Blog #3-Mike Derouin

I chose option 3 to make up my own hypothetical company that sells surfboards. The name of my company is Oahu Surf Co. and their target demographic is younger people aged 18-25 that are adventurous and enjoy surfing. We focus on intermediate level surfers that are focused on new and cool products that want to have an edge. The company was founded in late 2014 and is new to the industry. The company is currently focused on making the highest quality and best surfboard on the market, however we would like to expand over the next few years to include more surfing and extreme sports products. Oahu Surf Co. has a large presence in Hawaii, however our boards are manufactured in Pennsylvania.

A. Fixed Costs (In dollars per month):

            Equipment: 75,000

            Admin Costs: 20,000

            Overhead (Rent and Utilities): 15,000

            Total Fixed Costs=$110,000 per month

B. Variable Costs (In dollars per unit):

            Raw Materials: 105

            Labor: 20

            Shipping: 30

            Total Variable Costs per unit=$155

C. Cost Function:

            C(q)  = 110,000 + 155q

D. Price per Unit:

            =$750

            R(q) = 750q

E. Profit Function:

            P(q) = R(q) – C(q)

            P(q) = 750q – (110,000 + 155q)

            P(q) = 595q – 110,000

F. Breakeven Point:

            R(q) = C(q)

            750q = 110,000+155q

            750q - 155q = 110,00

            q = 110,000/595

            q = 185 units to break even

G.

H. Interpretation of the Cost & Revenue Function Graph

The breakeven point is the point where revenue=expenses. At this point, Oahu Surf Co. is neither making or losing money. This can be seen on the graph where the x value, which is the number of surfboards sold, equals 185 and the y value, which is total revenue of the sold surfboards, equals $138,500. Both the revenue function and the cost function have linear slopes.








I.

J. Interpretation of the Profit Function Graph


The profit function has a linear slope, because it demonstrates the direct relationship between number of sold surfboards and the revenue being made from the sold surfboards. It begins below the x axis to demonstrate a loss until enough units are sold that revenue is higher than expenses. This is shown by the break even point at 185 units. At that point revenue and expenses will be equal to one another.








K. Marginal Cost:

            Daily Units = 25 surfboards

            C(q) = 110,000+155q

            C’(q) = 155

The marginal cost will remain the same for every additional unit produced. For the 25th unit produced the marginal cost will remain $155 per unit.

L. Average Cost:

            A(q) = C(q) / q 

            A(q) =110,00 + 155*25 /25 = 29.39

            

M. Marginal Revenue

           R(q)=750-155

            = 595

            

1) The marginal revenue will be higher at q = n because both marginal revenue and marginal cost are constant.

2) The number of units sold on a daily basis will be below the break even point, because the break even point   takes into account all of the costs associated with the production of the surfboards. These costs are typically measured on yearly or quarterly basis, with this being said, the value the break even point does not matter on a daily basis.

3) An increase of one unit more unit per day will lead to an increase in profits for the company.

R(51)=595*51=30,345

C(51)=110,000+155(51)=117,905

Although it would appear that we are still losing money by increasing sales by one unit per day, this is just temporary. As each day passes we will be able to make more money with fixed costs remaining the same for the entire year. With this being said each additional unit produced will keep the fixed costs the same and we will make more money off of each unit produced. We should want to sell as many surfboards as possible.

4) An increase in production will decrease the average cost of producing each surfboard for Oahu Surf Co.

5) Decreasing the average cost of each surfboard will definitely be beneficial to Oahu Surf Co. because the less each board costs the more money the company will make off of each board.

Analysis for the future of Oahu Surf Co.

Based on the information gathered for this experiment I believe that Oahu Surf Co. has the ability to do very well over the next five years. Their fixed costs are relatively low as are their variable costs needed to produce their boards. Looking at the numbers Oahu Surf Co. is doing I believe the outlook is very positive. If they continue to sell 25 surfboards a day (which is a relatively conservative estimate) they would bring $30,000 in revenue a day with fixed costs at $110,000. This would leave plenty of room to make a profit. As long as they are able to market their products well and continue to evolve and diversify as a company, I believe they could do very well. I have also found that as the economy continues to improve more people have the ability to spend money on hobbies and sporting goods. So as long as the economy remains strong, the company continues to be a hip and likable brand, and they continue to focus on their target demographic Oahu Surf Co. has the ability to become a very successful and widely recognized brand in the young, extreme sports industry.

Blog #3 Paris Maragkos

Hypothetical Start up Company

Greek Gelato is a startup ice cream business owned by Maragkos brothers and currently operating only in the United States.  The ambitious brothers decided to start up this company and invested a lot of time, effort and money into it. Greek Gelato's fixed costs included monthly rent of $10,000, utilities of $2,000 and wages of $8,000. The company offered only one size of ice cream cup, at a cost of $0.50/ unit and the cost of the ice cream was $0.75/cup. The also, Greek Gelato was selling their ice cream for $4.50 per cup.

Part 2
Fixed Costs of the company:  10,000+2,000+8000 = $20,000
Variable Costs : 0.50+0.75 = $1.25/ice cream cup
Selling Price : $4.50 per unit
Cost function : C(q) = 1.25*q + 20,000
Revenue function : R(q) = 4.5*q
Profit function : P(q) = R(q) - C(q) = (4.5*q) - (1.25*q + 20,000)
Break Even Point : It would be where R(q)=C(q). That point is q=6,154 and $27,693

Break even point is the point where the cost of operating equals the revenue of the company, meaning that R(q) = C(q). That explains why that point is where the two function intercept. The slope of the revenue function has to do with the fact that the company earns $4.5 for every additional unit they sell. Likewise, the slope and the steepness of the cost function have to do with the fact that the company needs to spend $1.5 for every additional unit of ice cream they produce.

The break even point on the profit graph is the point where the profit function touches the x axis. At that point, the company has covered its expenses and any quantity above it will generate profits.
The graph starts at -20,000, because these are the company's fixed costs and then concaves down until the company starts to generate a few revenues out of its operations and then the graph starts concaving up as the company covers more and more of its variable expenses. When the graph hits the x-axis, the company has generate enough revenues to covers its costs and from now on any quantity will generate profits.






Part 3

Ice Creams produced every day: q=1,000

MC = C'(q)
MC = $1,25/ additional unit

C(1,000) = 1.25*(1,000) + 20,000
C(1,000) = 21,250

a (1,000) = 21,250/1,000 = $21,25

1) At q=1000 MR>MC, but the company is still operating at a loss area.
2) It is before the break even point, which means that the company should keep producing more units in order to meet it and begin earning profits.
3)Yes, the company would still make profits since R(1000+1)= $4,504.5 and C(1000+1)-C(1000)= 21,251.25 - 21,250 = $1.25
4) Since MC<AC then an increase in production would decrease the AC
5) A decrease in the AC would be better for the company because it will require them to pay less. This would be the case up to the point where MC=AC, because after an increase in production will result an increase in the company's costs.

Part 4

The company will do really good in the next 5 years because they will be operating in a profit area. As I showed earlier, the company will need approximately 7 days of the month in order to break even, and after that point they will constantly generating profits. Even if they increase their production, their revenues will exceed their costs and maybe a future expansion will be possible

Blog Post Three

Blog Post 3

Marley Kraft

Applied Calculus

March 30^th 2015

Option C;

Standard Micro

Standard Micro, a company founded in 2014, is a producer of microprocessors built for rockets and long-range ballistic missiles. Standard Micro sells primarily to governments and private space programs such as Space X. The two Killow Brothers, who had previously worked in the microprocessor division of Boeing before breaking off to start Standard Micro, founded the company. Standard Micro is incorporated in Delaware, but holds primary and factory operations in Alexandria, VA.

Fixed Costs ($);

            Property Plant and Equipment; 350,000

            Administrative Costs; 273,000

            Overhead; 100,000

Variable Costs ($);

            Raw Materials Per Unit; 2000

            Labor Per Unit; 1000

            Shipping Per Unit; 50

Cost Function;

            C(q)  = 723,000 + 3050q

Price per Unit; 9000

            R(q) = 9000q

Profit Function;

            P(q) = R(q) – C(q)

            P(q) = 9000q – (723,000 + 3050q)

            P(q) = 5950q – 723,000

Breakeven Point;

            R(q) = C(q)

            9000q = 723,000 + 3050q

            9000q - 3050q = 723,000

            q = 723,000/5950

            q = 121.512605 units to break even

Graph Interpretation;

            The Cost/Rev graph intersects at the break even point and from that point continue and further separate showing where profit is made. The Cost function slope is a constant meaning the marginal cost is similarly constant, and it that starts at the fixed costs of 723,000. The Revenue function starts at the original and takes off rapidly, with a constant marginal revenue/slope of 9000.

           

The Profit Function Graph;

            The Profit function graph begins at the origin point in this graph, though I have made the origin equal to the breakeven point this is too show that there is no profit below this quantity.

Marginal Cost;

            Daily Units = 50

            C(q) = 723000 + 3050q

            C’(q) = 3050

            Marginal Cost is constant, to produce the 50^th unit costs 3050

Average Cost

            A(q) = C(q) / q

            A(q) = 723000 + 3050*50 / 50 = 17510

           

Marginal Revenue  = 9000

            

Marginal Revenue is higher at q = n because both marginal revenue and marginal cost are constant.

Daily Units; The number of units sold daily is below the break even point but that only really matters if you look at it narrowly focused on a day by day basis. In reality fixed costs are, yearly, which means the cost function and the break-even point derived from it, are stated on a yearly basis. In three days time the company will produce/sell nearly 30 more units total than its break even point(150 – 121  = 29), so in reality the fact that we produce less than our breakeven point in a day is pointless, by the third day of the year we have surpassed our breakeven point and producing a net profit.

Increased By One Unit;

            

Regardless of the math I can tell you with certainty in the long run the company will continue to make money.

            R(51) = 9000*51

            R(51) = 459,000

            C(51) = 723,000 + 3050(51) = 878,550

Again in the short run it appears we’ve lost money and are only continuing to lose money, but because we only pay our fixed costs once per year, in three days time we will be making profit.

Increase in Production at q = n;

When marginal cost is less than Average cost increasing production decreases average cost, thus because that is the case here, increasing production will decrease average cost.

            

Decreasing average cost is always a good thing, the less we pay per unit the easier it is to produce more units, and thus the cheaper it is to produce those units and so on and so forth.

Future Analysis;

In the next five years, fixed costs will only amount to approximately 3 million dollars, while revenues will be over 820 million and with variable costs of around 278 million that’s nearly 539 million dollars in profit. This is of course all assuming that the business continues to sell and produce 50 units a day, even at average for the year. Large spikes in business or losses in business could throw a company with such high costs into turmoil, but with a solid profit margin if the company is able to survive more than a few years it will have profit to reinvest or hold incase of any damages.

Blog #3

C) I will be creating a hypothetical business model for a company that produces and sells tea. The company will be called Boston Tea Company (BTC).

Boston Tea Company is an American tea company that was founded at the end of 1773, a day after the Boston Tea Party, when its founder ran off with a few boxes of tea as opposed to throwing them into Boston Harbor. BTC sells different varieties of tea to the tea drinkers of America. They mostly sell tea to older people, however are looking into penetrating the market of younger Americans to have them drinking tea instead of coffee.

1- Fixed costs for BTC are warehouse rent ($5,000/month), utilities($1,000/month), salaries to executive employees ($10,000/month) and Guard dog up keep ($500/month)
2- Variable costs- tea purchases ($10 per pound).
3-  Tea sells for $25 a pound.
4- Cost function= 10q+ 16,500
5- Revenue function- R(q)= 25q
6- Profit function = P(q)= 25q- 10q+16,500
7- The break even point is- 1,100 units and $27,500
8-

Blog 2

Part A:

a.     While rafting at a falls, the speed of the boat was measured at 3-s intervals for 30 s.

b.     In the last summer vacation, I with 4 of my friends went for rafting at a falls. At that time, I measured the speed of our boat at a regular interval when it was sliding at the downward slope of the falls. The data was tabulated as follows. Find the acceleration of the boat at exactly 18 sec.

c.     The data of the speed of the boat

Time, t(s)	Velocity, V(cms-1)
0	1
3	1.906
6	3.6328
9	6.9240
12	13.1971
15	25.1536
18	47.9424
21	91.3776
24	174.1645
27	236.96
30	331.70

d.     The graph

The graphs shows the velocity of the boat at specific time. Three secants are drawn for three different intervals. The tangent shows the change of the velocity at the specific time 18 s.

e.     ARC calculation: For ARC calculation at the time 18 s, at least three secants are considered. These are, the slope between (18,21), (18,24) and (18,27).

The slope at (18, 21) = (91.3776 -47.9424) / (21-18) = 14.48

The slope at (18, 24) = (174.1645 -47.9424) / (24-18) = 19.37

The slope at (18, 27) = (236.96 -47.9424) / (27-18) = 21.002

The average change of the velocity is represented by the slope of the secants. In this application, it represents the average change of the velocity of the boat which is defined by the average acceleration of the boat. The unit of the slope is cm/s²

For the shorter time, the slope is highest. The average of the three slope values is 18.28 which is almost equal to the second slope. It means at this time the change of velocity, which is defined by the acceleration, is almost constant.

f.      Tangent line.

The graphs shows the velocity of the boat at specific time. The tangent shows the change of the velocity at the specific time 18 s.

Two points on the tangent line has been considered. These are

P(15, 11.1) and Q(24, 121.62)

The slope of the tangent is: (121.62 – 11.1)/(24-15) = 12.28

From this tangent, it represents the exact acceleration at the specific time 18 s. This is the instantaneous acceleration of the boat.

g.     Since the tangent considers only the smallest time at a specific point, the slope, rate of change of velocity, is the spontaneous acceleration. This is the actual acceleration. The unit is cm/s²

Appendix:

Matlab code to draw secant and tangent.

A.     Secant Draw

clear all; clc; close all;

x=0:3:30;

y=[1,1.9059,3.63278,6.924,13.1971,25.1535,47.9423,91.3775,164.1645,236.96,331.70];

format long;

idx = input('Enter index of X:');

plot(x,y,'.','LineWidth', 8); hold on

n = 10;

[p,~,mu] = polyfit(x,y,n); % degree 10 polynomial fit, centered and scaled

%

% Plot the fit with the data

xfit = linspace(min(x),max(x),100);

yfit = polyval(p,xfit,[],mu);

plot(xfit,yfit,'g','LineWidth', 1.5)

%% Slope at a given point

m=(y(idx+1)-y(idx))/(x(idx+1)-x(idx)) %Equation of forward difference

m1=(y(idx+2)-y(idx))/(x(idx+2)-x(idx)) %Equation of backward difference

m2=(y(idx+3)-y(idx))/(x(idx+3)-x(idx)) %Average of both, also equation of secant

% x(idx)

%%

ave = (m+m1+m2-18)/3;

fprintf('Equation of tangent \n')

fprintf('q = %4.2f*(p-%4.2f)+%4.2f \n',ave,x(idx),y(idx))

p = 12:0.01:30;

q = ave*(p-x(idx))+ y(idx);

hold on;

plot(p,q,'r', 'LineWidth', 2); grid on;

grid on; title('Veocity vs. time for the boat'); hold on;

plot(x,y,'x', 'LineWidth', 4);

xlabel('time, t(s)'); ylabel('Velocity, V (cm/s)');

p = 18:0.01:21;

q = m*(p-x(idx))+ y(idx);

hold on;

plot(p,q,'b', 'LineWidth', 1.5); grid on; hold on;

p = 18:0.01:24;

q = m1*(p-x(idx))+ y(idx);

hold on;

plot(p,q,'k', 'LineWidth', 1.5); grid on; hold on;

p = 18:0.01:27;

q = m2*(p-x(idx))+ y(idx);

hold on;

plot(p,q,'y', 'LineWidth', 2); grid on; hold on;

legend ('Measured Velocity','Poly Fit line','Tangent at Point of Interest', 'Measured Velocity (bold)', 'Secant 1', 'Secant 2', 'Secant 3')

%% Slope at a given point

m=(y(idx+1)-y(idx))/(x(idx+1)-x(idx)) %Equation of forward difference

m1=(y(idx)-y(idx-1))/(x(idx)-x(idx-1)) %Equation of backward difference

m2=(y(idx+1)-y(idx-1))/(x(idx+1)-x(idx-1)) %Average of both, also equation of secant

B.     Tangent Draw

clear all; clc; close all;

x=0:3:30;

y=[1,1.9059,3.63278,6.924,13.1971,25.1535,47.9423,91.3775,164.1645,236.96,331.70];

format long;

idx = input('Enter index of X:');

plot(x,y,'.','LineWidth', 8); hold on

n = 10;

[p,~,mu] = polyfit(x,y,n); % degree 10 polynomial fit, centered and scaled

%

% Plot the fit with the data

xfit = linspace(min(x),max(x),100);

yfit = polyval(p,xfit,[],mu);

plot(xfit,yfit,'g','LineWidth', 1.5)

%% Slope at a given point

m=(y(idx+1)-y(idx))/(x(idx+1)-x(idx)) %Equation of forward difference

m1=(y(idx+2)-y(idx))/(x(idx+2)-x(idx)) %Equation of backward difference

m2=(y(idx+3)-y(idx))/(x(idx+3)-x(idx)) %Average of both, also equation of secant

% x(idx)

%%

ave = (m+m1+m2-18)/3;

fprintf('Equation of tangent \n')

fprintf('q = %4.2f*(p-%4.2f)+%4.2f \n',ave,x(idx),y(idx))

p = 12:0.01:30;

q = ave*(p-x(idx))+ y(idx);

hold on;

plot(p,q,'r', 'LineWidth', 2); grid on;

grid on; title('Veocity vs. time for the boat'); hold on;

plot(x,y,'x', 'LineWidth', 4);

xlabel('time, t(s)'); ylabel('Velocity, V (cm/s)');

idx=6;

p1=15; q1=12.28*(p1-18)+ 47.94; hold on;

plot(p1,q1,'x','LineWidth', 6)

text(p1+2,q1-2,['P(' num2str(p1) ',' num2str(q1) ')']);

idx=9;

p2=x(idx); q2=12.28*(p2-18)+ 47.94; hold on;

plot(p2,q2,'x','LineWidth', 6)

text(p2+2,q2-2,['Q(' num2str(p2) ',' num2str(q2) ')']);