Simple word problem, wrong solution manual?


New member
Nov 4, 2006

Please take a look at the following word problem, it is from the Linear Programming section of my Math131 course:

A theatre is presenting a program on drinking and driving for students and parents. The proceeds will be donated to a alcohol information center. Admission is $2 for parents and $1 for students. There are two constraints; the theater can hold no more than 150 people, and every two parents must bring at least 1 student. How many parents and students should attend to raise the most money?

I know the answer is 100 parents and 50 students however I wasn't exactly sure how to put it in an algebric equation so I checked the solution manual which shows the following:

Let x = number of students
Let y = number of parents

x+y<=150 and 2x-y>=0

Answer is x+2y (100, 50) or 100 students and 50 parents.

I believe this to be wrong, that would equal $200. It should be the other way around! 100 parents and 50 students meets all the constraints and equals $250.

Can someone please confirm and give the correct formula(s). Wouldn't x+2y<=150 meet both constraints?

Thank you


Super Moderator
Staff member
Feb 4, 2004
Since the max/min points occur at corners of the feasibility region, you know from your graph that (x, y) = (100, 50) can be neither a maximum nor a minimum.

I would guess that they typoed the answer, reversing the x- and y-coordinates. The point (x, y) = (50, 100) is a corner point, and clearly provides a more optimal solution.

You might want to point out this error to your instructor, so he can pass it along to the rest of the class.



Senior Member
Feb 17, 2004
trip20 said:
Answer is x+2y (100, 50) or 100 students and 50 parents.
That's obviously "or 50 students and 100 parents"; typo perhaps?

Easy to see that maximum is made up of "2 parents + 1 student" combos,
or 3 persons for $5; 150/3 = 50; 50 * $5 = $250

By the way, minimum would be 2 parents bringing 148 students, right? :wink: