You use the information in the situation to arrive at the Objective Function which is the profit for any given day. You also use the information to arrive at the constraints for the number of scientific and graphing calculators.
x: # of Scientific Calculators
y: # of Graphing Calculators
Objective Function: Profit = -5x + 9y
Constraints
x > 70 x < 100
y > 50 y < 80
x + y > 130
You graph each inequality on the coordinate plan shading each one correctly and then the overlapping shaded regions are the feasibility solution set.
To find the intersection points, solve each system of the two lines individually for example
x = 70 and y = 80 gives the intersections point of (70, 80)
x + y = 130 and x = 70 gives the intersection (70, 60)
You asked for the minimum cost; however, there is no information about cost in the problem; therefore, I cannot provide an answer about the minimum cost.
To find the maximum profit, you evaluate the Objective Function at each intersection point.
@ (70, 60) Profit = -5(70) + 9(60) = 190
@ (80, 50) Profit = -5(80) + 9(50) = 50
@ (100, 50) Profit = -5(100) + 9(50) = -50
@ (100, 80) Profit = -5(100) + 9(80) = 220
@ (70, 80) Profit = -5(70) + 9(80) = 370
Therefore, the maximum profit will happen when 70 scientific calculators and 80 graphing calculators are shipped and sold.
Dale W.
02/23/23