Q4. a fish processing company in charlotteville processes three types of products: type a, type b and type c. its operations involve three basic activities: cleaning, cutting and packaging. type a products require 4 minutes of cleaning 2 minutes of cutting and 2 minutes of packaging time. type b products require 6 minutes of cleaning 4 minutes of cutting and 2 minutes of packaging time. type c products require 8 minutes of cleaning 2 minutes of cutting and 4 minutes of packaging time.
given that the total time available for cleaning, cutting and packaging is 3.5hours,2.5 hours and 1.5 hours respectively. a. clearly defining your variables, show the model which represents the information given in this problem. state any restrictions on the variables in the problem.
q5. given that weekly demand curve of a local wine producer is p=50−0.1q, and that the total cost function is c(q)=1500+10q, where q bottles are produced each day and sold at a price of $p per unit.
a. express the weekly profit as a function of price p. b. at what price must a bottle of wine be sold to realise this maximum profit? c. what is the maximum profit that can be made by the producer? q6. a company is trying to expose as many people as possible to its new brand through social media advertising targeting 4 million possible viewers. a model for the number of people p, in millions, who are aware of the new brand after t days of advertising is estimated to be = 4(1 – −0.037) a. how many persons can they expect to meet in 1 week? b. how many days, to the nearest day, will the advertising campaign be launched that 60% of the possible viewers will be aware of the new brand?

Q4. Product types A , B , C ≥ = 0

Step-by-step explanation:

Defining variables

let type A product = a

type B = b

type C = c

Model for the information

For cleaning, the total time is

4a + 6b + 8c = 3.5

For cutting, the total time is

2a + 4b + 2c = 2.5

For packaging, the total time is

2a + 2b + 4c = 1.5

The restrictions are

x , y , z >= 0

Q5.

Price per unit is defined as p=50−0.1q and revenue R(q)= qp(q)

R(q)= q(50-0.1q)                                                      Equation 1

R(q)= 50q−0.1q²                                                       Equation 2

The cost function as provided is

C(q)=1500+10q                                                     Equation 3

The profit function is given by

P(q)= R(q)- C(q)                                                     Equation 4 (a)

      ={50q−0.1q²} - {1500+10q}

      =50q−0.1q² - 1500-10q

      =40q−0.1q² - 1500                                                      Equation 4(b)

Since we already have a price function as  p=50−0.1q then multiplying each side by 10 and re-arranging and making q the subject we have

q=500-10p

Equation 4(b) can now be written while substituting q with 500-10p hence we obtain equation 5

40(500-10p)−0.1(500-10p)² - 150

20000-400p-0.1(250000-10000p+100p²)-1500

20000-400p-25000+1000p-10p²-1500

-6500+600p-10p²                                                    Equation 5

Therefore, the weekly profit is expressed as -10p²+600p-6500

b) Equation 5 is a parabolic equation which when compared with typical ax²+bx+c=0 hence in our case a=-10, b=600 and c=-6500 from equation 5

The maximum point is at

Therefore, our

=

Therefore, our vertex is (30, 2500)

The price of a bottle to realize maximum profit is therefore $30

c)

As already shown in equations for part b, the maximum profit is $2500

Q6.

Since P is given as 4(1 – (Exp)-0.037t) then in one week t is 7 days for 1 week

hence P will be 4(1 – (Exp)-0.037*7)= 0.912 million

b)

The target is 4 million hence to get 60% means 60/100 of 4 million =2.4 million people

Substituting 2.4 where there's P our equation is

2.4= 4(1 – (Exp)-0.037t)

Dividing both sides of the above equation by 4 we obtain

0.6= (1 – (Exp)-0.037t)

Rearranging the above equation we obtain

(Exp)-0.037t= 1-0.6

(Exp)-0.037t= 0.4

t=24.76 days

t= 25 days the the nearest day

Keywords:

social media advertising targeting, advertising campaign, possible viewers, demand curve, Charlotteville

step-by-step explanation:

30% of $420 is 0.30($420) = $126 and $420 + $126 = $546.

or, 130% of $420 is 1.3($420) = $546.

notice that this is not the $600 original list price with which we started.

why doesn't this work?

the method does not work because 30% of $420 is not the same amount as the 30% of $600. it may seem obvious when stated this way, but it is a very common error made on problems of this type. we cannot just use the same percent and get back to the original amount because we are taking that percent of a different value.

to work problems of this type correctly, it is generally best to write an equation for the situation and then solve.

example: since (100% – 30%)(original price) = (sale price)

70% of the   (original price) is $420.

0.7x   = 420

equation image indicator

x = 600

the original cost was $600.

we may also solve this problem using the two-step method. working the problem in two steps requires us to use more algebra skills. we have to know that x = 1x (identity property of multiplicationtext annotation indicator) and that 1x – 0.3x = 0.7x (subtracting like terms). then we solve the problem as follows:

(original price) – 30%(original price) = sale price

x – 0.3x = 420

0.7x = $420

equation image indicator

x = 600

the original cost was $600.  

note that with both of the above methods we do obtain the correct original amount.

example: after a 5% pay raise, hermione is earning $22,680 per year. what was she earning before the pay raise?

105% of (original salary) = current salary

1.05x = 22,680

equation image indicator

x = 21,600

hermione earned $21,600 per year before the pay raise.

this is an example so maybe you will get how to at least do it!