Help to solve this question, please

Question

A theater is presenting a program on drinking and driving for students and their parents or other responsible adults. The proceeds will be donated to a local alcohol information center. Admission is $6.00 for adults and $3.00 for students. However, this situation has two constraints: the theater can hold no more than 150 people and for every two adults, there must be at least one student. How many adults can students should attend to raise the maximum amount of money?

 

To raise the maximum amount of money, ___adults and ___students should attend.

Answer 1

Let A = number of adults

Let S = number of students

 

You want to maximize the revenue R = 6*A + 3*S subject to the constraints 

 

A + S <= 150  ==>  S <= 150 - A

(A/2) <= S

A >= 0

S >=0

 

Graph these 4 inequalities with A on the x-axis and S on the y-axis.  The graph should show that the common solution of the 4 inequalities form a triangle in the first quadrant with one vertex at the origin (0,0), one vertex at (0,150), and the third vertex where S = 150 - A and S = A/2 intersect.  Solving for this last vertex 

 

150 - A = A/2

(3/2)*A = 150

A = 100     ==>   S = 50   ==> Vertex is at (100,50)

 

The extrema (maximum and minimum) of the revenue function will occur at the vertices of the triangle, so plug those values into the revenue equation to see which is largest 

 

R(0,0) = 6*0 + 3*0 = 0 + 0 = 0

R(0,150) = 6*0 + 3*150 = 0 + 450 = 450

R(100,50) = 6*100 + 3*50 = 600 + 150 = 750  

 

The largest value is 750, so 100 adults and 50 students should attend