Saayed Alam

Bridge Workshop - Math

Week 1 Assignment

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.

$f(x)=-2x^{2}+100x-800$

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

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

Q2 <- function(x){8*x^3 + 7*x^2 - 5}  
Q2(3) ## Step 1

## [1] 274

Q2(-2)  ## Step 2

## [1] -41

Q2(x+c) = 8(x+c)^3 +7(x+c)^2 - 5 ## Step 3

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

Step 1. Find $lim_{x->1^-} f(x)$
Step 2. Find $lim_{x->1^+} f(x)$
Step 3. Find $lim_{x->1} f(x)$

0 ## Step 1
-5 ## Step 2
Does Not Exist ## Step 3

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

library(Deriv)
Q4 <- function(x){(-2)*x^3}
Deriv(Q4)

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

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

Q5 <- function(x){(-8)/(x^2)}
Deriv(Q5)

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

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

Q6 <- function(x){5*x^(1/3)}
Deriv(Q6)

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

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

Q7 <- function(x){-2*x^(9/8)}
Deriv(Q7)

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

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.

library(MASS)
## The average rate of change:
y2 <- 35 
y1 <- 40 
x1 <- 4 
x2 <- 0

m <- ((y2-y1)/(x2-x1))
fractions(m)

## [1] 5/4

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

# Average cost = Total Cost / Number of Items
Q9 <- function(x){(630+2.4*x)/x}
print(Q9)

## function(x){(630+2.4*x)/x}

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

Q10 <- function(x){(-2*x^(-2)+1)*(-5*x+9)}
Deriv(Q10)

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

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

Q11 <- function(x){(5*x^(1/2)+7)/(-x^3+1)}
Deriv(Q11)

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

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

Q12 <- function(x){(3 * x^(-3) -8 * x + 6)^(4/3)}
Deriv(Q12)

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

$f'(x)=\frac{4}{3}(-9x^{-4}-8)(3x^{-3}-8x+6)^\frac{1}{2}$

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 occured. How fast is the level changing after 3 days? Round to the nearest whole number.

Q13 <- function(t){(550*t^2)/( t^2  + 15)^(1/2)}
round(Deriv(Q13)(3),0)

## [1] 547

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 $(0 \le {t} \le 14)$ 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.

Q14 <- function(t){1000*(6 + 0.1*t)^(1/2)}
round(Q14(3),0) ## Step 1

## [1] 2510

round(Deriv(Q14)(3),0) ## Step 2

## [1] 20

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

Step 1. Use implicit differentiation to find $\frac{dy}{dx}$.
Step 2. Find the slope of the tangent line at $(3, -1)$.

x <- 3
y <- -1
dydx <- (-3*x^2) / (4*y^2) ## Step 1
fractions(dydx) ## Step 2

## [1] -27/4

Find the intervals on which the following function is increasing and on which it is decreasing. $f(x) = \frac{x+3}{x-8}$.

$f'(x)=-\frac{11}{(x-8)^{2}}$ ## First Derivative
Since the denominator is squared, a positive number, $f'(x)$ will always be negative and decreasing.

A frozen pizza is placed in the oven at $t = 0$. The function $F(t)=14+\frac{367t^{2}}{t^{2}+100}$ pproximates 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?

$F'(t)=\frac{73400t}{(t^{2}+100)^{2}}$ ## First Derivative
$F'(t)=0$
$t=0$

$F''(t)=\frac{73400(-3t^{2}+100)}{(t^{2}+100)^{3}}$ ## Second Derivative

Evaluating $t=0$,
$F''(0)=\frac{73400(-3*0^{2}+100)}{(0^{2}+100)^{3}}$
$F''(0)=\frac{367}{50}$ ## Since $F''(0)>0$, $F(0)$ is a minimum with $t=0$.

Temperature increases when $F(t)>0$ for $t>0$.
Temperature decreases when $F(t)<0$ for $t<0$.

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 $1<x<30$ 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.

$f'(x)=\frac {3x^{2}}{3340000}-\frac{14x}{9475}+\frac{42417727}{1265860000}$ ## First Derivative
$f'(x)=0$
$x=23, x=\frac{1844249}{1137}$

$f''(x)=\frac {3x}{1670000}-\frac{14}{9475}$ ## Second Derivative
$f''(23)=-\frac{909049}{632930000}$ ## Since $f''(23)<0$, most packages were delivered on 23rd of November
$f(23)=0.41$ ## The maximum number of packages delievered is 0.41 million.

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

$f'(x)=14x+28$ ## First Derivative
$f'(x)=0$
$14x+28=0$
$x=-2$

$f''(x)=14$ ## Second Derivative
$f''(-2)=14$ Since $f''(-2)>0$, $x=-2, f(-2)=-63$ is the minimum.
$(-2,-63)$ ## Local Extrema

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

$f'(x)=-18x^{2}+54x+180$ ## First Derivative
$f'(x)=0$
$-18x^{2}+54x+180=0$
$x=-2, x=5$

$f''(x)=-36x+54$ ## Second Derivative
$f''(-2)=126$ Since $f''(-2)>0$, $x=-2, f(-2)=-240$ is the minimum.
$f''(5)=-126$ Since $f''(5)<0$, $x=5, f(5)=825$ is the maximum.
$(-2,204)$ and $(5,825)$ ## Local Extrema

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,
$x$ number of orders per year
$\frac{120}{x}$ lot size divided by the number of order per year
$\frac{120}{x}*1.60$ storage fees
$x*(6+4.5*\frac{120}{x})$ reorder fees
$f(x)=\frac{120}{x}*1.60+x*(6+4.5*\frac{120}{x})$ ## inventory cost function

$f'(x)=6-\frac{192}{x^{2}}$ ## First Derivative
$f'(x)=0$
$x=5.65$

$f''(x)=\frac{384}{x^{3}}$ ## Second Derivative
$f''(5.65)=607.88$
$f''(x)=0$
$x=20$ ## In 20 lot size and 6 times per year an order should be placed to minimize inventory costs

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.

Let,
$x=$ dimension for square base of the crate
$y=$ height of the crate
$3(4(x+y))+3(x^{2})$ cost of top and sides material
$5(x^{2})$ cost of bottom material
$v=18432ft^{3}$

Total cost,
$f(x)=(3(4(x+y))+3(x^{2}))+(5(x^{2}))$

Height of the crate,
$V=x*x*y$
$18432ft^{3}=x^{2}y$
$y=\frac{18432}{x^{2}}$

Total cost function in terms of $x$,
$f(x)=\frac{221184}{x}+8x^{2}$

$f'(x)=16x-\frac{221184}{x^{2}}$ ## First Derivative
$f'(x)=0$
$16x-\frac{221184}{x^{2}}=0$
$x=24$

To find y,
$y=\frac{18432}{x^{2}}$
$y=\frac{18432}{24^{2}}$
$y=32$
## To minimize the cost of material, the dimensions of the crate must be 24ft x 24ft x 32ft.

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.

Let,
$x=$ wide side of the fence
$y=$ length side of then fence
$14.4(2x+2y)$ cost of the exterior fence material
$12(2x)$ cost of the interior fence material
$A=1056yards^{2}$

Total cost,
$f(x)=(14.4(2x+2y))+(12(2x))$

Length of the pen,
$A=1056yards^{2}$
$1056=xy$
$y=\frac{1056}{x}$

Total cost function in terms of $x$,
$f(x)=52.8x+\frac{30412.8}{x}$

$f'(x)=52.8-\frac{30412.8}{x^{2}}$ ## First Derivative
$f'(x)=0$
$52.8-\frac{30412.8}{x^{2}}=0$
$x=24$

To find y,
$y=\frac{1056}{x}$
$y=\frac{1056}{24}$
$y=44$
## To minimize the cost of material, the dimensions of the rectangular must be 24 yards x 44 yards.

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,
$t_{0}=37000$ price of machine 7 years ago
$t_{1}=67000$ price of machine now

Using exponential decay formula,
$f(t_{0})=f(t_{1})*e^{kt}$
$67000=37000*e^{7k}$
$k=-0.0848$ ## decay rate of machine’s value

To find t in 9 years,
$P(t_{2})=37000*e^{9k}$
$P(t_{2})=17245.27$ ## The value of the machine 9 years from now will be $17,245.27

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.

Let,
$x=$ units of television
$p(x)=23.2-0.4x$ price of television

Since revenue = units * price,
$D(x)=23.2x-0.4x^{2}$

$D'(x)=23.2-0.8x$ ## First Derivative
$D'(x)=0$
$23.2-0.8x=0$
$x=29$
## To maximize the revenue, the level of production must be 29 units.

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.

Let,
$t_{0}=0$ cost of goods and services in year 1995
$t_{1}=11$ cost of goods and services in year 2006
$t_{2}=22$ cost of goods and services in year 2017

Using exponential growth formula,
$P(t_{1})=P(t_{0})*e^{kt}$
$426.80=400*e^{11k}$
$k=0.00589$ ## Growth rate of dollar value

To find t in year 2017,
$P(t_{2})=400*e^{0.00589*22}$
$P(t_{2})=455.34$ ## The cost of same goods and services in year 2017 is $455.34

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

Marginal profit function as given,
$P'(x)=380-4x$

Therefore, the profit function by taking integral,
$P(x)=380x-2x^{2}+C$

To find c,
$1700=380(38)-2(38)^{2}+C$
$C=-9852$

Step 1,
$P(x)=380x-2x^{2}-9852$ ## The profit function

Step 2,
$P(x)=380*56-2*56^{2}-9852$
$P=5156$ ## The profit from the sale of 56 clock radios is $5,156.00.

Use integration by substitution to solve the integral below. $\int \frac{-5(ln(y))^{3}}{y}dy$.

Using integration by substituion,
$\int f(g(x))*g'(x)dx$
$\int f(u)du$
$u=g(x)$

Set,
$u=ln(y)$
$\frac{du}{dy}=\frac{1}{y}$
$du=\frac{1}{y}dy$

Replace u,
$-5\int u^{3}du=\frac{-5}{4}u^{4}+C$
$-\frac{5}{4}ln(y)^{4}+C$ ## Final Answer

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.

To find the population function take the integral,
$P´(t)=75-9t^\frac{1}{2}$
$P(t) = 75t-6t^\frac{3}{2}+C$

Population of wild animals on the 9th year,
$P(9)=75(9)-6(9)^\frac{3}{2}$
$P(9)=513$
$P(0)=0$
$P(9)-P(0)=513$ ## The population increased by 513 wild animals during the first 9 years.

Find the area of the region bounded by the graphs of the given equations. $y=6x^{2},y=6\sqrt{x}$.

To find points of interaction, solve the equations:
$6x^{2}=6\sqrt{x}$
$x(x^{3}-1)=0$
$x=0, x=1$

Calculating definite integrals,
$\int_{0}^{1}6\sqrt{x}dx = 4$
$\int_{0}^{1}6x^{2}dx = 2$
$4-2=2$ ## The area of the region bounded by the graphs is 2.

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

$\frac{dy}{dx}=x^{3}y$
$dy=x^{3}ydx$
$\frac{dy}{y}=x^{3}dx$
$\int\frac{dy}{y}=\int x^{3}dx$
$ln(y)+C_{1}=\frac{x^{4}}{4}+C_{2}$
$ln(y)=\frac{x^{4}}{4}+C_{2}-C_{1}$
$ln(y)=\frac{x^{4}}{4}+C$
$y=e^{\frac{x^{4}}{4}}+C$
$y=e^{\frac{x^{4}}{4}+C}$ ## Answer

Use integration by parts to evaluate the definite integral below. $\int^{2}_{-7}x\sqrt{x+7}dx$. Write your answer as a fraction.

Q32 <- function(x){x*sqrt(x+7)}
Q32a <- integrate(Q32, -7, 2)
fractions(Q32a$value)

## [1] -144/5

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

Let,
$x$ number of bounces
$f(x)=46*(0.22)^{x}$

To find the sum of the finite geometric sequence:
$\sum_{x=1}^{20} 46(0.22)^{x}=12.97$

To find the vertical distance of the 20th bounce:
$46+2*12.97=71.95$ ## Step 1

To find the vertical distance of indefinite bounce:
$d(x)=46+2*\sum_{x=1}^{20} 46(0.22)^{x}=71.95$ ## Step 2

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

$P(x)=3e^{17}+15e^{17}(x-4)+\frac{72}{2}e^{17}(x-4)^{2}+\frac{125}{2}e^{17}(x-4)^{3}+\frac{625}{8}e^{17}(x-4)^{4}+\frac{628}{8}e^{17}(x-4)^{5}$