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.

Let \(n\) be the number of dolls the retailer can sell per week. Thus, \(n = 80 - 2x\).

Let \(p\) be the profit gained for all dolls sold. Thus, \(p = (x-10) * n\)

Substituting from above for \(n\) we get: \((x-10)(-2x+80)\). Which we can then multiply to get: \(-2x^2+100x-800\). Expressed in an R function as:

profit <- function(x)
{
  p <- -2*x^2 + 100*x - 800
  return(p)
}

So, selling dolls at $20 we’d sell 40 dolls (\(80 - 2(20) = 40\)). We’d expect $400 in total profit ($20 - $10 for each doll times 40 dolls). Let’s check:

profit(20)
## [1] 400

Problem 2

Given the following function: \(f(x) = 8x^3+7x^2-5\)

  1. Find \(f(3)\)
  2. Find \(f(-2)\)
  3. Find \(f(x+c)\)

We can set up this function in R quite easily:

p2 <- function(x)
{
  ans <- 8*(x^3)+7*(x^2)-5
  return(ans)
}

Now we can evaluate #1:

p2(3) 
## [1] 274

and #2:

p2(-2)
## [1] -41

However, because #3 asks us to solve symbolically, we’ll use a manual method: \[8(x+c)^3+7(x+c)^2-5\] We can expand the terms: \[8(c^3+3c^2+3cx^2+x^3) + 7(c^2+2cx+x^2) - 5\] Multiply through: \[8c^3+24c^2+24cx^2+8x^3+7c^2+14cx+7x^2-5\] …and group terms: \[8c^3+8x^3+31c^2+24cx^2+7x^2+14cx-5\]

Problem 3

  1. We’re asked to evaluate \(\lim\limits_{x \to 1^{-}}f(x)\) for the given graph. As \(x\) gets arbitrarily close to 1 from the left side, \(f(x)\) get’s arbitrarily close to 2. Despite the continuity at \(f(1)\) the function is still approaching 2.

  2. Next we’re asked to evaluate the other limit. That is \(\lim\limits_{x \to 1^{+}}f(x)\). This time, we see that \(f(x)\) gets close to -5 as \(f(x)\) approaches 1 from the right side.

  3. Finally, we evaluate \(\lim\limits_{x \to 1}\). We know that \(\lim\limits_{x \to 1^{-}}f(x) = 2\) and \(\lim\limits_{x \to 1^{+}}f(x) = -5\). Because the left and right limits differ, there is a discontinuity and no general limit.

Problem 4

Find the derivative for the following function: \(f(x) = -2x^3\)

# Define our function
p4 <- function(x)-2*x^3

#Find the derivative
Deriv(p4)
## function (x) 
## -(6 * x^2)

…which matches what we expect from the power rule.

Problem 5

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

This is also an application of the power rule. It becomes obvious once we re-write it to: \(-8x^{-2}\).

We’d expect R to give us the answer of \(16x^{-3}\)

p5 <- function(x)-8/(x^2)

Deriv(p5)
## function (x) 
## 16/x^3

…which it does.

Problem 6

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

As before, this is an application of the power rule, which can be seen when the equation is re-written as \(5x^{\frac{1}{3}}\)

We should see the derivative as \(\frac{5}{3}x^{\frac{-2}{3}}\)

p6 <- function(x)5*(x^(1/3))

Deriv(p6)
## function (x) 
## 1.66666666666667/x^0.666666666666667

Problem 7

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

Same as above, we can use R to compute the derivative:

p7 <- function(x)-2*x^(9/8)

Deriv(p7)
## function (x) 
## -(2.25 * x^0.125)

Problem 8

Given a 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 as an integer or simplified fraction.

At \(x=0\) the graph shows \(f(x)=40\) and at \(x=4\) the value of \(f(x)=35\). The average rate of change is equal to: \[\frac{\Delta f(x)}{\Delta x}\] In this particular case that is equal to \(\frac{5}{4}\).

Problem 9

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

Average cost is the total Cost \(C\) divided by the number of items \(x\). So here it would be \(\frac{630 + 2.4x}{x}\).

Problem 10

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

The product rule is, in general, \(\frac{\mathrm d}{\mathrm {dx}} (f(x) \cdot g(x)) = \frac{\mathrm d}{\mathrm {dx}}f(x)\cdot g(x) + \frac{\mathrm d}{\mathrm {dx}}g(x)\cdot f(x)\)

So, using the product rule: \(\frac{\mathrm d}{\mathrm {dx}} f(x)=(4x^{-3})(-5x+9)+(-5)(-2x^{-2}+1)\)

…multiplying through, we get: \(-20x^{-2}+36x^{-3}+10x^{-2}-5\)

We combine like terms and simplify to get: \(36x^{-3}-10x^{-2}-5\)

Problem 11

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

The quotient rule is, in general, \[\frac{\mathrm d}{\mathrm {dx}} \frac{f(x)}{g(x)} = \frac{\frac{\mathrm d}{\mathrm {dx}} f(x)\cdot g(x) - \frac{\mathrm d}{\mathrm {dx}}g(x)\cdot f(x)}{g(x)^2}\]

In this instance, that means: \[\frac{(\frac{5}{2}x^{\frac{-1}{2}})(-x^3+1) - (-3x^2)(5x^{\frac{1}{2}}+7)}{(-x^3+1)^2}\]

We can simplify a bit to get: \[\frac{\frac{-5}{2}x^{\frac{5}{2}} + \frac{5}{2}x^{\frac{-1}{2}} + 15x^{\frac{5}{2}}+21x^2}{(-x^3+1)^2}\]

Problem 12

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

Here we can use the chain rule.

Let \(u = 3x^{-3} - 8x + 6\), the inside of the equation. So, via the chain rule, our derivative would be: \[\frac{4}{3}(3x^{-3}-8x+6)(-9x^{-4}-8)\]

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.

To solve this, we need to find the derivative of the function \(f(x)\) and evaluate it at \(f^\prime (3)\).

If we re-write the equation, we can use the product rule here: \[f(t)=(550t^2)(t^2+15)^{\frac{-1}{2}}\]

\[\frac{\mathrm d}{\mathrm {dx}}f(x)=(1100t)(t^2+15)^{\frac{-1}{2}}+\frac{1}{2}(t^2+15)^{\frac{-3}{2}}(2t)(500t^2)\] We can then evaluate for \(t=3\):

p13 <- function(t)
{
  ans <- (1100*t)*(t^2+15)^(-1/2)+(1/2)*(t^2+15)^(-3/2)*(2*t)*(500*t^2)
  return(ans)
}
p13(3)
## [1] 788.4295

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.

We evaluate for \(N(3)\):

P14_1 <- function(t)
{
  ans <- 1000*(6 + 0.1*t)^(1/2)
  return(ans)
}

P14_1(3)
## [1] 2509.98

Step 2. Determine \(N^{\prime}(5)\) and interpret its meaning. Round your answer to the nearest whole number.

First, we find the derivative:

\[N^{\prime}(t) = 1000(\frac{1}{2}(6+0.1t)^{\frac{1}{2}}(0.1))\] Simplified: \[N^{\prime}(t) = \frac{50}{(6+0.1t)^{\frac{1}{2}}}\] We evaluate for \(t=5\)…

P14_2 <- function(t)
{
  ans <- (50)/((6+0.1*t)^(1/2))
  return(ans)
}

round(P14_2(5),0)
## [1] 20

…and find that attendance is increasing by 20 people per week as of week 5.

Problem 15

Consider the following function: \(3x^3+4y^3=77\)

Step 1. Use implicit differentiation to find \(\frac{\mathrm dy}{\mathrm {dx}}\).

\[\frac{\mathrm d}{\mathrm {dx}}3x^3+\frac{\mathrm d}{\mathrm {dx}}4y^3=\frac{\mathrm d}{\mathrm {dx}}77\] Differentiate: \[9x^2+12y\frac{\mathrm dy}{\mathrm {dx}}=0\] Then we solve for \(\frac{\mathrm dy}{\mathrm {dx}}\): \[\frac{\mathrm dy}{\mathrm {dx}}=\frac{-9x^2}{12y}\]

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

\[\frac{-9(3)^2}{12(-1)} = \frac{-81}{-12} = \frac{27}{4}\]

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: \[f^{\prime}(x) = \frac{-11}{(x-8)^2}\] Then, we can see that the derivative will never be positive (as the denominator will never be negative to offset the numerator). We can confirm this visually by plotting the derivative:

a = seq(-40,40,by=0.1)

p16 <- function(x)
{
  ans <- -11/((x-8)^2)
  return(ans)
}

fa <- p16(a)

plot(fa~a,type="l",ylab="f(x)",xlab="x",ylim=c(-0.5,0.5))

We see the discontinuity at \(x=8\) approaching as the denominator approaches zero, \(f^{\prime}(x)\) approaches \(-\infty\).

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.

Using the quotient rule, we find the derivative to be:

\[F^{\prime}(t)=\frac{73400}{(t^2+100)^2}\]

Looking at the \(F^{\prime}(t)\) equation, we can tell that it will never be negative. So, the temperature is increasing from \([0-\infty]\).

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

So what we want to know is: \[\lim_{x\to\infty} 14+\frac{367t^2}{t^2 + 100}\] …which we can see is 381

Problem 18

A study says that the package flow in the East during the month of November follows: \[f(x) = \frac{x^3}{3340000} - \frac{7x^2}{9475} + \frac{42417727x}{1265860000} + \frac{1}{33}, 1 \leq x \leq 30\] What is the maximum number of packages delivered in November?

To find the maximum, we need to find a maxima within the domain \(1 \leq x \leq 30\).

p18 <- function(x)
{
  ans <- ((x^3)/(3340000) - (7*x^2)/(9475) + (42417727*x)/(1265860000) + (1/33))
  #return(ans)
}

# We can use the optimize function to find the maximum within the range
op <- optimize(p18,interval=c(1,30),maximum=TRUE)
op
## $maximum
## [1] 23
## 
## $objective
## [1] 0.4138353

…which gives us a maximum of 0.41 million packages on day 23

To test this, we can plot the function and see if it visually agrees with our output:

a = seq(0, 30, by=0.1)
fa = p18(a)
plot(fa ~ a, type="l",ylab="f(x)",xlab="x")

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: \(7x^2+28x-35\)

First we find \(f^\prime(x)\) and then \(f^{\prime\prime}(x)\) using the power rule:

\[f^\prime(x) = 14x + 28\] \[f^{\prime\prime}(x) = 14\] Our critical values will be where \(f^\prime(x) = 0\). That is obviously \(f^\prime(-2)\). We would then evaluate \(f^{\prime\prime}(-2)\) but since \(f^{\prime\prime}(x)\) is a positive constant, it is concave up and a local minimum at \((-2,f(-2))\) or \((-2,-63)\).

We can check by plotting the function:

p19 <- function(x)
{
  ans <- 7*x^2+28*x-35
  return(ans)
}

x <- seq(-10,10,by=0.1)
fx <- p19(x)
plot(fx ~ x, type="l")
points(-2,-63,col="red",pch=19) # our predicted minimum

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

Using the power rule, we find: \[f^\prime(x)=-18x^2+54x+180\] \[f^{\prime\prime}(x)=-36x+54\] Critical values occur at \(f^\prime(x)=0\) which would be at \(x=-2\) and \(x=5\). We then evaluate \(f^{\prime\prime}(x)\) of those values: \(f^{\prime\prime}(-2) = 126\) and \(f^{\prime\prime}(5) = -126\).

So, the point \((-2,-204)\) is a minimum and the point \((5,825)\) is a maximum. The plot shows this quite nicely:

p20 <- function(x)
{
  ans <- -6*x^3+27*x^2+180*x
  return(ans)
}
x <- seq(-5,7,by=0.1)
fx <- p20(x)
plot(fx ~ x, type="l")
points(c(-2,5),c(-204,825),pch=19,col=c("dark green","red"))
text(c(-2,5),c(-204,825),c("Local Minimum","Local Maximum"),pos=c(3,1))

Problem 21

A beauty supply 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?

Let \(h\) equal the number of irons on-hand and \(k=120-h\) be the number of irons ordered, where \(h+k=120\) to meet demand.

Assuming a constant demand, our cost for irons on-hand would be: \[C_h = 1.6h\] and our cost for irons ordered (in terms of h): \[C_k = 4.5(120-h) + 6\] If we sum these functions into a total cost, we can see where the minimum is. The function for total cost is: \[C_t = -2.9h+546\] Since its linear, the minimum is the 120 quantity on-hand.

Question 22

Question 23

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

Here we can figure the total cost function by looking at the price of each segment and adding them based on the dimension (\(x\) or \(y\)). Here is the function with the external \(x\) walls, the internal \(x\) walls, and the external \(y\) walls, respectively:

\[C(x,y) = 28.8x+24x+28.8y\] Then we re-write \(y\) in terms of \(x\). Since we know \(A=1056=xy\) then \(y=\frac{1056}{x}\), we’ll now have a function of just one variable: \[C(x)=28.8x+24x+28.8(\frac{1056}{x})\] …which simplifies to: \[C(x) = \frac{30412.8}{x}+52.8x\] We’ll find the minimum of this cost function in R by finding the critical value of the first derivative:

cTotal <- function(x){(30412.8/x)+52.8*x}
cPrime <- Deriv(cTotal)
str(cPrime)
## function (x)
cv <- uniroot.all(cPrime,c(1,1000))
cv
## [1] 24

We can test the second derivative at this \(x\) value to see if it is a minimum or a maximum by looking at the sign or looking at the plot:

x <- seq(1,1055,by=0.1)
cX <- cTotal(x)
plot(cX ~ x, type="l", col="blue")
points(24,cTotal(24),col="red")

Problem 24

It is determined that the value of a piece of machinery declines exponentially. A machine that was purchased 7 years ago for $67000 is worth $37000 today. What will be the value of the machine 9 years from now? Round your answer to the nearest cent.

Let \(t\) = number of years since the purchase of the equipment. The function \(v(t)\) is the value of the equipment after \(t\) years. We are given \(v(7)=37000\) and \(v(0)=67000\).

Exponential decay is defined as: \[v(t)=a(1-r)^t\] By using \(v(7)=37000\) we can plug in and find the rate \(r\) of decay is 0.0813361. Now, we can build our formula properly: \[v(t)=67000(1-0.0813361)^t\]

v <- function(t)
{
  return(67000*(1-0.0813361)^t)
}
v(7)
## [1] 36997.41
v(16) #9 years later
## [1] 17241.74

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 revenue is maximized.

Revenue = Price * Quantity

and we know that \[p=23.2-0.4x\] So, \(R(x)=(23.2-0.4x)x\) or \(R(x)=(23.2x-0.4x^2)\)

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.

Exponential growth is defined as: \[v(t)=a(1+r)^t\] By using \(v(11)=426.80\) we can plug in and find the rate \(r\) of growth is 0.0059129. Now, we can build our formula properly: \[v(t)=400(1+0.006)^t\]

v <- function(t)
{
  return(400*(1+0.0059129)^t)
}
round(v(11),2)
## [1] 426.8

Estimating another 11 years in 2017:

round(v(22),2)
## [1] 455.4

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.

To get the profit function, we integrate the marginal profit function: \[\int {380-4x}\space dx= 380x - 2x^2 + c\] To find the value of \(c\) we use the given values of \(380(38)-2(38)^2 + c = 1700\).Solving for \(c\) we find that \(c=-11552\).

So the profit function is: \[P(x) = 380x-2x^2-11552\]

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

P <- function(x)
{
  return(380*x-2*x^2-11552)
}

P(56)
## [1] 3456

Problem 28

Use integration by substitution to solve the integral: \[\int \frac{-5(ln(y))^3}{4} \space \mathrm dy\]

First, rearrange the equation: \[-5 \int \frac{1}{x}(ln(y))^3\] Let \(u = ln(y)\). Thus \(\frac{1}{x} \space dx = du\). Now we rewrite the equation as: \[-5 \int u^3\space du\] Then we solve: \[-\frac{5}{4}u^4+c\] Now, we substitute back: \[-\frac{5}{4}(ln(y))^4 + c\]

Problem 29

It was discovered that after \(t\) years a certain population of wild animals will increase at a rate of \(P^\prime(t) = 75 - 9t^{\frac{1}{2}}\) 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.

To get the original function, we integrate: \[\int_{0}^{9} 75-9t^{\frac{1}{2}} \space dt\] …which gives us: \[75t-6t^{\frac{3}{2}} + c\] Evaluating from \(\mid_0^9\): \[75(9)-6(9)^{\frac{3}{2}} - 75(0)-6(0)^{\frac{3}{2}} = 675 - 162 = 513\]

Problem 30

Find the area of the region bounded by the graphs of the given equations: \[y=6x^2 \space\space y=6\sqrt{x}\]

To find the bounds of the region, we look for where the functions are equal: \[6x^2 = 6\sqrt{x}\] The functions are equal only when \(x=1\) and \(x=0\).

Next we find the integrals of each:

\[\int6x^2\space dx = 2x^3 + c\] \[\int6\sqrt{x}\space dx = 6\int x^\frac{1}{2}\space dx = \frac{12}{3}x^{\frac{3}{2}}+c\] We now evaluate each from 0 to 1: \[2(1)^3 - 2(0)^3 = 2\] \[\frac{12}{3}1^{\frac{3}{2}} - \frac{12}{3}0^{\frac{3}{2}} = \frac{12}{3}\] Now we subtract: \[\frac{12}{3}-2 = 2\] We can visualize this too:

p30_1 <- function(x)
{return(6*x^2)}

p30_2 <- function(x)
{return(6*x^(1/2))}

x <- seq(-1,1,by=0.1)
p30_1x <- p30_1(x)
p30_2x <- p30_2(x)

plot(c(p30_1x,p30_2x)~c(x,x),type="p")

Problem 31

Solve the differential equation given: \(\frac{dy}{dx}=x^3y\)$

\[ dy=x^3y\space dx\] \[\frac{1}{y}\space dy = x^3 \space dx\]

\[\int \frac{1}{y}\space dy = \int x^3 \space dx\] \[\ln(y) + c = \frac{1}{4}x^4 + c\]

\[y = e^{\frac{x^4}{4}}+C\]

Problem 32

Use integration by parts to evaluate the definite integral: \[\int_{-7}^{2} x\sqrt{x+7}\space dx\]

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 20^th time?

ball1 <- function(n)
{
  return(46*(1-(22/100)^n)/(1-(22/100)))
}

round(ball1(20),2)
## [1] 58.97

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

46/(1-(22/100))
## [1] 58.97436

Problem 34

Find the Taylor polynomial of degree 5 near \(x = 4\) for the following function. \(y = 3e^{(5x - 3)}\)

\[\frac{3}{e^3} + \frac{15x}{e^3} + \frac{75x^2}{2e^3} + \frac{125x^3}{2e^3} + \frac{625x^4}{8e^3} + \frac{625x^5}{8e^3}\]