Question 574513: 1. A calculator company produces a scientific calculator and a graphing calculator. Long term projections indicate an expected demand of at least 100 scientific and 80 graphing calculators each day. Because of limitations on production capacity, no more than 200 scientific and 170 graphing calculators can be made daily. To satisfy a contract a total of at least 200 must be shipped each day. If each scientific calculator sold results in a $2 loss, but each graphing calculator produces a $5 profit, how many of each type should be made daily to maximize net profits.
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! The book will say this is a linear optimization problem, where you are trying to maximize the function profit. You find the boundaries of your feasibility region, which will be a convex polygon; calculate the coordinates of the polygon's vertices, and calculate the value of the function at each vertex to see where the maximum occurs and what that maximum value is.
THE SIMPLE ANSWER (Just ask your little sister, who has not learned that math is scary yet)
If each scientific calculator made by the factory results in a $2 loss, why make more than the 100 needed? If each graphing calculator made results in a profit, why not make the 170 maximum that can be made? If you ship the 170 graphing calculators and 100 scientific calculators, your total is 270, complying with the minimum 200 in your contract. The factory should make 100 scientific calculators and 170 graphing calculators.
THE SHORT (expected) ANSWER:
You can do what the book says, and write out the expected answer pretty fast:
x=number of scientific calculators made daily
y=number of scientific calculators made daily
P=daily profit, in $
P=-2x+5y
The constraints are:
 ,  ,  ,  , and  .
The boundary lines are:
x=100, y=80, x=200, y=170, and x+y=200
 --> (x,y)=(100,100)
and  --> (x,y)=(120,80)
The feasibility region is the pentagon with vertices at the intersection of boundary lines:
(100,100), (120,180), (100,170), (200,80), and (200,170).
Intersection point (100,80) is outside the feasibility region because it does not satisfy . (The teacher may or may not expect you to graph/sketch that pentagon).
The values of P at the vertices of the pentagon are:
At (100,100), P=-2(100)+5(100)=-200+500=300
At (120,80), P=-2(120)+5(80)=-240+400=120
At (100,170), P=-2(100)+5(170)=-200+850=650
At (200,80), P=-2(200)+5(80)=-400+400=0
At (200,170), P=-2(200)+5(170)=-400+850=450
The maximum P is  , which happens at (100,170) so  scientific calculators and  graphing calculators should be made daily.
THE LONG EXPLANATION
I'll add pictures and narrative in case you want to enjoy the journey, and not just get to the solution.
Let's name variables:
Let x be the number of scientific calculators, and y be the number of graphing calculator produced each day.
Let P, a function of x and y be the daily profit (If it is negative we will still call it profit).
Let's put the constraints into inequalities:
and <-- at least 100 scientific and 80 graphing calculators will be needed daily
and <-- no more than 200 scientific and 170 graphing calculators can be made daily
<-- a total of at least 200 must be shipped each day.
Your boundary lines are the equations you get when you use equal signs in all those inequalities:
x=100, y=80, x<=200, y<=170, and x+y>=200
We can graph those boundaries to map out the "feasible region" that we will be working in.
 The green lines are minimum variable values  ,  ,  that we must meet.
The red lines are maximum values due to production capacity limitations.
The slanted line represents and means on or above that line.
The pentagon with two red sides and three green sides is our feasible region.
VERTICES OF THE POLYGON:
You can determine the coordinates of the corners of that pentagon. They are the intersection points of the five boundary lines. The easy ones are the intersection of vertical and horizontal lines (100,170), (200,170), and (200,80). For the other two, we need to find the intersections of with and . So we solve the very simple systems of linear equations
and
to get (100,100) and (120,80).
The graph shows you that the point (100,80) (where vertical green line and horizontal green line cross) is outside your feasibility region and does not matter.
Let's write the profit function. In $, P is:
That is our objective function, and our objective is to maximize its value. P is a linear function.
With just two dimensions, we could write the x-y relation for various values of P and graph the resulting x-y relations to get a contour map of P. For example, for a profit of $200 as  , and we could graph it as the blue line below:
 We could set other values for P and graph the x-y lines. For example, for P=600, we get the linear equation 200=-2x+5y, and we can add the line to our graph to get:
 Our contour map ends up being a boring collection of parallel lines. As we increase P, the line moves up, parallel to all the other lines.
Seeing the plots, it is obvious that the maximum will happen at (100,170), and at that point P=-2(100)+5(170)=-200+850=650.
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