Problem 1

It costs a toy retailer $10 to purchase a certain doll. He estimates that, if he charges $x$ dollars per doll, he can sell $80 - 2x$ dolls per week. Find a function for his weekly profit.

The profit function $P$ is a function is $x$, the price of dolls sold to consumers. Profit is revenue - cost, represented by $R(x)$ and $C(x)$, respectively. Quantity sold to consumers is $Q(x) = 80 - 2x$. Revenue is quntity sold times price:

$$R(x) = x \times Q(x) = x(80 - 2x) = 80x - 2x^2$$ $$C(x) = 10 \times Q(x) = 10(80 - 2x) = 800 - 20x$$

So the profit function is:

$$P(x) = R(x) - C(x) = (80x - 2x^2) - (800 - 20x)$$

$$= 80x - 2x^2 - 800 + 20x = 2x^2 - 100x + 800$$

Answer. $P(x) = 2x^2 - 100x + 800$

Problem 2

Given the following function $f(x) = 8x^3 + 7x^2 - 5$, find $f(3)$, $f(-2)$, and $f(x + c)$.

Define the function in R:

f <- function(x) 8*x^3 + 7*x^2 - 5

$$f(3) = 274$$ $$f(-2) = -41$$ $$f(x + c) = 8(x+c)^3 + 7(x+c)^2 -5$$

Problem 3

Use the graph to find the indicated limits. If there is no limit, state ``Does not exist.’’

Step 1. Find $\lim_{x \rightarrow 1^-} f(x)$.

They key here is the negative sign in the notation, which tells us we want to approach the limit from negative side (``below’’). Without that notation, the limit would not exist.

Answer. -5

Step 2. Find $\lim_{x \rightarrow 1^+} f(x)$.

Like Step 1, but approach the limit from the positive side (``above’’). $f(1)$ is undefined, but since we’re only concerned with $f$(near 1), we can still point out a specific limit.

Answer. 2

Step 3. Find Find $\lim_{x \rightarrow 1} f(x)$.

As noted above, this limit does not exist because $\lim_{x \rightarrow 1^+} f(x) \neq \lim_{x \rightarrow 1} f(x)$.

Answer. Does not exist

Problem 4

Find the derivative for the following function $f(x) = -2x^3$.

Set constant $2$ aside and use rule to evaluate the derivative of $x^3$:

$$f'(x) = \frac{d}{dx} (-2x^3) = -2 \left(\frac{d}{dx}x^3 \right) = -2 \times 3x^2 = -6x^2$$

Using R:

f <- function(x) -2*x^3
Deriv(f)

## function (x) 
## -(6 * x^2)

Problem 5

Find the derivative for the following function: $f(x) = \frac{-8}{x^2}$.

Set constant $-8$ aside and find derivative of $\frac{1}{x^2}$ by taking advantage of the fact $\frac{1}{x^2} = x^{-2}$ and the usual derivative rule of exponents.

$$f'(x) = \frac{d}{dx} \frac{-8}{x^2} = -8 \left(\frac{d}{dx} \frac{1}{x^2} \right) = -8 \left(\frac{d}{dx} x^{-2} \right) = -8 \times -2x^{-3} = 16x^{-3}$$

Using R:

f <- function(x) -8 / (x^2)
Deriv(f)

## function (x) 
## 16/x^3

Problem 6

Find the derivative for the following function: $g(x) = 5 \sqrt[3]{x}$.

Set constant $5$ aside and convert remaining to exponent form:

$$g'(x) = \frac{d}{dx} 5 \sqrt[3]{x} = 5 \left(\frac{d}{dx} \sqrt[3]{x} \right) = 5 \left(\frac{d}{dx} x^{1/3} \right) = 5 \times \frac{1}{3} x^{-\frac{2}{3}} = \frac{5}{3} \frac{1}{x^{2/3}}$$

Using R:

g <- function(x) 5 * nth_root(x, 3)
Deriv(g)

## function (x) 
## 1.66666666666667/x^0.666666666666667

Problem 7

Find the derivative for the following function: $y = -2x^{\frac{9}{8}}$.

Set constant $-2$ aside:

$$y' = \frac{d}{dx} \left( -2x^{\frac{9}{8}} \right) = -2 \left(\frac{d}{dx} x^{\frac{9}{8}} \right) = -2 \times \frac{9}{8}x^{\frac{1}{8}} = -\frac{18}{8} x^{\frac{1}{8}}$$

Using R:

y <- function(x) -2 * x^(9/8)
Deriv(y)

## function (x) 
## -(2.25 * x^0.125)

Problem 8

Consider the graph of $f(x)$. What is the average rate of change of $f(x)$ from $x_1 = 0$ to $x_2 = 4$? Please write your answer an integer or simplified fraction.

The average rate of change between two points is the slope of the secant line connecting them. This can be calculated using the usual rise-over-run function:

$$AROC = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 0}{35 -40} = \frac{4}{-5}$$

Answer. $-\frac{4}{5}$

Problem 9

The cost of producing $x$ baskets is given by $C(x) = 630 + 2.4x$. Determine the average cost function.

Mock up the function, pick an arbitrary integer for $x$, compute $C(x)/x$ to get the average cost of $8.70/basket.

C <- function(x) 630 + 2.4*x
x <- 100
C(100) / x

## [1] 8.7

Problem 10

Use the Product Rule or Quotient Rule to find the derivative: $f(x) = (-2x^{-2} + 1)(-5x + 9)$.

Break down the $f(x)$ for suitable input into the product rule:

$$f(x) = f_{1}(x) f_{2}(x)$$ $$f_{1}(x) = -2x^{-2} + 1$$ $$f_{2}(x) = -5x + 9$$

So:

$$f'(x) = f_{1}'f_{2} + f_{1}f_{2}'$$

Take the derivative of each function seperately:

$$f_{1}'(x) = \frac{d}{dx}(1) -2(\frac{d}{dx} (x^{-2})) = 0 - 2(-2x^{-3}) = 4x^{-3}$$

$$f_{2}'(x) = \frac{d}{dx}(9 - 5x) = \frac{d}{dx}(9) - 5(\frac{d}{dx}(x)) = 0 - 5(1) = -5$$

Putting it all together and multiplying through:

$$f'(x) = 4x^{-3}(-5x + 9) - 5(-2x^{-2} + 1)$$ $$= 4(\frac{1}{x^3})(-5x + 9) + (10x^{-2} - 5)$$ $$= 4 \frac{-5x + 9}{x^3} + (\frac{10}{x^2} - 5)$$

Using R:

f <- function(x) (-2*x^(-2) + 1) * (-5*x + 9)
Deriv(f)

## function (x) 
## 4 * ((9 - 5 * x)/x^3) - 5 * (1 - 2/x^2)

Problem 11

Use the Product Rule or Quotient Rule to find the derivative: $$f(x) = \frac{5x^{\frac{1}{2}} + 7}{-x^3 + 1}$$

Break down $f(x)$ into its components for the Quotient rule:

$$f_1(x) = 5x^{\frac{1}{2}} + 7$$ $$f_2(x) = -x^3 + 1$$ and

$$f'(x) = \frac{f_{1}'(x) f_2(x) - f_1(x) f_2'(x)}{f_2(x)^2}$$

First, take derivative of both components:

$$f_{1}'(x) = 5(\frac{d}{dx}(x^{\frac{1}{2}})) + \frac{d}{dx}(7) = 5(\frac{1}{2}x^\frac{-1}{2}) + 0 = \frac{5}{2} x^\frac{-1}{2}$$

$$f_{2}'(x) = \frac{d}{dx}(-x^3) + \frac{d}{dx}(1) = -3x^2 + 0 = -3x^2$$

Finally resulting in this monster of an expression:

$$\frac{(\frac{5}{2} x^\frac{-1}{2}) (-x^3 + 1) - (5x^{\frac{1}{2}} + 7) (-3x^2)}{(-x^3 + 1)^2}$$

Using R:

f <- function(x) (5*x^(1/2) + 7) / (-x^3 + 1)
Deriv(f)

## function (x) 
## {
##     .e1 <- 1 - x^3
##     .e2 <- sqrt(x)
##     (2.5/.e2 + 3 * (x^2 * (5 * .e2 + 7)/.e1))/.e1
## }

Problem 12

Find the derivative for the given funciton. Write your answer using positive and negative exponents and fractional exponents instead of radicals: $f(x) = (3x^{-3} - 8x + 6)^{\frac{4}{3}}$.

Use the product rule, where:

$$f_1(x) = x^{\frac{4}{3}}$$ $$f_2(x) = 3x^{-3} - 8x + 6$$

So:

$$f_1'(x) = \frac{d}{dx} x^{\frac{4}{3}} = \frac{4}{3}x^{\frac{1}{3}}$$

$$f_2'(x) = 3(\frac{d}{dx} x^{-3}) - 8(\frac{d}{dx} x) + (\frac{d}{dx} 6) = 3(-3x^{-4}) - 8(1) + 0 = -9x^{-4} - 8$$

Putting it together:

$$f'(x) = (\frac{4}{3}x^{\frac{1}{3}})(3x^{-3} - 8x + 6) + (x^{\frac{4}{3}})(-9x^{-4} - 8)$$

Using R:

f <- function(x) (3*x^(-3) - 8*x + 6)^(4/3)
Deriv(f)

## function (x) 
## -(1.33333333333333 * ((3/x^3 + 6 - 8 * x)^0.333333333333333 * 
##     (8 + 9/x^4)))

Problem 13

After a sewage spill, the level of pollution in Sootville is estimated by $f(t) = \frac{550t^2}{\sqrt{t^2 + 15}}$, where $t$ is the time in days since the spill occurred. How fast is the level changing after 3 days? Round to the nearest whole number.

Get the solution by taking the derivative of $f(t)$ with the quotient rule and then the answer is $f'(3)$. Let $f_1(x)$ and $f_2(x)$ represent the numerator and denominator, respectively.

$$f_1'(t) = \frac{d}{dt} 500t^2 = 550(2t) = 1100t$$

To get $f_2'(t)$, we must use the chain rule $F'(x) = u'(v) v'$:

$$u' = \frac{d}{dx} \sqrt{x} = \frac{1}{2}x^{-\frac{1}{2}}$$ $$v' = \frac{d}{dt} \left( t^2 + 15 \right) = 2t$$ Putting it together:

$$f_2'(t) = \frac{1}{2}(t^2 + 15)^{-\frac{1}{2}} \times 2t = t \times \frac{1}{(t^2 + 15)^{\frac{1}{2}}} = \frac{t}{\sqrt{t^2 + 15}}$$

Then use the quotient rule to put it together like:

$$\frac{f_1'(x) f_2(x) - f_1(x) f_2'(x)}{f_2(x)^2}$$

The result is a quite massive expression; so, using R:

f <- function(t) (550*t^2) / (sqrt(t^2 + 15))
f_tick <- Deriv(f)
f_tick

## function (t) 
## {
##     .e1 <- t^2
##     .e2 <- 15 + .e1
##     550 * (t * (2 - .e1/.e2)/sqrt(.e2))
## }

Answer. The level of sewage $f'(3) = 547.3078644$ units per day after 3 days

Problem 14

The average home attendance per week at a Class AA baseball park varied according to the formula $N(t) = 1000(6 + 0.1t)^{\frac{1}{2}}$ where $t$ is the number of weeks into the season $(0 \leq t \leq 14)$ and $N$ represents the number of people.

Step 1. What was the attendance during the third week into the season? Round your answer to the nearest whole number.

Find $N(x)$ where $x = 3$:

N <- function(t) 1000 * (6 + 0.1*t)^(1/2)
round(N(3))

## [1] 2510

Step 2. Determine $N'(5)$ and interpret its meaning. Round your answer to the nearest whole number.

Find $N'(t)$ and input 5:

Begin by factoring out constant:

$$N'(t) = 1000 (\frac{d}{dt} (\sqrt{6 + 0.1t}))$$ Continue by using the chain rule:

$$= 1000 \times \frac{1}{2}(6 + 0.1t)^{\frac{-1}{2}} \times \frac{d}{dt}(6 + 0.1t) = 1000 \times \frac{1}{2}(6 + 0.1t)^{\frac{-1}{2}} \times 0.1$$

Simplifying:

$$N'(t) = 500 \times 0.1 \times \frac{1}{\sqrt{6 + 0.1t}} = \frac{50}{\sqrt{6 + 0.1t}}$$

The answer is $N'(5)$; using R:

N_tick <- Deriv(N)
round(N_tick(5))

## [1] 20

This means that that the fifth week had 20 more persons in attendance than the previous week. Compare to $N(5) - N(4) =$ 19.6876287.

Answer. 20

Problem 15

Consider the following function: $3x^3 + 4y^3 = 77$.

Step 1. Use implicit differentiation to find $\frac{dy}{dx}$.

First, differentiate both sides of the equation:

$$\frac{d}{dx} (3x^3) + \frac{d}{dx} (4y^3) = \frac{d}{dx} (77)$$ $$\equiv 9x^2 + 12y^2y' = 0$$

Now get $y'$ by itself:

$$12y^2y' = -9x^2$$ $$\equiv y' = \frac{-9x^2}{12y^2} = -\frac{3x^2}{4y^2}$$

Step 2. Find the slope of the tangent line at $(3, -1)$.

Using the above result, plug in 3 for $x$ and $-1$ for $y$:

y_tick <- function(x, y) (3*x^2) / (4*y^2)
y_tick(3, -1)

## [1] 6.75

Problem 16

Find the intervals on which the following function is increasing and on which it is decreasing:

$$f(x) = \frac{x + 3}{x - 8}$$

First, find the derivative of the function:

f <- function(x) (x + 3) / (x -8)
f_tick <- Deriv(f)
f_tick

## function (x) 
## {
##     .e1 <- x - 8
##     (1 - (3 + x)/.e1)/.e1
## }

$$f'(x) = \frac{1 - \frac{3 + x}{x - 8}}{x -8}$$ Now find the critical points, the $x$’s where f(x) = 0. I’ll select an interval $I$ between -100 and 100, and see if there are any such $x$’s:

s <- seq(-100, 100, 1)
f_s <- sapply(s, f_tick)
0 %in% f_s

## [1] FALSE

Since zero is not in f_s, we know there aren’t any critical points, meaning the entire function is either increasing or decreasing. Since $f'(1) = -0.2244898$ is negative, we can say $f(x)$ is decreasing across the entire function.

Answer. $f(x)$ is decreasing across the $(\infty, \infty)$

Problem 17

A frozen pizza is placed in the oven at $t = 0$. The function $F(t) = 14 + \frac{367t^2}{t^2 + 100}$ approximates the temperature (in degrees Fahrenheit) of the pizza at time $t$.

Step 1. Determine the interval for which the temperature is increasing and the interval for which it is decreasing. Please express your answers as open intervals.

First find the derivative of $F(t)$:

$$F'(t) = \frac{d}{dt} 14 + \frac{d}{dt} \left( \frac{367t^2}{t^2 + 100} \right) = \frac{d}{dt} \left( \frac{367t^2}{t^2 + 100} \right)$$

Using quotient rule:

$$F_1'(t) = \frac{d}{dt} 367t^2 = 367(2t) = 734t$$

$$F_2'(t) = \frac{d}{dt} (t^2 + 100) = 2t$$

$$F'(t) = \frac{(734t)(t^2 + 100) - (2t)(367t^2)}{ (t^2 + 100)^2 } = \frac{734t^3 + 73400t - 767t^3 }{(t^2 + 100)^2} = \frac{73400t}{(t^2 + 100)^2}$$

Now let’s find the critical points, i.e., the $t$ that ensures $F'(t) = 0$. Just looking at the equation, we can see that the only critical point is $t = 0$: The numerator becomes 0 when $t = 0$. Since this is an active oven, this makes sense—we only expect the oven to be 0 degrees (presumably above to room temperature) just as the gas is turned on. ``Negative time’’ is not in our domain so I will not consider $t < 0$.

Answer. $F(t)$ is increasing over $(0, \infty)$

Step 2. Over time, what temperature is the pizza approaching?

F <- function(t) 14 + (367*t^2)/(t^2 + 100)
s <- seq(0, 500, 1)
k <- sapply(s, F)

Answer. The pizza appears to be approaching about 380 degrees

Problem 18

A study says that the package flow in the East during the month of November follows $f(x) = \frac{x^{3}}{3340000} - \frac{7x^2}{9475} + \frac{42417727x}{1265860000} + \frac{1}{33}$, where $1 \leq x \leq 30$ is the day of the month and $f(x)$ in in millions of packages. What is the maximum number of packages delivered in November? On which day are the most packages delivered? Round your final answer to the nearest hundredth?

Mock up the function in R, use deriv() to find the derivative, max() to find the maximum number of packages delivered, and uniroot() to find the day the maximum occurred on.

f <- function(x) (x^3/3340000) - ((7*x^2)/9475) + ((42417727*x)/1265860000) + (1/33)
s <- seq(1, 30, 1)
a <- f(s)
a_max <- max(a)
f_tick <- Deriv(f)
t_max <- uniroot.all(f_tick, c(1, 30))

Answer. t_max = November 23 delivered the most packages, about 413,835

Problem 19

Use the Second Derivative Test to find all local extrema, if the test applies. Otherwise, use the First Derivative Test. Write any local extrema as an ordered pair:

$$f(x) = 7x^2 + 28x - 35$$

The derivative of $f(x)$ is $f'(x) = 14x + 28$. The only critical value it has is -2. However, we can’t use the second derivative test, because $f''(x) = 14$, so we have to rely on the first derivative test.

$f'(-3) = -14$ and $f'(-1) = 14$, so we know that $f'(x)$ changes from negative to positive at -2, so there is a local minima at $(-2, -63)$.

Answer. Local minima at $(-2, -63)$

Problem 20

Use the Second Derivative Test to find all local extrema, if the test applies. Otherwise, use the First Derivative Test. Write any local extrema as an ordered pair:

$$f(x) = -6x^3 + 27x^2 + 180x$$

We know from the form of the equation there are two extrema. Let’s find the derivative and the critical points with R:

f <- function(x) -6*x^3 + 27*x^2 + 180*x
f_tick <- Deriv(f)
r <- uniroot.all(f_tick, c(-10, 10))
r

## [1] -2  5

Use the second derivative test to see if these are local mins or maxs or neither:

f_ticktick <- Deriv(f_tick)
f_ticktick(r[1])

## [1] 126

f_ticktick(r[2])

## [1] -126

Answer. There is a local minimum at $(-2, f(-2)=-204)$ and a local maximum at $(5, f(5)=825)$

Problem 21

A beauty store expects to sell 120 flat irons during the next year. It costs $1.60 to store one flat iron for one year. To reorder, there is a fixed cost of $6, plus $4.50 for each flat iron ordered. In what lot size and how many times per year should an order be placed to minimize inventory costs?

The question is asking us to find the number of orders to minimimize a cost function $C$ which is subject to the cost of the flatirons itself, a flat order fee, and storage costs.

First, let’s break down the costs in terms of $x$, the number of flatirons per order such that $\frac{120}{x} =$ number of orders. $C(x) = $ cost of (annual) storage $C_s(x) +$ cost of the orders (annual) $C_r(x)$. Assuming that flat irons are sold at a constant rate, the average stock will be $\frac{x}{2}$:

$$C_s(x) = storage\_cost \times stock = 1.6 \frac{x}{2}$$

$$C_r(x) = order\_cost \times qty\_orders = (4.5x + 6) \times \frac{120}{x}$$

So:

$$C(x) = 1.6 \frac{x}{2} + 120 \left( \frac{4.5x + 6}{x} \right)$$

Now let’s find the $x$ that minimizes $C(x)$ by finding $C'(x)$, setting it to 0, and solving for $x$:

C <- function(x) (1.6 * (x/2)) + 120 * ((4.5*x + 6) / x)
Deriv(C)

## function (x) 
## 0.8 + 120 * ((4.5 - (4.5 * x + 6)/x)/x)

Simplify the derivative:

$$C'(x) = .8 - \frac{720}{x^2}$$

And solving:

$$0 = .8 - \frac{720}{x^2} \equiv .8 = \frac{720}{x^2} \equiv x^2 = 800 \equiv x = 30$$

Since $\frac{120}{30} = 4$:

Answer. The optimum number of orders is 4 at 30 flatirons/order, and the optimal storage facility will hold 30 flatirons

Problem 22

A shipping company must design a closed rectangular shipping crate with a square base. The volume is 18432 ft cubed. The material for the top and side costs $3 per sqare foot and the material for the bottom costs $5 per square foot. Find the dimensions of the create that will minimize the total cost of material.

First, write an intelligible cost function. We know what the volume of this crate is $w \times l \times h$, and because the bottom (and top) are square, that $w = l$, making volume $w^2h$.

The desired volume is 18432 ft cubed, we can get that $h = \frac{18432}{w^2}$, where $w^2$ can be interpreted as the area of the top and bottom. The area of each side is $wh$, so total area of the sides is $4wh$.

The cost function is the sum of the cost of top, bottom, and sides:

$$C = 3w^2 + 5w^2 + 3(4wh)$$

Substituting $\frac{18432}{w^2}$ for $h$:

$$C = 3w^2 + 5w^2 + 12w \frac{18432}{w^2}$$

Find the derivative of C, set to 0, and solve for $w$:

C <- function(w) 3*w^2 + 5*w^2 + 12*w*(18432/w^2)
Deriv(C)

## function (w) 
## 16 * w - 221184/w^2

Using the fact that $16x = \frac{16w^3}{16w^2}$:

$$\frac{16w^3 - 221184}{w^2} = 0 \equiv 16w^3 - 221184 = 0 \equiv 16w^3 = 221184 \equiv w^3 = 13824 \equiv w = 24$$

Our dimensions are:

$W = 24$ ft. $L = 24$ ft $H = \frac{18432}{24^2} = 32$ ft.

Verify the area is still 18432 ft. cubed:

24*24*32

## [1] 18432

Problem 23

A farmer wants to build a rectangular pen and then divide it with two interior fences. The total area inside of th epen will be 1056 square yards. The exterior fencing costs $14.40 per yard and the interior fencing costs $12.00 per year. Find the dimensions of the pen that will minimize the cost.

Since we know total area, we know that $xy = 1056$. We also know the cost functions of respective fencing:

$$C_i(x) = 2(12x)$$ $$C_l(y) = 2(14.4y)$$ $$C_w(x) = 2(14.4x)$$

The total cost function is the sum of the three preceeding equations:

$$C(x,y) = 28.8y + 24x + 28.8x$$

We can substitute $y = \frac{1056}{x}$ and simplify:

$$C(x) = 52.8x + \frac{30412.8}{x}$$

Take derivative, set to 0, and find $x$:

C <- function(x) 52.8*x + (30412.8/x)
Deriv(C)

## function (x) 
## 52.8 - 30412.8/x^2

$$0 = 52.8 - \frac{30412.8}{x^2}$$ $$52.8 = \frac{30412.8}{x^2}$$ $$x^2 = 576$$ $$x = 24$$

Answer. Width $x = 24$ yards and length $y = 44$ yards

Problem 24

It is determined that the value of a piece of machinery declines exponentially. A machine was purchased 7 years ago for $67,000 is worth $37,000 today. What will be the value of the machine 9 years from now? Round your answer to the nearest cent.

We need to estimate an exponential function $f(t) = ab^t$ from the two points we were given, (0, 67000) and (7, 37000), and then find $f(7+9)$.

We can use 67,000 for $a$ and set the result to $37000$ and $t$ to $7$, and then solve for $b$:

$$f(t) = 67000b^t$$

$$37000 = 67000b^7 \equiv \frac{37}{67} = b^7 \equiv b = 0.9187$$

Plugging $7+9=16$ into $f(t)$ gives us the value of the machine in another 9 years:

$$f(16) = 67000 * 0.9187^16 = 17252.59$$

Answer. The value of the machine 9 years from now is $17,252.59

Problem 25

The demand function for a television is given by $p = D(x) = 23.2 - 0.4x$ dollars. Find the level of production for which the reveue is maximized.

This is a linear function so it has no critical points, therefore, it can have no maxima.

Answer. No maxima

Problem 26

The amount of goods and services that costs $400 on January 1, 1995 costs $426.80 on January 1, 2006. Estimate the cost of the same goods and services on January 1, 2017. Assume the cost is growing exponentially. Round your answer to the nearest cent.

We have two points $(0, 400)$ and $(11, 426.80)$, and we need to find $y$ in $(22, y)$. Use 400 for $a$, 426.80 for the result of the function, and 11 for $t$ to get $b$:

$$426.80 = 400b^11 \equiv 1.067 = b^11 \equiv b = 1.005913$$

Using this equation:

$$f(22) = 400(1.005913)^22 = 455.396$$

Answer. The basket of goods costs $455.40 in January 1, 2017

Problem 27

A manufacturer has determined that the marginal profit from the production and sale of $x$ clock radios is approximately $380 - 4x$ dollars per clock radio.

Step 1. Find the profit function if the profit from the production and sale of 38 clock radios is $1700.

We are given the marginal profit function $P'(x)$, which is the derivative of the profit function $P(x)$. We need to find an antiderivative of $P'(x)$ such that when $x = 38$ profit $p = 1700$.

First, let’s find an antiderivate:

$$P(x) = \int P'(x) = \int 380 - 4x~dx = \int 380~dx - \int 4x~dx = (380x + C) - 4 \int x~dx$$ # continue $$= (380x + C) - 4(\frac{x^2}{2} + C) = 380x - 2x^2 + C$$

Finish the profit function with $C$ by finding the $C$ such that $P(38) = 1700$:

$$1700 = 380(38) - 2(38)^2 + C = 11552 + C$$

Thus $CC = -9852$, and the final equation is:

$$P(x) = 380x - 2x^2 - 9852$$ .

Step 2. What is the profit from the sale of 56 clock radios?

P <- function(x) 380*x - 2*x^2 - 9852
P(56)

## [1] 5156

Problem 28

Use integration by substitution to solve the integral below.

$$\int \frac{-5(ln (y))^3}{y}~dy$$

First, rearrange to get into the form of $\int f(g(x)) g'(x)~dx$:

$$= 5 \int ln(y)^3 \times \frac{1}{y}~dy$$

Substitute $u = ln(y)$:

$$5 \int u^ du$$ $$= 5 \times \frac{u^4}{4} + C$$

Add $u$ back in for the solution:

$$\int \frac{-5(ln (y))^3}{y}~dy = 5 \times \frac{ln(y)^3}{4} + C$$

Problem 29

It was discovered that after $t$ years a certain population of wild animals will increase at a rate of $P(t) = 75 - 9t^{\frac{1}{2}}$ animals per year. Find the increase in the population during the first 9 years after the rate was discovered. Round your answer to the nearest whole animal.

We need to find the area $A$ underneath $P(t)$ from $t = a = 0$ through to $t = b = 9$. First, find the derivative of the function (ignoring $C$ for now):

$$\int_{0}^{9} P(t)~dt= \int_{0}^{9} 75 - 9t^{\frac{1}{2}}~dt = \int_{0}^{9} 75~dt - \int_{0}^{9} 9t^{\frac{1}{2}}~dt$$

$$= 75t - 9\int_{0}^{9} t^{\frac{1}{2}}~dt$$

$$=75t - 9 \left( \frac{2t^{\frac{3}{2}}}{3} \right)$$ $$= 75t - 6t^{\frac{3}{2}}$$

Now evaluate the integral between 0 and 9:

$$A = 75t - 6t^{\frac{3}{2}} \Big|_0^9 = \left( 75(9) - 6(9)^{\frac{3}{2}} \right) - \left( 75(0) - 6(0)^{\frac{3}{2}} \right)$$

A = function(t) 75*t - 6*t^(3/2)
A(9) - A(0)

## [1] 513

Problem 30

Find the area of the region bounded by the graphs of the given equations.

$y_1 = 6x^2, y_2 = 6 \sqrt{x}$

Let’s take a look at the graph. It’s hard to see precisely the area visually, but it’s small, smaller than half of the total area of the graph $6 \times 1 = 6$ units.

f1 <- function(x) 6*x^2
f2 <- function(x) 6 * sqrt(x)
x <- seq(0, 1.05, .01)
y1 <- sapply(x, f1)
y2 <- sapply(x, f2)
plot(x, y1 ,type='l', col='blue', xlab='x', ylab='y_i')
lines(x, y2, col='green')
grid(nx=1, ny=6)

Since $y_2$ is ``on top’’ we know that:

$$A = \int_{0}^{1} 6 \sqrt{x}~dx - \int_{0}^{1} 6x^2~dx$$

First find the integral of $y_2$:

$$\int_{0}^{1} y(x)_2~dx= \int_{0}^{1} 6 \sqrt{x}~dx = 6 \int_{0}^{1} \sqrt{x}~dx$$

And

$$6 \left( \frac{2x^{\frac{3}{2}}}{3} \right) = \frac{12x^{\frac{3}{2}}}{3} = 4x^{\frac{3}{2}}$$

Evaluate:

$$4(1)^{\frac{3}{2}} - 4(0)^{\frac{3}{2}} = 4 - 0 = 4$$

Now find the integral of $y_1$:

$$\int_{0}^{1} y(x)_1 = \int_{0}^{1} 6x^2~dx = 6 \int_{0}^{1} x^2~dx$$

$$= 6 \frac{x^3}{3} = 2x^3$$

and now evalute it over 0 and 1:

$$2(1)^3 - 2(0)^2 = 2 - 0 = 2$$

Subtract the two evaluations and $4-2 = 2$. Looking at the graph, that seems reasonable.

Answer. $2$

Problem 31

Solve the differential equation given below: $$\frac{dy}{dx} x^3y$$ .

First seperate out the variables so that all the $x$’s and $y$’s are on different sides of the equality sign:

Problem 32

Use integration by parts to evaluate the definite integral below.

$$\int_{-7}^2 x \sqrt{x + 7}~dx$$

Let $u = x$ and $dv = \sqrt{x + 7}$. So:

$$du = \frac{d}{dx} x = 1~dx$$ $$v = \int dv = \int \sqrt{x + 7}~dx$$

Now put these all back together using the parts formula:

$$= x \times (\sqrt{x + 7}) \Big|_{-7}^2 - \int_{-7}^2 \sqrt{x + 7} \times 1~dx$$

Solve the integral on the right hand side of the equal:

$$\int \sqrt{x + 7} \times 1~dx = \frac{2(x+7)^{\frac{3}{2}}}{3}$$

Now let’s find the areas under both, and subtract the right-hand from the left-hand side of the equations:

left <- function(x) x * (sqrt(x + 7))
right <- function(x) (2*(x+7)^(3/2)) / 3    
(left(2) - left(-7)) - (right(2) - right(-7))

## [1] -12

Answer. -12

Problem 33

The following can be answered by finding the sum of a finite or infinite geometric sequence. Round the solution to 2 decimal places.

A rubber ball is dropped from a height of 46 meters, and on each bounce it rebounds up 22% of its previous height.

Step 1. How far has it traveled vertically at the moment when it hits the ground for the 20th time?

Use the sum formula:

$$S_n = \frac{a_1(1 - r^n)}{1 - r}$$

where $a_1$ is the initial starting point, 46 meters, and $r = .22$:

$$S_20 = \frac{46(1 - .22^20)}{1 - .22} = 58.97436$$

Answer. The ball has travelled 58.97436 meters vertically

Step 2. If we assume it bounces indefinitely, what is the total vertical distance traveled?

The sum of an infinite geometric sequence is

$$S = \frac{a_1}{1 - r}$$

So:

$$S = \frac{46}{1 - .22} = 58.97436$$

In fact, the balls stops bouncing up and down on the 12th bounce.

Answer. Total vertical distance traveled is 58.97436 meters

Problem 34

Find the Taylor polynomial of degree 5 near x = 4 for the following function:

$$y = 3e^{5x - 3}$$

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