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 shipping contract, a total of at lease 200 calculators 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 if each type should be made daily to maximize net profits?
A) Define the objective equation.
B)Define the constraint equations.
C)Graph the associated constraint equations.
D) Determine the vertices of each of the graphed equations.
E) Determine how many of each type of calculators should be made daily to maximize the net profits and the amount of net profit.
maximize p = -2x+5y subject to
x <= 200
y <= 170
x+y >= 200
looks like 30 scientific, 170 graphing
How did you get 30 and 170? If the equation is -2x+5y?
Well, just intuitively, if you lose money on each scientific calculator, you want to produce as few as possible. Ideally, zero.
Unfortunately, you have to produce at least 200 boxes a day, and can only produce 170 graphing calculators. So, you gotta do at least 30 of the losers.
To solve this problem, we can use linear programming techniques. Linear programming involves maximizing or minimizing an objective function subject to a set of constraints.
A) Define the objective equation:
Let x be the number of scientific calculators produced daily, and y be the number of graphing calculators produced daily. The objective is to maximize net profits, which can be represented by the equation:
Net profit = 5y - 2x
B) Define the constraint equations:
i) Daily demand:
Scientific calculators: x ≥ 100
Graphing calculators: y ≥ 80
ii) Production capacity:
Scientific calculators: x ≤ 200
Graphing calculators: y ≤ 170
iii) Shipping contract:
Total calculators: x + y ≥ 200
C) Graph the associated constraint equations:
To graph these constraint equations, we'll plot the inequalities on a graph.
iv) x ≥ 100:
This is a horizontal line on the graph, passing through the point (100, 0).
v) y ≥ 80:
This is a vertical line on the graph, passing through the point (0, 80).
vi) x ≤ 200:
This is a vertical line on the graph, passing through the point (200, 0).
vii) y ≤ 170:
This is a horizontal line on the graph, passing through the point (0, 170).
viii) x + y ≥ 200:
This is a diagonal line on the graph, with slope -1, passing through the points (100, 100) and (200, 0).
D) Determine the vertices of each of the graphed equations:
To find the vertices, we need to find the intersection points of the lines formed by the constraint equations:
- Intersection of (iv) and (v): (100, 80)
- Intersection of (v) and (vi): (100, 170)
- Intersection of (vi) and (vii): (200, 170)
- Intersection of (v) and (viii): (100, 100)
- Intersection of (iv) and (viii): (200, 0)
E) Determine how many of each type of calculator should be made daily to maximize net profits and the amount of net profit:
Now, we need to plug in the vertices into the objective function to find the net profits at each vertex:
- At (100, 80): Net profit = 5(80) - 2(100) = 240
- At (100, 170): Net profit = 5(170) - 2(100) = 740
- At (200, 170): Net profit = 5(170) - 2(200) = 540
- At (100, 100): Net profit = 5(100) - 2(100) = 300
- At (200, 0): Net profit = 5(0) - 2(200) = -400
Based on these calculations, it can be concluded that to maximize net profits, 100 scientific calculators and 170 graphing calculators should be made daily. The maximum net profit achievable is $740.