Question

Please help me I will do anything I have been stuck on this problem for honestly 3 weeks.

A local grocery store has agreed to sell your homemade bread. You will use the following information along with some ideas from Chapter 3 to decide how many loaves should be manufactured each week and what price should be charged.

After tracking weekly sales at several different prices, you get the following data:
Loaves Sold, x
Price, p
355/$1.50
320/$2.00
265/$2.50
235/$3.00
180/$3.50
125/$4.00

In order to increase manufacturing capacity, you’ve taken out a loan to buy an industrial sized oven for $4000. The new oven will allow you to make a maximum of about 400 loaves of bread per week. The loan is to be paid back monthly over two years at an annual interest rate of 10% compounded monthly. The monthly payments are $203.40. (You can check these numbers after section 5.7.) The ingredients for two loaves of bread are given in the table below. The $1.182 is the cost of the ingredients for a single loaf of bread.

Demand Equation. Make a scatter plot of the six data points (using the number sold as the x-coordinate.) Does the relationship appear to be linear? Use regression analysis to find the line of best fit. This line will be your demand equation. How strong is the correlation?

Revenue Function. Find R(x), the weekly revenue as a function of loaves sold, x. (Note that R(x) is an equation not a single value.)

Cost Function. Find C(x), the weekly cost for producing x loaves of bread. Be sure to include both the cost of the oven and the ingredients. What is the domain of the cost function?

Profit Function. Find P(x), the weekly profit for producing and selling x loaves of bread. (Hint: profit = revenue – cost.)

Maximum Revenue. Find the number of loaves that should be sold in order to maximize revenue. What is the maximum revenue? What price should be charged in order to maximize revenue?

Maximum Profit. Find the number of loaves that should be produced and sold in order to maximize the profit. What is the maximum profit? What price should be used to maximize profit?

Conclusion. How many loaves of bread will you produce each week and how much will you charge for each loaf? Why?

Answer 1

this looks like a pain

Answer 2

haha yes i have been stumped for a week, im just hoping that someone can help me out some.

Answer 3

here is a picture they wanted of the points plotted
http://www.wolframalpha.com/input/?i=%28355,1.50%29+%28320,2.00%29+%28265,2.50%29+%28235,3.00%29+%28180,3.50%29+%28125,4.00%29

Answer 4

Demand Equation: y=-0.01x+5.4287, this is what i got for the demand equation

Answer 5

here is the line
http://www.wolframalpha.com/input/?i=best+fit+line+%28355,1.50%29+%28320,2.00%29+%28265,2.50%29+%28235,3.00%29+%28180,3.50%29+%28125,4.00%29

Answer 6

\[y=5.42873-0.0108597 x \]

Answer 7

probably more decimals than you need

Answer 8

so for the revenue function would i use the demand equation to find that?

Answer 9

i actually have no idea
it is the number of loaves sold times the price per loaf

Answer 10

a quick google search provides the answer

Answer 11

$3.00 is the best price point but i need to also find the max revenue and max profit for loaves of bread and that could be any number from 1-400

Answer 12

can you post the link?

Answer 13

the demand function is what you just got, \[y=5.42873-0.0108597 x\] hold on i got confused, let me send you an example

Answer 14

oh i see

Answer 15

y in our case was the demand, x was the price
so the revenue funtion is the demand times the price, i.e. \(xy=5.42873x-0.0108597 x^2\)

Answer 16

that is \[R(x)=5.429x-0.0109x^2\] here i rounded a bit

Answer 17

or you can round like you did if you prefer
in any case the revenue is that times x

Answer 18

how we doing so far?

Answer 19

good i think that the cost function is C(x)=50.85+1.182*x

Answer 20

yeah looks good, divided the monthly cost by 4

Answer 21

yes that is what i did then for the profit function P(x)=5.429x-50.85+1.182*x right?

Answer 22

profit is revenue minus cost right?

Answer 23

yes

Answer 24

something screwy cause revenue had an \(x^2\) term in it which got lost somewhere

Answer 25

\[5.429x-0.0109x^2-(50.85+1.182x )\]

Answer 26

you will of course want to multiply out and combine like terms

Answer 27

that is for the profit right because it was literally just the revenue equation- the profit equation.

Answer 28

that is profit since it is "revenue minus cost"

Answer 29

i used wolfram to deal the the decimals and got this \[P(x)=-0.0109 x^2+4.247 x-50.85\]

Answer 30

http://www.wolframalpha.com/input/?i=5.429x-0.0109x^2-%2850.85%2B1.182x+%29

Answer 31

All right and then to find the Maximum rev and profit dont you just make that equation equal to 0?

Answer 32

oh no

Answer 33

that will tell you when you get zero profit or zero revenue
you certainly don't want that !!

Answer 34

which one do you want to find, the max of the revenue function or that max of the profit function (or both)?

Answer 35

well i have to find both of them for this project but we can do max revenue first.

Answer 36

ok the max of the revenue function is the second coordinate if the vertex
you find it by finding the first coordinate of the vertex, and plugging it it to get the second coordinate
in this case the first coordinate represents the number of loaves and the second represents the max revenue

Answer 37

the first coordinate of the vertex of \(y=ax^2+bx+c\) is \(-\frac{b}{2a}\)

Answer 38

\[R(x)=5.429x-0.0109x^2\]
\[a=-0.0109,b=5.429\] use a calculator to compute \[\frac{5.429}{2\times 0.0109}\]

Answer 39

Ok so i got 249.04 for that

Answer 40

yeah i got that too, i would go with 249 as it is loaves of bread

Answer 41

then to get the revenue, plug it in

Answer 42

i would cheat

Answer 43

so for revenue it would be 5.429(249)-0.0109x^2 right

Answer 44

scroll down to where it says "global max" then click on "approximate form" you will see that it is about 696 when x is about 249

Answer 45

no you have to replace both x's by 249

Answer 46

\[5.429(249)-0.0109\times 249^2 \]

Answer 47

oops forgot to send the link
http://www.wolframalpha.com/input/?i=5.429x-0.0109x^2

Answer 48

So i got 676 for the maximum rev but you said 696 earlier did i do something wrong?

Answer 49

maybe i misread it

Answer 50

yes, you are right and i am wrong 676

Answer 51

oh ok and then to find the Maximum profit what should i do?

Answer 52

same thing

Answer 53

but this time for the profit function instead of the revenue function
compute \(-\frac{b}{2a}\) for \[P(x)=-0.0109 x^2+4.247 x-50.85\]

Answer 54

would that be -50.85/2(-0.0109)

Answer 55

no

Answer 56

\(b=4.247\) not \(-50.85\)

Answer 57

and there is a minus sign in the formula, so the minus signs will go and the answer will be positive (good thing too)

Answer 58

\[4.247\div(2\times 0.0109)\]

Answer 59

yeah that one

Answer 60

alright for that i got 194.82

Answer 61

me too
might want to round to 195 since they are whole loaves

Answer 62

and now i have to go, hope you are goo from there
at least it did not take us three weeks to do this!

Answer 63

more than an hour, but not three weeks!

Answer 64

haha thank you so much you don't even realize how much trouble I have had with this you are a life saver.

Answer 65

yw