SOLUTION: A factory makes two types of beds, type A and type B. Each month, a number a number of type A and a number of type B are produced. The following constraints control monthly product

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Question 1198959: A factory makes two types of beds, type A and type B. Each month, a number a number of type A and a number of type B are produced. The following constraints control monthly production:
No more than 50 beds of Type A and no more than 40 beds of type B can be made.
At least 60 beds in all must be made.
The maximum number of beds that can be produced is 80.
The profit on type A is Php300 and on type B is Php150. How many beds on both types must be produced to maximize the profit? What is the minimum profit?
Found 3 solutions by math_tutor2020, greenestamps, ikleyn:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Given Facts:
  1. No more than 50 beds of Type A can be made.
  2. No more than 40 beds of type B can be made.
  3. At least 60 beds in all must be made.
  4. The maximum number of beds that can be produced is 80.
  5. The profit on type A is Php300.
  6. The profit on type B is Php150.
x = number of beds of type A
y = number of beds of type B
where x,y are nonnegative integers

Fact 1 leads to the inequality
Since x is nonnegative, we can further clarify that (i.e. x is between 0 and 50 inclusive of each endpoint)
x is in the set {0,1,2,...,49,50}

Fact 2 leads to , then we can clarify to get
y is in the set {0,1,2,...,39,40}

Fact 3 means because "at least 60" means "60 or more".

Then fact 4 says to put a ceiling on the production amount.
The most beds that can be made is 80.

Fact 5 tells us 300x is the profit for just the type A beds.
Fact 6 tells us 150y is the profit for just the type B beds.
Combine those facts to get 300x+150y as the total profit for both beds.

The goal is to max out P = 300x+150y

System of inequalities

Graph

The blue trapezoidal region represents the set of (x,y) points that satisfy all of the inequalities mentioned in the system above.
Points on the boundary are part of the shaded solution set.

The corner points are
A = (20, 40)
B = (40, 40)
C = (50, 30)
D = (50, 10)
Each corner point can be determined using algebra.
For instance, intersect the line y = 40 and x+y = 60 to determine the location of point A(20,40)

After getting those corner points, we then will plug each into the profit function to see which yields the largest value of P.

If we tried the x and y coordinates of point A, then,
P = 300x+150y
P = 300*20+150*40
P = 12000
Repeat for points B through D

These are the results you should get
PointCoordinatesProfit (Php)
A(20,40)12000
B(40,40)18000
C(50,30)19500
D(50,10)16500
Point C is the winner in terms of max profit of Php19500
The min profit happens at point A.

===============================================

Answer:

How many beds on both types must be produced to maximize the profit? 50 of type A and 30 of type B

What is the minimum profit? Php 12000

Edit: just in case you made a typo and are asking for the maximum profit, then that max profit would be Php 19500
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The very good response from the other tutor solves the problem in detail, using the standard process shown in most references: determine the corners of the feasibility region and evaluate the objective function at each corner.

In fact it is almost always NOT necessary to evaluate the objective function at every corner of the feasibility region. You can determine where the maximum value of the objective function will be obtained by comparing the slopes of the constraint functions and the objective function.

The slanted constraint boundary lines are and ; both of those lines have a slope of -1.

The objective function is ; that line has a slope of -2.

The maximum (and minimum!) values of the objective function will occur where a line with slope -2 just touches the feasibility region.

In this problem, a quick sketch showing the feasibility region will easily show that the maximum value of the objective function will occur where the constraint boundary lines x=50 and x+y=80 intersect, at (50,30).

So the maximum profit is 300(50)+150(30) = 15000+4500 = 19500, when 50 beds of type A and 30 of type B are produced.

ANSWERS: 50 type A, 30 type B; profit 19500
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

It is strange to read the last question in your post

        - What is the minimum profit?


Are you sure that it not an error ?


/////////////////


In any case, in this problem, both the solution and the answer can be obtained   
for both versions of the major question, since they both are  ,  from the common sense point of view.

        (1)  The solution for minimum profit


To get the minimum profit, make the minimum possible total amount of beds (60), 
and among this amount, make the maximum possible number of B-beds (40) 
since they give the minimal profit (Php 150).  

    So, the answer in this case is  (20 A and 40 B).

        (2)  The solution for maximum profit


The solution for maximum profit is OBVIOUS, too: apply the most aggressive strategy 
and make as many A-beds as possible under given restrictions, i.e. 50 A-beds (since they create the maximum profit)
and complement it by 30 B-beds, making them as many as possible, still remaining
under the given restrictions. 

    So, the answer in this case is (50 A and 30 B).


        To me,  this problem looks more as a kind of joke problems,
        than a subject to develop serious reasoning or teaching.


Actually,  in the  " entertaining Math ",  there is such type of problems,  " false  LP-method problems ".
It is the kind of problems that look like  " LP-method problems ",  but  ACTUALLY  they are
for entertainment purposes,  ONLY,  and their major intention is

        - (a)  to check if the reader has enough common sense,     and
        - (b)  to check if the reader has enough sense of humor to make a distinction,     and
        - (c)  to entertain,  of course.



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