Equations for Math

Question

Imagine that you're a production manager at a calculator company. Your company makes two types of calculators, a scientific calculator and a graphing calculator.

1. Each model uses the same plastic case and the same circuits. However, the graphing calculator requires 20 circuits and the scientific calculator requires 10. The company has has 240 plastic cases and 3200 circuits in stock. Graph the system of inequalities that represents these constraints.
2. The profit on a scientific calculator is $8.00, while the profit on a graphing calculator is $16.00. Write an equation that describes the companies profit from calculator sales.
3. How many of each type of calculator should the company produce to maximize profit using the stock on hand?

Answer 1

Let x=the number of graphing calculators

Let y=the number of scientific calculators

 

1. x+y≤240 This inequality models the constraint on the total number calculators the company can make.

  20x+10y≤3200 This inequality models the constraint on the total amount of circuits can be used to make both types of calculators.

 

Determine the x and y intercepts of each equation in which for the first equation are (0,240) and (240,0). Connect the points with a solid line, and the shade goes below the line.

Divide 10 from the second equation to simplify the arithmetic.

2x+y≤320 Determine the x and y intercepts in which are (0,320) and (320/2,0)=(160,0). Connect the dots with a solid line.

 

Determine the point in which the lines intersect by solving the system of equations.

 

 x+y=240

2x+y=320 Subtract the two equations

 

     -x=-80  x=80

 80+y=240

-80      -80

       y=160

Hence, the point of intersection is (80,160). Draw the shade below this point to satisfy the two inequalities. 

 

2. P(x,y)=16x+8y Is the profit equation

 

3. P(0,240)=16(0)+8(240)=$1920

    p(80,160)=16(80)+8(160)=1280+1280=$2560

    p(160,0)=16(160)+8(0)=$2560

 

Hence, the company needs to make  80  graphing and 160 scientific calculators to maximize the profits using the stock on hand. Also, the company could just make 160 graphing and zero scientific calculators to make the same maximized profit. However, its better to have the variety.

Answer 2

Hi, Natalie,

 

These types of problems are usually referred to as "linear programming", and they usually follow the same pattern. We generally have two different competing variables, and we want to minimize or maximize something based on them.

 

1) In this case, we are producing two different kinds of calculators - scientific and graphing. Let's say that "x" is the hypothetical number of scientific calculators we produce, and "y" is the hypothetical number of graphing calculators that we produce.

 

We are told that each scientific calculator requires 10 circuits and 1 case, while each graphing calculator requires 20 circuits and 1 case. If we make "x" scientific calculators, that would require 10*x circuits. If we make "y" graphing calculators, that would require 20*y circuits, etc...

 

I think it helps to make a table out of this information, and we can then use it to come up with equations/inequalities:

 

 

                                     circuits                       cases

 

# scientific (x)                  10x                            x

 

# graphing (y)                  20y                            y

 

 

So, the total number of circuits required for our entire production would be 10x+20y, and the total number of cases would be x+y. We are limited in both the total number of circuits and total number of cases, so we can make inequalities out of each of these relationships:

 

    10x + 20y ≤ 3200

 

    x + y ≤ 240

 

We can rearrange each of these inequalities to put them in the usual slope-intercept format:

 

    y ≤ (-1/2)x + 160

 

    y ≤ -x + 240

 

We can graph these two inequalities together to find the "feasibility region": the area under BOTH of these lines, which represents the combinations of x and y that satisfy those requirements. 

 

2) If the profit is $8/scientific and $16/graphing, the total profit would be:

 

    8x + 16y

 

3) To find the right combination that maximizes profit, we want to test the x/y combinations at the "corners" of the feasibility region. We can find these by looking at the graph, or by setting the inequalities equal to each other. I found "corners" at (x,y) coordinates of (0,160), (160,80), and (240,0).

 

So, the total profit at each of these points would be:

 

    At (0,160):       8(0) + 16(160) = $2560

 

    At (160,80):     8(160) + 16(80) = $2560

 

    At (240,0):       8(240) + 16(0) = $1920

 

So, it looks like profit can be maximized if the company produces either 0 scientific and 160 graphing calculators, or if they produce 160 scientific and 80 graphing calculators, since the profit is the same in both cases.

 

I hope this helps!