SOLUTION: The Halloween Express is making pre-packaged bags of candy for customers to give out to trick-or-treaters. The store has 20 tootsie rolls and 18 suckers. Package A requires 1 toots

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: The Halloween Express is making pre-packaged bags of candy for customers to give out to trick-or-treaters. The store has 20 tootsie rolls and 18 suckers. Package A requires 1 toots      Log On


   



Question 1185076: The Halloween Express is making pre-packaged bags of candy for customers to give out to trick-or-treaters. The store has 20 tootsie rolls and 18 suckers. Package A requires 1 tootsie roll and 3 suckers. Package B requires 5 tootsie rolls and 1 sucker. If package A sells for $2 and package B sells for $3, how many of each package should they make to maximize profit?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Not enough information....

Profit can't be maximized knowing only the sales price; we also need to know the cost.

Re-post with all the required information.


Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
The Halloween Express is making pre-packaged bags of candy for customers to give out to trick-or-treaters.
The store has 20 tootsie rolls and 18 suckers.
Package A requires 1 tootsie roll and 3 suckers.
Package B requires 5 tootsie rolls and 1 sucker.
If package A sells for $2 and package B sells for $3, how many of each package should they make to maximize cross%28profit%29 revenue ?
~~~~~~~~~~~~~~~~~~


            It is quite obvious,  that at the given circumstances,  the problem's question
            should be about the  REVENUE  and not about the profit.

            With this editing,  see my solution below.


Let X be the number of packages A  and Y be the number of packages B.


Then we want maximize the revenue function

    R(X,Y) = 2X + 3Y


under the following restrictions

     X + 5Y <= 20      (toorsie rolls)

    3X +  Y <= 18      (suckers)

     X >= 0,  Y >= 0


The plot of the feasibility domain is shown in the Figure below.


    


    Plots y = %2820-x%29%2F5  (red) and y = 18-3x (green)



It is a quadrilateral in QI with the vertices  (X,Y) = (0,0), (0,4), (5,3), (6,0).


We apply the standard Linear Programming method in its geometric interpretation.


The solution is one of these 4 points, where the objective function (profit) has a maximum.


You calculate the values of the objective function  R(X,Y)  (revenue)  at listed points


    P(0,0) = 2*0 + 3*0 =    0

    P(0,4) = 2*0 + 3*4 =   12

    P(5,3) = 2*5 + 3*3 =   19

    P(6,0) = 2*6 + 3*0 =   12.


Comparing these values, you find the optimal point.

It is  (X,Y) = (5,3),  which means  5 boxes A and 3 boxes B, providing maximum REVENUE of 19 dollars.

Solved.

--------------

In this site, there is a lesson
    - Solving minimax problems by the Linear Programming method
which explains, for beginners, metodology of solving such problems in more details.