

### Step 4: Find the minimum value

To find the absolute minimum value, we substitute $x = e^{\frac{1}{5}}$ back into the original function $f(x)$:

$$
f(e^{\frac{1}{5}}) = 5e^{\frac{1}{5}} \ln(e^{\frac{1}{5}}) - 6e^{\frac{1}{5}}
$$

Since $\ln(e^{\frac{1}{5}}) = \frac{1}{5}$, we have:

$$
f(e^{\frac{1}{5}}) = 5e^{\frac{1}{5}} \cdot \frac{1}{5} - 6e^{\frac{1}{5}} = e^{\frac{1}{5}} - 6e^{\frac{1}{5}} = -5e^{\frac{1}{5}}
$$

Therefore, the absolute minimum value is:

$$
f(e^{\frac{1}{5}}) = -5e^{\frac{1}{5}} \approx -6.105
$$

### Final Answer

The absolute minimum value of the function on $(0, \infty)$ is approximately $-6.105$.

</details>

# 4.6.21 Maximum revenue and profit. A company manufactures and sells x television sets per month. The monthly cost and price–demand equations are $$C(x)=72000+60x$$ $$p=200-\dfrac{x}{30}; \qquad 0 \leq x \leq 6000$$

## a. Find the maximum revenue.

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

We are given the following:

- **Cost function**:
$$
C(x) = 67000 + 50x
$$

- **Price–demand equation**:
$$
p = 300 - \frac{x}{20}; \quad 0 \leq x \leq 6000
$$

Where:
- $x$ represents the number of television sets sold per month.
- $p$ is the price at which each set is sold.

We are tasked with finding the maximum **revenue** and **profit**.

### Step 1: Revenue function $R(x)$

The revenue $R(x)$ is given by the product of the number of units sold ($x$) and the price per unit ($p$):

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

Substitute the price–demand equation $p = 300 - \frac{x}{20}$ into the revenue equation:

$$
R(x) = x \left(300 - \frac{x}{20}\right)
$$

Simplify:

$$
R(x) = 300x - \frac{x^2}{20}
$$

### Step 2: Find the first derivative $R'(x)$ to maximize revenue

To find the critical points for maximum revenue, we differentiate the revenue function:

$$
R'(x) = \frac{d}{dx}\left(300x - \frac{x^2}{20}\right) = 300 - \frac{2x}{20}
$$

Simplify:

$$
R'(x) = 300 - \frac{x}{10}
$$

### Step 3: Set $R'(x) = 0$ to find critical points

Set the first derivative equal to 0 to find the value of $x$ that maximizes revenue:

$$
300 - \frac{x}{10} = 0
$$

Solve for $x$:

$$
x = 3000
$$

### Step 4: Verify maximum revenue by checking endpoints

We need to check whether $x = 3000$ gives the maximum revenue by comparing it with the boundary values of $x = 0$ and $x = 6000$.

1. **At $x = 0$**:
$$
R(0) = 300(0) - \frac{0^2}{20} = 0
$$

2. **At $x = 3000$**:
$$
R(3000) = 300(3000) - \frac{3000^2}{20} = 900000 - 450000 = 450000
$$

3. **At $x = 6000**:
$$
R(6000) = 300(6000) - \frac{6000^2}{20} = 1800000 - 1800000 = 0
$$

Thus, the maximum revenue is $450,000$ when $x = 3000$ television sets are sold.

### Final Answer

- The maximum **revenue** is $450,000$ when 3,000 units are sold.

</details>

## b. Find the maximum profit, the production level that will realize the maximum profit, and the price the company should charge for each television set.

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

We are given the following:

- **Cost function**:
$$
C(x) = 67000 + 50x
$$

- **Price–demand equation**:
$$
p = 300 - \frac{x}{20}; \quad 0 \leq x \leq 6000
$$

Where:
- $x$ represents the number of television sets sold per month.
- $p$ is the price at which each set is sold.

We are tasked with finding:
1. The **maximum profit**.
2. The **production level** that will realize the maximum profit.
3. The **price** the company should charge for each television set.

### Step 1: Revenue function $R(x)$

The revenue $R(x)$ is given by the product of the number of units sold ($x$) and the price per unit ($p$):

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

Substitute the price–demand equation $p = 300 - \frac{x}{20}$ into the revenue equation:

$$
R(x) = x \left(300 - \frac{x}{20}\right)
$$

Simplify:

$$
R(x) = 300x - \frac{x^2}{20}
$$

### Step 2: Profit function $P(x)$

The profit function $P(x)$ is the difference between the revenue and the cost:

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

Substitute $R(x) = 300x - \frac{x^2}{20}$ and $C(x) = 67000 + 50x$:

$$
P(x) = \left(300x - \frac{x^2}{20}\right) - \left(67000 + 50x\right)
$$

Simplify:

$$
P(x) = 300x - \frac{x^2}{20} - 67000 - 50x
$$

$$
P(x) = 250x - \frac{x^2}{20} - 67000
$$

### Step 3: Find the first derivative $P'(x)$ to maximize profit

To find the critical points for maximum profit, differentiate the profit function $P(x)$:

$$
P'(x) = \frac{d}{dx}\left(250x - \frac{x^2}{20} - 67000\right) = 250 - \frac{2x}{20}
$$

Simplify:

$$
P'(x) = 250 - \frac{x}{10}
$$

### Step 4: Set $P'(x) = 0$ to find critical points

Set the first derivative equal to 0 to find the value of $x$ that maximizes profit:

$$
250 - \frac{x}{10} = 0
$$

Solve for $x$:

$$
\frac{x}{10} = 250
$$

$$
x = 2500
$$

So, the production level that maximizes profit is $x = 2500$ television sets per month.

### Step 5: Verify the maximum by checking endpoints

The problem specifies the interval $0 \leq x \leq 6000$. We can verify that the maximum profit occurs at $x = 2500$ by evaluating the profit at the endpoints and the critical point $x = 2500$.

1. **At $x = 0$**:
$$
P(0) = 250(0) - \frac{(0)^2}{20} - 67000 = -67000
$$

2. **At $x = 2500$**:
$$
P(2500) = 250(2500) - \frac{(2500)^2}{20} - 67000 = 625000 - 312500 - 67000 = 245500
$$

3. **At $x = 6000$**:
$$
P(6000) = 250(6000) - \frac{(6000)^2}{20} - 67000 = 1500000 - 1800000 - 67000 = -367000
$$

Thus, the maximum profit is **$245,500** when $x = 2500$ television sets are sold.

### Step 6: Find the price for maximum profit

To find the price the company should charge at the production level of $x = 2500$, we use the price–demand equation:

$$
p = 300 - \frac{2500}{20}
$$

Simplify:

$$
p = 300 - 125 = 175
$$

So, the company should charge **$175** per television set to maximize profit.

### Final Answer

- The **maximum profit** is **$245,500**.
- The **production level** that maximizes profit is **2,500** television sets per month.
- The **price** the company should charge for each television set is **$175**.

</details>

## c. If the government decides to tax the company $5 for each set it produces, how many sets should the company manufacture each month to maximize its profit? What is the maximum profit? What should the company charge for each set?

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

We are given the following:

- **Cost function** with tax:
$$
C(x) = 67000 + (50 + 4)x = 67000 + 54x
$$

- **Price–demand equation**:
$$
p = 300 - \frac{x}{20}; \quad 0 \leq x \leq 6000
$$

Where:
- $x$ represents the number of television sets sold per month.
- $p$ is the price at which each set is sold.

The government imposes a tax of **$4** for each television set produced, which is included in the updated cost function.

We are tasked with finding:
1. The **maximum profit**.
2. The **production level** that will maximize the profit.
3. The **price** the company should charge for each television set.

### Step 1: Revenue function $R(x)$

The revenue $R(x)$ is given by the product of the number of units sold ($x$) and the price per unit ($p$):

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

Substitute the price–demand equation $p = 300 - \frac{x}{20}$ into the revenue equation:

$$
R(x) = x \left(300 - \frac{x}{20}\right)
$$

Simplify:

$$
R(x) = 300x - \frac{x^2}{20}
$$

### Step 2: Profit function $P(x)$

The profit function $P(x)$ is the difference between the revenue and the cost:

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

Substitute $R(x) = 300x - \frac{x^2}{20}$ and $C(x) = 67000 + 54x$ (which includes the $4 tax per unit):

$$
P(x) = \left(300x - \frac{x^2}{20}\right) - \left(67000 + 54x\right)
$$

Simplify:

$$
P(x) = 300x - \frac{x^2}{20} - 67000 - 54x
$$

$$
P(x) = 246x - \frac{x^2}{20} - 67000
$$

### Step 3: Find the first derivative $P'(x)$ to maximize profit

To find the critical points for maximum profit, differentiate the profit function $P(x)$:

$$
P'(x) = \frac{d}{dx}\left(246x - \frac{x^2}{20} - 67000\right) = 246 - \frac{2x}{20}
$$

Simplify:

$$
P'(x) = 246 - \frac{x}{10}
$$

### Step 4: Set $P'(x) = 0$ to find critical points

Set the first derivative equal to 0 to find the value of $x$ that maximizes profit:

$$
246 - \frac{x}{10} = 0
$$

Solve for $x$:

$$
\frac{x}{10} = 246
$$

$$
x = 2460
$$

So, the production level that maximizes profit is $x = 2460$ television sets per month.

### Step 5: Verify the maximum by checking endpoints

The problem specifies the interval $0 \leq x \leq 6000$. We can verify that the maximum profit occurs at $x = 2460$ by evaluating the profit at the endpoints and the critical point $x = 2460$.

1. **At $x = 0$**:
$$
P(0) = 246(0) - \frac{(0)^2}{20} - 67000 = -67000
$$

2. **At $x = 2460$**:
$$
P(2460) = 246(2460) - \frac{(2460)^2}{20} - 67000 = 605160 - 302436 - 67000 = 235724
$$

3. **At $x = 6000**:
$$
P(6000) = 246(6000) - \frac{(6000)^2}{20} - 67000 = 1476000 - 1800000 - 67000 = -385000
$$

Thus, the maximum profit is **$235,724** when $x = 2460$ television sets are sold.

### Step 6: Find the price for maximum profit

To find the price the company should charge at the production level of $x = 2460$, we use the price–demand equation:

$$
p = 300 - \frac{2460}{20}
$$

Simplify:

$$
p = 300 - 123 = 177
$$

So, the company should charge **$177** per television set to maximize profit.

### Final Answer

- The **maximum profit** is **$235,724**.
- The **production level** that maximizes profit is **2,460** television sets per month.
- The **price** the company should charge for each television set is **$177**.

</details>

# 4.6.26 A university student center sells 1,600 cups of coffee per day at a price of $2.40.
## a. A market survey shows that for every $0.05 reduction in price, 50 more cups of coffee will be sold. How much should the student center charge for a cup of coffee in order to maximize revenue?
## b. A different market survey shows that for every $0.10 reduction in the original $2.40 price, 60 more cups of coffee will be sold. Now how much should the student center charge for a cup of coffee in order to maximize revenue?

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

We are given that a university student center sells 1,500 cups of coffee per day at a price of $3.40. We are tasked with finding the price that maximizes revenue under two different market survey conditions.

### Part A

#### Given:
- **Current price**: $3.40
- **Current sales**: 1,500 cups/day
- **For every $0.05 reduction**, 40 more cups of coffee are sold.

#### Step 1: Define the variables

Let:
- $x$ be the number of $0.05$ reductions in price.
- The new price of coffee is $3.40 - 0.05x$.
- The number of cups sold is $1,500 + 40x$.

#### Step 2: Revenue function $R(x)$

Revenue $R(x)$ is the product of the price per cup and the number of cups sold:

$$
R(x) = (3.40 - 0.05x)(1500 + 40x)
$$

#### Step 3: Expand the revenue function

Expand the expression:

$$
R(x) = (3.40 - 0.05x)(1500 + 40x) = 3.40(1500 + 40x) - 0.05x(1500 + 40x)
$$

Simplify:

$$
R(x) = 5100 + 136x - 75x - 2x^2
$$

$$
R(x) = 5100 + 61x - 2x^2
$$

#### Step 4: Find the first derivative $R'(x)$

Differentiate $R(x)$ with respect to $x$ to find the critical points:

$$
R'(x) = 61 - 4x
$$

#### Step 5: Set $R'(x) = 0$ to find critical points

Set the derivative equal to zero:

$$
61 - 4x = 0
$$

Solve for $x$:

$$
x = \frac{61}{4} = 15.25
$$

#### Step 6: Find the price that maximizes revenue

Substitute $x = 15.25$ into the new price equation:

$$
\text{New price} = 3.40 - 0.05(15.25) = 3.40 - 0.7625 = 2.6375
$$

So, the student center should charge **$2.64** per cup to maximize revenue in this case.

---

### Part B

#### Given:
- **Current price**: $3.40
- **Current sales**: 1,500 cups/day
- **For every $0.10 reduction**, 60 more cups of coffee are sold.

#### Step 1: Define the variables

Let:
- $x$ be the number of $0.10$ reductions in price.
- The new price of coffee is $3.40 - 0.10x$.
- The number of cups sold is $1,500 + 60x$.

#### Step 2: Revenue function $R(x)$

Revenue $R(x)$ is the product of the price per cup and the number of cups sold:

$$
R(x) = (3.40 - 0.10x)(1500 + 60x)
$$

#### Step 3: Expand the revenue function

Expand the expression:

$$
R(x) = (3.40)(1500 + 60x) - (0.10x)(1500 + 60x)
$$

Simplify:

$$
R(x) = 5100 + 204x - 150x - 6x^2
$$

$$
R(x) = 5100 + 54x - 6x^2
$$

#### Step 4: Find the first derivative $R'(x)$

Differentiate $R(x)$ with respect to $x$ to find the critical points:

$$
R'(x) = 54 - 12x
$$

#### Step 5: Set $R'(x) = 0$ to find critical points

Set the derivative equal to zero:

$$
54 - 12x = 0
$$

Solve for $x$:

$$
x = \frac{54}{12} = 4.5
$$

#### Step 6: Find the price that maximizes revenue

Substitute $x = 4.5$ into the new price equation:

$$
\text{New price} = 3.40 - 0.10(4.5) = 3.40 - 0.45 = 2.95
$$

So, the student center should charge **$2.95** per cup to maximize revenue in this case.

### Final Answer

- **Part A**: The student center should charge **$2.64** per cup to maximize revenue.
- **Part B**: The student center should charge **$2.95** per cup to maximize revenue.

</details>

# 5.1.84 The graph of the marginal revenue function from the sale of x smart watches is given in the figure.
## a. Using the graph shown, describe the shape of the graph of the revenue function $R(x)$ as $x$ increases from 0 to 1,000.

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

</details>

## b. Find the equation of the marginal revenue function (the linear function shown in the figure).


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

</details>

## c. Find the equation of the revenue function that satisfies $R(0)=0$. Graph the revenue function over the interval $[0,1000]$. [Check the shape of the graph relative to the analysis in part (A).]

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

</details>

# 5.4.74 For a new employee in Problem 73, use left and right sums to estimate the area under the graph of $N(t)$ from $t = 20$ to $t = 100$. Use four equal subintervals for each. Replace the question marks with the values of $L_4$ or $R_4$ as appropriate:
| $t$ | 0 | 20 | 40 | 60 | 80 | 100 | 120 |
|-------|----|-----|-----|-----|-----|-----|-----|
| $N(t)$| 10 | 51 | 68 | 76 | 81 | 84 | 86 |

$$
? \leq \int_{20}^{100} N(t) \, dt \leq ?
$$

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

For a new employee in Problem 73, use left and right sums to estimate the area under the graph of $N(t)$ from $t = 30$ to $t = 150$. Use four equal subintervals for each. Replace the question marks with the values of $L_4$ or $R_4$ as appropriate:

| $t$ | 0 | 30 | 60 | 90 | 120 | 150 | 180 |
|-------|----|-----|-----|-----|------|------|-----|
| $N(t)$| 12 | 55 | 70 | 78 | 85 | 88 | 90 |

$$
? \leq \int_{30}^{150} N(t) \, dt \leq ?
$$

### Step 1: Divide the interval into four equal subintervals
The interval $[30, 150]$ is divided into four equal subintervals. The length of each subinterval is:

$$
\Delta t = \frac{150 - 30}{4} = 30.
$$

So, the subintervals are:
- $[30, 60]$
- $[60, 90]$
- $[90, 120]$
- $[120, 150]$

### Step 2: Left Riemann Sum ($L_4$)
For the left Riemann sum, you use the left endpoint of each subinterval to approximate the area:

$$
L_4 = \Delta t \left[ N(30) + N(60) + N(90) + N(120) \right].
$$

Substituting the values from the table:

$$
L_4 = 30 \left[ 55 + 70 + 78 + 85 \right] = 30 \times 288 = 8640.
$$

### Step 3: Right Riemann Sum ($R_4$)
For the right Riemann sum, you use the right endpoint of each subinterval:

$$
R_4 = \Delta t \left[ N(60) + N(90) + N(120) + N(150) \right].
$$

Substituting the values from the table:

$$
R_4 = 30 \left[ 70 + 78 + 85 + 88 \right] = 30 \times 321 = 9630.
$$

### Step 4: Bounds for the integral
Since the left and right sums provide under- and over-estimations, we know that:

$$
L_4 \leq \int_{30}^{150} N(t) \, dt \leq R_4.
$$

Thus, the estimate for the integral is:

$$
8640 \leq \int_{30}^{150} N(t) \, dt \leq 9630.
$$

</details>

# 5.2.68 Find the indefinite integral. Check by differentiating. $$\int \dfrac{e^x}{2e^x-1} \, dx$$

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

Let's solve the integral

$$
\int \dfrac{e^x}{3e^x - 2} \, dx.
$$

### Step 1: Substitution
Let
$$
u = 3e^x - 2.
$$
Then, differentiating with respect to $x$:
$$
du = 3e^x \, dx.
$$

Now, solve for $dx$:
$$
dx = \frac{du}{3e^x}.
$$

We also note that $e^x = \frac{u + 2}{3}$ from the substitution. Substituting into the integral:

$$
\int \frac{e^x}{3e^x - 2} \, dx = \int \frac{e^x}{u} \cdot \frac{du}{3e^x} = \frac{1}{3} \int \frac{du}{u}.
$$

### Step 2: Solve the integral
The integral of $\frac{1}{u}$ is $\ln |u|$, so:

$$
\frac{1}{3} \ln |u| + C.
$$

### Step 3: Substituting $u$ back
Substitute $u = 3e^x - 2$ back into the expression:

$$
\frac{1}{3} \ln |3e^x - 2| + C.
$$

### Step 4: Differentiation to check the result
To verify, differentiate:

$$
\frac{d}{dx} \left( \frac{1}{3} \ln |3e^x - 2| \right) = \frac{1}{3} \cdot \frac{1}{3e^x - 2} \cdot 3e^x = \frac{e^x}{3e^x - 2}.
$$

This matches the original integrand, so the solution is correct.

Thus, the indefinite integral is:

$$
\boxed{\frac{1}{3} \ln |3e^x - 2| + C}.
$$

</details>

# 5.5.60 Find the definite integral. $$\int_1^2 \dfrac{x+1}{2x^2+4x+4} \, dx$$

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

To solve the integral

$$
\int_1^2 \dfrac{x+2}{3x^2+12x+5} \, dx,
$$

let's follow the substitution:

### Step 1: Substitution
Let
$$
u = 3x^2 + 12x + 5.
$$

Taking the derivative of $u$ with respect to $x$:

$$
du = (6x + 12) \, dx = 6(x + 2) \, dx.
$$

This allows us to rewrite the integral:

$$
\int_1^2 \frac{x+2}{3x^2+12x+5} \, dx = \int_1^2 \frac{x+2}{u} \, dx.
$$

Using the differential $du = 6(x + 2) \, dx$, we get:

$$
dx = \frac{du}{6(x+2)}.
$$

Substitute this into the integral:

$$
\int_1^2 \frac{x+2}{u} \cdot \frac{du}{6(x+2)} = \frac{1}{6} \int \frac{du}{u}.
$$

### Step 2: Solve the integral
This simplifies to:

$$
\frac{1}{6} \ln |u| + C.
$$

### Step 3: Evaluate the integral
Now, substitute $u = 3x^2 + 12x + 5$ back into the expression:

$$
\frac{1}{6} \ln |3x^2 + 12x + 5|.
$$

Evaluate this from $x = 1$ to $x = 2$:

$$
\frac{1}{6} \left[ \ln(3(2)^2 + 12(2) + 5) - \ln(3(1)^2 + 12(1) + 5) \right].
$$

Simplify each term:

\begin{align}
&= \frac{1}{6} \left[ \ln(3(4) + 24 + 5) - \ln(3(1) + 12 + 5) \right] \\
&= \frac{1}{6} \left[ \ln(12 + 24 + 5) - \ln(3 + 12 + 5) \right] \\
&= \frac{1}{6} \left[ \ln 41 - \ln 20 \right].
\end{align}

Using the logarithmic property $\ln a - \ln b = \ln \frac{a}{b}$, this becomes:

$$
\frac{1}{6} \ln \frac{41}{20}.
$$

Thus, the final result is:

$$
\boxed{\frac{1}{6} \ln \frac{41}{20}}.
$$

</details>

# 6.1.65 Find the area bound by the graphs. Round to three decimal places. $$y=x^3-3x^2-9x+12 ; \qquad y=x+12$$

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

</details>

# 6.2.78 Find the consumers’ surplus and the produc-ers’ surplus at the equilibrium price level for the given price–demand and price–supply equations. Include a graph that identifies the consumers’ surplus and the producers’ surplus. Round all values to the nearest integer. $$p=D(x)=25-0.004x^2 ; \quad p=S(x)=5+0.004x^2$$

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

</details>

# 6.3.44 Integrate by parts. $$\int_1^2 x^3 e^{x^2} \, dx$$

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

Integrate by parts. $$\int_1^2 x^5 e^{x^3} \, dx$$

We are given the integral:

$$
\int_1^2 x^5 e^{x^3} \, dx
$$

This integral can be solved by using **substitution**, rather than integration by parts. Here's the process:

### Step 1: Substitution

Let:

$$
u = x^3
$$

Then, the derivative of $u$ with respect to $x$ is:

$$
du = 3x^2 \, dx
$$

Thus, we can express $x^5 \, dx$ as follows:

$$
x^5 \, dx = x^3 \cdot x^2 \, dx = u \cdot \frac{du}{3}
$$

### Step 2: Change the limits of integration

When $x = 1$:

$$
u = 1^3 = 1
$$

When $x = 2$:

$$
u = 2^3 = 8
$$

### Step 3: Rewrite the integral

Now, substitute everything into the original integral:

$$
\int_1^2 x^5 e^{x^3} \, dx = \int_1^8 u \cdot \frac{e^u}{3} \, du
$$

This simplifies to:

$$
\frac{1}{3} \int_1^8 u e^u \, du
$$

### Step 4: Integration by parts

Now we will apply integration by parts. Recall the formula for integration by parts:

$$
\int u \, dv = uv - \int v \, du
$$

Let:
- $v = e^u$, so $dv = e^u \, du$
- $u$ stays as is, $du = du$

Now apply the formula:

$$
\int u e^u \, du = u e^u - \int e^u \, du
$$

The integral of $e^u$ is simply $e^u$. Therefore:

$$
\int u e^u \, du = u e^u - e^u
$$

### Step 5: Apply limits of integration

Now, substitute back into the expression:

$$
\frac{1}{3} \left[ u e^u - e^u \right]_1^8
$$

Substitute the limits $u = 8$ and $u = 1$:

$$
= \frac{1}{3} \left[ 8 e^8 - e^8 - (1 e^1 - e^1) \right]
$$

Simplify:

$$
= \frac{1}{3} \left[ 7 e^8 - 0 \right] = \frac{7 e^8}{3}
$$

### Final Answer:

The value of the integral is:

$$
\frac{7 e^8}{3}
$$

</details>