Most of the students are don’t realize what is Universal Set U
is all about ,Let’s see an example problem which explains us more ….

Question:-

Out of forty students, 14 are taking English Composition and 29 are taking Chemistry. If five students are in both classes, how many students are in neither class? How many are in either class? What is the probability that a randomly-chosen student from this group is taking only the Chemistry class?

Answer:-

This online help with math gives the solution of the above problem.
Here the total number of students in class is what is Universal Set U ,which is given as 40 students

Here is the venn diagram template for the complete question

Two students are taking neither class.
There are 38 students in at least one of the classes.
There is a 24/40 = 0.6 = 60%  probability that a randomly-chosen student in this
group is taking Chemistry but not English.

Topic : Word Problem on Determining the Revenue Function.

Question : A cable television firm presently serves 5000 households and charges $20 per month. A marketing survey indicates that each decrease of $1 in the monthly charge will result in 500 new customers. Let R(x) denote the total monthly revenue when the monthly charge is x dollars. Determine the revenue function R. find the value of x that results in maximum monthly revenue.

Answer :

Total monthly revenue = number of customers x monthly rent
for $20 rent : Total monthly revenue = 5000 x 20 = $100000
for every $1 decrease in the rent , there is an increase of 500 customers.
So , the relation is formulated as follows:

decrease of $1 : monthly rent = $(20-1) = $19
new customers = 500 = 500 x 1 ,
total number of customers = 5000 + 500 = 5500 = 5000 + 500 x 1
total monthly revenue = (5000 + 500 x 1) x (20 – 1)

decrease of $2 : monthly rent = $(20-2) = $18
new customers = 500 + 500 = 1000 = 500 x 2 ,
total number of customers = 5000 + 1000 = 6000 = 5000 + 500 x 2
total monthly revenue = (5000 + 500 x 2) x (20 – 1)

decrease of $3 : monthly rent = $(20-3) = $17
new customers = 500 + 500 + 500 = 1500 ,
total number of customers = 5000 + 1500 = 6500 = 5000 + 500 x 3
total monthly revenue = (5000 + 500 x 3) x (20 – 3)
……………………..
the relation is now : R(x) = (5000 + 500 x X) x ( 20 – x) = 500 (10 + X)(20-X)
to find the max. monthly revenue we have to find the first derivative of R(x) and equate it to 0.
d/dx ( R(x)) = 500 { (10 + X)(-1) + (1)(20 – X) } = 500 { 10 – 2X }
we should equate that to 0 .
so, 10 – 2x = 0 , so , X = 5
So , the total monthly revenue will be maximum if the monthly rent is decreased by $5.
Finally the answer is :
Revenue Function is : R(x) = (5000 + 500 x X) x ( 20 – x)
The total monthly revenue will be maximum if the monthly rent is decreased by $5 ; i.e if monthly rent is $15 the total monthly revenue will be maximum.