Solve linear programming using the graphical method: 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 least 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 of each type should be made daily to maximize net profits?
The Correct Answer and Explanation is:
Correct Answer:
Produce 100 scientific calculators and 170 graphing calculators daily.
Maximum net profit = $650
Explanation
This is a linear programming problem with two decision variables:
- Let x = number of scientific calculators produced per day
- Let y = number of graphing calculators produced per day
Objective Function:
We aim to maximize profit, given by:
Profit (P) = -2x + 5y
(Each scientific calculator results in a $2 loss → -2x, each graphing calculator gives a $5 profit → +5y)
Constraints:
- Demand constraints (minimum production):
- x ≥ 100 (at least 100 scientific calculators)
- y ≥ 80 (at least 80 graphing calculators)
- Capacity constraints (maximum production):
- x ≤ 200
- y ≤ 170
- Shipping constraint:
- x + y ≥ 200 (at least 200 calculators total)
Feasible Region:
Graph these constraints on the x-y plane and find the intersection area (feasible region). Check corner points of this region — these give potential optimal solutions.
The relevant corner points (from graphical analysis) are:
- (100, 170)
- (200, 170)
- (200, 80)
- (120, 80) — intersection of x + y = 200 and y = 80
- (100, 100) — intersection of x + y = 200 and x = 100
Evaluate the objective function at each:
- (100, 170): P = -2(100) + 5(170) = -200 + 850 = 650 ✅
- (200, 170): P = -400 + 850 = 450
- (200, 80): P = -400 + 400 = 0
- (120, 80): P = -240 + 400 = 160
- (100, 100): P = -200 + 500 = 300
Conclusion:
The maximum net profit is $650, achieved by producing 100 scientific calculators and 170 graphing calculators daily. This satisfies all constraints and gives the best outcome.
