{r setup, include=FALSE} knitr::opts_chunk$set(echo = TRUE)
Test Name: homework1(Test)
Submitted by Sudhan Maharjan
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. Here: Cost price per doll = $10.00 Price per doll to be sold = x Profit per doll = x - $10.00 Number of dolls sold per week = 80 - 2x
Profit = x * (number of dolls sold in a week) - 10 * (number of dolls sold in a week) Therefore, the function will be p(x) = x * (80- 2x) - 10 * (80 - 2x) p(x) = 80x - 2x2 - 800 + 20x p(x) = -2x2 + 100x - 800
Given the following function: f (x) = 8x^3 + 7x^2 - 5
Step 1. Find f (3).
Ans: = 8 * 33 + 7 * 32 - 5 = 216 + 63 - 5 = 274
Step 2. Find f (-2).
Ans: = 8 * (-2)3 + 7 * (-2)2 - 5 = -64 + 28 - 5 = -41
Step 3. Find f (x + c). Ans: = 8 * (x + c)3 = 7 * (x + c)2 - 5 = 8x3 + 7x2 +24cx2 + 24c2x +8c3 + 14cx + 7c2 - 5
Step 1. Find (lim)???(???x???1???^- )a??? f (x)???.
Ans : (lim)???(???x???1???^- )a??? f (x)??? = 2
Step 2. Find (lim)???(???x???1???^+ )a??? f (x)???.
Ans: (lim)???(???x???1???^+ )a??? f (x)??? = -5
Step 3. Find (lim)???(x???1)a??? f (x)???.
Ans: (lim)???(x???1)a??? f (x)???
The limit does not exist.
Find the derivative for the following function. f (x) = -2x^3 Ans:
f’ (x) = -6x^2
Find the derivative for the following function. f (x) = (-8)/x^2
Ans: f’ (x) = 16/x^3
Ans: g’(x) = 5/???3x???^(2/3)
Find the derivative for the following function. y = -2x^(9/8) Ans: y^’= -(9x^(1/8))/4
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.
Here, x1 = 0, y1 = 40
x2 = 4, y2 = 35
m = (y2-y1)/(x2-x1)
=(35-40)/(4-0) = - 5/4The cost of producing x baskets is given by C(x) = 630 + 2.4x. Determine the average cost function.
Here, Cost of producing x baskets C(x) = 630 + 2.4x
Average Cost A(x) = (Total Cost)/(Number of Units)
= (630+2.4x)/xUse the Product Rule or Quotient Rule to find the derivative.Ans: f^’ (x) = -10x-2 +36x-3 -5
Use the Product Rule or Quotient Rule to find the derivative.f (x) = (5x^(1/2) + 7)/(-x^3 + 1)
Ans: f^' (x) = (25x^3+42x^(5/2) + 5)/(2x^(1/2) ???(x^3- 1)???^2 )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)
Ans: f' (x) = 4/3 (-9x-4 - 8) (3x^(-3) - 8x + 6)^(1/3) 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. Here, f (t) = (550t^2)/???( &t^2 + 15)
t = 3 Now, f(3) =(550 *3^2)/???( &3^2 + 15) = 4950/4.9 = 1010.20
The answer is 1010.
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. Here, N(t) = 1000(6 + 0.1t)^(1/2) Since, t = 3 N(3) = 1000(6 + 0.13)^(1/2) = 1000 2.51 = 2510
Step 2. Determine N ´(5) and interpret its meaning. Round your answer to the nearest whole number.
Here, For N(t)= 1000(6 + 0.1t)^(1/2) N’(5)= 19.60
Ans: The answer is 20. Since, the number is positive and increasing, the attendance is also increasing.
Consider the following function:Step 1. Use implicit differentiation to finddy/dx.
Here, 33x2 +34y2 *dy/dx = 0
dy/dx=(-3x^2)/( 4y^2 )Step 2. Find the slope of the tangent line at (3,-1).
Here, x = 3, y = -1 slope = (-3x^2)/( 4y^2 ) = (-3???3???^2)/( 4???-1???^2 )
= (-27)/( 4)Find the intervals on which the following function is increasing and on which it is decreasing.Here, f^’ (x)=- 11/???(x-8)???^2
In f^' (x) when x < 8, the result is negative; f(x) will be decreasing for x < 0.
In f^' (x) when x > 8, the result is negative; f(x) will be decreasing for x > 0.
In f^' (x) when x = 8, the result is undefined because something divided by 0 is undefined. A frozen pizza is placed in the oven at t = 0. The function F(t) = 14 + (367t^2)/(t^2 + 100) approximates the temperature (in degrees Fahrenheit) of the pizza at time . 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.
Ans: Here, When F’(t) = 0, we get t = 0, 0 = 14 + (367t2)/(t2 + 100)
Again F''(t) = (73400 (-3t^2+100))/???(t^2+1000)???^3
When t = 0,
We get F''(t) = 367/50Therefore, when temperature increases F(t)>0 and t > 0. when temperature decreases F(t)<0 and t < 0.
Step 2. Over time, what temperature is the pizza approaching?
Here,
We get check the temperature by using the limit.lim???(t??????)a???(14 + (367t2)/(t2 + 100))^ ???= 381 Therefore, the temperature of the pizza is 381 degree Fahrenheit.
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. Here, Lets choose x = 3 and x = 25.
f (x) = x^3/3340000 - (7x^2)/9475 + 42417727x/1265860000 + 1/33
f '(x) = ???3x???^2/3340000 - (14x^ )/9475 + 42417727/1265860000 When x = 3, f^‘’ (x) = 3x/1670000- 14/9475
f^’‘(3) = -0.001471 When x = 25, f^’‘(x) = 3x/1670000- 14/9475 f^’’ (25) = -0.001432 Since, x = 25 has the greater value than x = 3 The most packages made in November is Day 25th.
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 Here, First Derivative: f’(x) = 14x + 28 When f ’(x) =0 0=14*x+28 x= -2 Second Derivative: f ‘(x) =14 f’(-2) =14 Since, f ’(-2) =0 we substitute x= -2 on f (x) = 7x^2 + 28x - 35 f (-2) = -63 which is the minimum.
Therefore, the answer is Local Extrema = (-2, -63).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
First Derivative: f^‘(x) = -18x^2 + 54x+180 When f’(x) = 0; 0 = -18x^2 + 54x+180 x=5 or x= -2 Second Derivative: f ‘’(x) = 36x + 54 Since x = -2 or x = 5 f’‘(-2) = 126 Since f’‘(-2) > 0, we substitute x = -2 on f (x) = -6x^3 + 27x^2 + 180x f (-2) = -204 is the minimum f’‘(5) = 126 Since f’’(5) > 0, we substitute x = 5 on f (x) = -6x^3 + 27x^2 + 180x f (5) = 825 is the maximum Therefore, Local Extrema is (-2, 204) and (5, 825).
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? Now, f (x) = Storage fees + Reorder fees f (x) = Lot size X 1.60 + Number of orders per year X (6 + 4.5 * (Lot size)) f (x) = 120/x * 1.60 + x(6+4.5 120/x)
Now, f^’ (x)= (-192)/(???-x???^2 )+6
When f^’ (x)=0
0= (-192)/(???-x???^2 )+6
x= 5.65 or x= -5.65 Since x cannot be negative x=5.65
f^’’ (x)= 384/(x^3 )
Now, If f^‘’ (x)=5.65, we get f^’‘(5.65)=607.88, then f^’‘(5.65)>0, therefore x=5.65 Since the order cannot be decimal we will round the value X = 6. f^’’ (6)=608 Therefore, when we place 6 orders during the year with 20 items per each lot, the store will minimize the inventory cost.
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 of square base of the crate. y = height of the crate Volume=18432 ft3 Cost of Material for sides + Top = 3 (4(x ??? y)) + 3(x ??? x)
Cost of Material for base = 5(x ??? x)
The Total Cost = (Cost of Material for sides + Top) + (Cost of Material for base) C(x) = (3 (4(x ??? y)) + 3(x ??? x)) + 5(x ??? x)
Volume: V = x ??? x ??? y
By solving the equation in terms of y, we obtain: 18432 = x2y 18432/(x^2 ) = y
Now, cost function in terms of x: C(x) = (3(4(x ( 18432/(x^2 ) ))) + 3(x * x )) + (5(x * x))
C(x) = 221184/(x ) + 8x^2
Now, by calculating first and second derivatives, we get: C’(x) = 16x- 221184/(x^2 )
C’’(x) = 16+ 442368/(x^3 )
We get the following results: x = 24 and y = 32
Answer: The minimum cost can be achieved by having a square base of 24 ft. per side and a height of 32 ft.
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 rectangle y = length of the rectangle Area = 1056 yards2 Cost of Material for the exterior fence = 14.40(y + x + y + x) Cost of Material for the interior fence = 12(x + x)
Total Cost for the fence C(x) = (Cost of Material for the exterior fence) + (Cost of Material for the interior fence) C(x) = 14.40(2x + 2y) + 12(2x) Now, Area = x * y
Solving the equation in terms of y, we get: 1056 = x ??? y y = 1056/x Again, Cost function in terms of x will be: C (x) = 14.40(2x + 2( 1056/(x ) ) + 12(2x) C (x) = 52.8x + 30412.8/(x ) Now, by calculating first and second derivatives, we get: x = 24 and y = 44 Answer: The minimum cost of material can be achieved by having x = 24 yards and y = 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.After equation solving:
f^’ (t)= 67000*( 37/(67 )) (t/7)/ f(16)= 17244.50
The value of the machine after 9 years will be $17244.50.
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. Since, Revenue = units * Price R(x) = x * (23.2???0.4x) R(x) = 23.2x???0.4x2 Now, by calculating first and second derivatives, we get: x = 29 In order 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: January 1, 1995 t = 0, t in years. January 1, 2006 t = 11, t in years. January 1, 2017 t = 22, t in years. Since we have (t1, f(t1)), (t2,f(t2)): (0, 400), (11, 426.80); we can find y = abt
After solving equation: f(t)= 400*( (426..08)/(400 )) (t/11)/ Again, when t = 22, we get, f(22)=455.40 Therefore, the value of the machine after 22 years will be $455.40.
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. Here, We have marginal profit function: P’(x) = 380???4x; We can get profit function P(x) = 380x ??? 2x2 + C When x = 38, P(x) = 1700 P(x) = 380x ??? 2x2 + C 1700 = 380(38) ??? 2(38)2 + C C = -9852 Therefore, the profit function will be: P(x) = 380x ??? 2x2 ??? 9852
Step 2. What is the profit from the sale of 56 clock radios? x = 56 P(x) = 380x ??? 2x2 ??? 9852 The profit for selling 56 clock radios will be 5156. 28. Use integration by substitution to solve the integral below.
????????? (-5(lna(y) )^3)/y???dy
Here, By applying integral substation we get: ?????????f(g(x))g^’ (x)dx= ??? ?????????f(u)du ??? having u=g(x) u = lna???(y)du/dy???= 1/y
du = 1/ydy -5?????????u^3 du= (-5)/4 u^4+C??? After reverse substitution we have: (-5)/4 ???ln(y)???^4+C
It was discovered that after t years a certain population of wild animals will increase at a rate of P´(t) = 75 - 9t^(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. Now, t = 9 and t = 0 P(9) - P(0) We get 513 Therefore, the population increase in 9 years is 513 animals.
Find the area of the region bounded by the graphs of the given equations.Enter your answer below. Here: 6x2 = 6???x x2 = ???x x4 = x x(x3 - 1) = 0 x = 0 and x = 1 Now, ???_0^1??????6???xdx??? ???_01??????6x2 dx??? = 2
The area for the bounded region is 2.
Solve the differential equation given below.dy/dx = x^3 y
dy = x^3 ydx
dy/y = x^3 dx
??????dy/y= ?????????x^3 dx???
ln(y) + C1 = x2/4 + C2
y = e^(x^4/4) + C1Use integration by parts to evaluate the definite integral below.???_(-7)^2?????? x???( &x + 7) dx???
Write your answer as a fraction.
???_(-7)^2?????? x???( &x + 7) dx??? = -144/5 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?
x = number of bounces f(x) = 46 * (0.22)x ???_(x=1)^20??????f(x)???
???_(x=1)20??????46(???0.22)???x )= ??? 12.97 Since it bounces up and down, the distance travelled should be multiplied by 2. Distance travelled = 46 + 2 * 12.97 = 71.95
Step 2. If we assume it bounces indefinitely, what is the total vertical distance traveled?
Find the Taylor polynomial of degree 5 near x = 4 for the following function.P(x) = 3e^17 + 15e^17 ???(x-4)???^+ 36e^17 ???(x-4)???^2+ 125/2 e^17 ???(x-4)???^3+625/8 e^17 ???(x-4)???^4+ 625/8 e^17 ???(x-4)???^5
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