MathAssignment1
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January 11, 2017
Questioni 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.
Answer
The weekly profit (at x dollars per doll) is:
Selling Price (SP) - Cost Price (CP)
SP(x) = The amount of dolls * cost per doll = (80 - 2x) x CP(x) = The amount of dolls $10 = (80 - 2x) * 10
SP - CP = (80 - 2x)x - 10(80-2x) = 80x - 2x^2 - 800 + 20x = -2x^2 + 100x - 800
Question 2
Given the following function: f (x) = 8x^3 + 7x^2 - 5
Step 1. Find f (3).
Step 2. Find f (-2).
Step 3. Find f (x + c).
Answer
q2 <- function(x){8*x^3 + 7*x^2 - 5}
step1 <- q2(3)
step2 <- q2(-2)Step 1 Answer is: 274 Step 2 Answer is: -41
Question 3
Answer
Step 1 = 0
Step 2 = -5
Step 3 - Does not exist
Question 4
Find the derivative for the following function. f (x) = -2x^3
Answer
q4 <- function(x){-2*x^3}
q4ans <- Deriv(q4)
q4ans## function (x)
## -(6 * x^2)Question 5
Find the derivative for the following function. f (x) = (-8)/x^2
Answer
q5 <- function(x){(-8)/x^2}
q5ans <- Deriv(q5)
q5ans## function (x)
## 16/x^3Question 6
Find the derivative for the following function. g(x) = 5*x^(1/3)
Answer
q6 <- function(x){5*x^(1/3)}
q6ans <- Deriv(q6)
q6ans## function (x)
## 1.66666666666667/x^0.666666666666667Answer as a fraction is:
Question 7
Find the derivative for the following function. y = -2x^(9/8)
Answer
q7 <- function(x){-2*x^(9/8)}
q7ans <- Deriv(q7)
q7ans## function (x)
## -(2.25 * x^0.125)Question 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 as an integer or simplified fraction.
Answer
(35 - 40)/(4-0) = 1 1/4
Question 9
The cost of producing x baskets is given by C(x) = 630 + 2.4x. Determine the average cost function.
Answer
Average Cost = (Total Cost)/Quantity
= (630 + 2.4x)/x
Question 10
Use the Product Rule or Quotient Rule to find the derivative. f (x) = (-2x^(-2) + 1) (-5x + 9)
Answer
q10 <- function(x){(-2*x^(-2) + 1) * (-5*x + 9)}
q10ans <- Deriv(q10)
q10ans## function (x)
## 4 * ((9 - 5 * x)/x^3) - 5 * (1 - 2/x^2)Question 11
Use the Product Rule or Quotient Rule to find the derivative. f (x) = (5x^(1/2) + 7)/(-x^3 + 1)
Answer
q11 <- function(x){(5*x^(1/2) + 7)/(-x^3 + 1)}
q11ans <- Deriv(q11)
q11ans## function (x)
## {
## .e1 <- 1 - x^3
## .e2 <- sqrt(x)
## (2.5/.e2 + 3 * (x^2 * (5 * .e2 + 7)/.e1))/.e1
## }Another representation of the answer is: 
Question 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)^(4/3)
Answer
q12 <- function(x){(3*x^(-3) - 8*x + 6)^(4/3)}
q12ans <- Deriv(q12)
q12ans## function (x)
## -(1.33333333333333 * ((3/x^3 + 6 - 8 * x)^0.333333333333333 *
## (8 + 9/x^4)))Calculation by hand (and representation using exponents rather than radicals) gives a seemingly different answer:
Question 13
After a sewage spill, the level of pollution in Sootville is estimated by f (t) = (550t^2)/???( 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.
Answer
If f(t) represents the level of pollution to find how fast the level is changing we have to find the rate of change represented by the derivative of f(t).
q13 <- function(t){(550*t^2)/( t^2 + 15)^(1/2)}
q13ans <- Deriv(q13)
q13ans## function (t)
## {
## .e1 <- t^2
## .e2 <- 15 + .e1
## 550 * (t * (2 - .e1/.e2)/sqrt(.e2))
## }To find how fast the level is changing after 3 days we have to find f’(3)
round(q13ans(3),0)## [1] 547Question 14
The average home attendance per week at a Class AA baseball park varied according to the formula N(t) = 1000(6 + 0.1t)^(1/2) where t is the number of weeks into the season (0 £ t £ 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.
Step 2. Determine N ´(5) and interpret its meaning. Round your answer to the nearest whole number.
Answer Step 1
N(3)
q14 <- function(t){1000*(6 + 0.1*t)^(1/2)}
q14ans <- q14(3)
q14ans## [1] 2509.98Answer Step 2
q14ans2 <- Deriv(q14)
q14ans2## function (t)
## 50/sqrt(0.1 * t + 6)q14ans2(5)## [1] 19.61161At 5 week attendance increases by 19 persons
Question 15
Consider the following function: 3x^3 + 4y^3 = 77
Step 1. Use implicit differentiation to find dy/dx.
Step 2. Find the slope of the tangent line at (3,-1).
Answer
Step 1 
Step 2 Slope = y’ at (3, -1)
Therefore = -(33)/2(-1-1) = 9/2 = 4 1/2
Question 16
Find the intervals on which the following function is increasing and on which it is decreasing. f (x) = (x + 3)/(x - 8)
Answer
First we need to find the derivative
q16 <- function(x){(x + 3)/(x - 8)}
q16ans <- Deriv(q16)
q16ans## function (x)
## {
## .e1 <- x - 8
## (1 - (3 + x)/.e1)/.e1
## }Then we need to find the critical points, when f’(x) = 0 or undefined.
However, f’(x) is undefined when x = 8. Therefore the intervals are (-infinity, 8) and (8, +infinity)
Question 17
A frozen pizza is placed in the oven at t = 0. The function F(t) = 14 + (367t2)/(t2 + 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.
Step 2. Over time, what temperature is the pizza approaching?
Answer
Step 1
q17 <- function(t){14 + (367*t^2)/(t^2 + 100)}
q17ans <- Deriv(q17)
q17ans## function (t)
## {
## .e1 <- t^2
## .e2 <- 100 + .e1
## 367 * (t * (2 - 2 * (.e1/.e2))/.e2)
## }root17 <- uniroot.all(q17ans, c(-1000,1000))
trend1 <- q17ans(root17-1)
trend2 <- q17ans(root17)
trend3 <- q17ans(root17+1)
trend1## [1] -7.195373trend2## [1] 0trend3## [1] 7.195373Therefore, when t < 0 the temperature is decreasing (trend1 is negative) and when t > 0 the temperature is increasing (trend3 is positive)
Question 18
A study says that the package flow in the East during the month of November follows f (x) = x^3/3340000 - (7x^2)/9475 + 42417727x/1265860000 + 1/33, where 1 <= x <= 30 is the day of the month and f (x) is 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.
Answer
Question 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
Answer
Question 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
Answer
Question 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?
Question 22
A shipping company must design a closed rectangular shipping crate with a square base. The volume is 18432 ft^3. The material for the top and sides costs $3 per square foot and the material for the bottom costs $5 per square foot. Find the dimensions of the crate that will minimize the total cost of material.
Answer
Let s be the length of one side of the square base and h be the height of the crate.
Now we can find the minimum value of the total cost function
If the minimum value of s is 24 the the minimum value of h = 18432/s^2 = 32
The dimensions of the crate that will minimize the cost of material is: 24 x 24 x 32
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.
Area = xy = 1056, y = 1056/x
Exterior fencing cost = (2x + 2y)*14.40
Interior fencing cost = 2x*12
Total Cost = Cost of Interior + Exterior fencing
= 24x + 28.8x + 28.8(1056/x) = 52.8x + 30412.8x^-1
TC’(x) = 52.8 - 30412.8*x^-2
when TC’(x) = 0: -30412.8/x^2 = -52.8 x^2 = -30412.8/-52.8 x = sqrt(576) = 24
When x = 24, y = 1056/24 = 44
Therefore at x = 24 and y = 44 the dimensions of the fence in minimised.
Question 31
Solve the differential equation given below. dy/dx = x^3* y
Answer
y dy = x^3 dx
Now integrate both sides:
(y^2)2 = (x^4)/4
y^2 = 2(x^4)/4
y^2 = (x^4)/2
Linear Algebra Question
- Using matrix operations, describe the solutions for the following family of equations: x + 2y - 3z = 5 2x + y - 3z = 13 -x + y + 2z= -8
- Find the inverse of the above 3x3 (non-augmented) matrix.
- Solve for the solution using R.
- Modify the 3x3 matrix such that there exists only one non-zero variable in the solution set.
- Consider the matrix, q=matrix(c(3,1,4,4,3,3,2,3,2),nrow=3). Let b=c(1,4,5). Use Cramer’s rule and R to determine the solution, x, to qx=b, if one exists. Show all determinants.
Answer a
A <- matrix(c(1,2,-1,2,1,1,-3,-3,2),3,3)
solve(A)## [,1] [,2] [,3]
## [1,] -0.8333333 1.1666667 0.5
## [2,] 0.1666667 0.1666667 0.5
## [3,] -0.5000000 0.5000000 0.5Answer b
b = c(5,13,-8)
solve(A)%*%b## [,1]
## [1,] 7
## [2,] -1
## [3,] 0