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A manufacturer produces two models of mountain bikes.
Model A requires 5 hours of assembly time and 2 hours of painting time,
and Model B requires 4 hours of assembly time and 3 hours of painting time.
The maximum total weekly hours available in the assembly department and the painting department are
200 hours and 108 hours, respectively.
The profits per unit are $25 for Model A and $15 for Model B. How many of each type should be produced to maximize profit?
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Let X be the number of bikes Model A;
Y be the number of bikes Model B.
Then the constraints are
5X + 4Y <= 200, (1) (Assembly time constraint of 200 hours)
2X + 3Y <= 108. (2) (Painting time constraint of 108 hours)
X >= 0, Y >= 0. (3) (Non-negativity constraint)
The profit function is
P(X,Y) = 25X + 15Y. (4)
The feasibility domain is shown in the Figure below.
It is the quadrilateral in the Quadrant I below two straight lines that are constraints.
Plots 5x + 4y = 200 (red line) and 2x + 3y = 108 (green line).
The vertices of the quadrilateral are
P1 = (0,36) (y-intersept of the green line)
P2 = (40,0) (x-intercept of the red line)
P3 = (24,20) (the intersection point of the red and green lines) <<<---=== You may find it by solving the system of equations
P4 = (0,0) (the origin of the coordinate system)
According to the Linear Programming method, you need to find the values of the profit function in the vertices of the feasibility domain,
and then select the vertex, where the value of the profit function is maximal.
These calculations re shown below:
at P1: P(X,Y) = P(0,36) = 0*25 + 36*15 = 540;
at P2: P(X,Y) = P(40,0) = 40*25 + 0*15 = 1000;
at P3: P(X,Y) = P(24,20) = 24*25 + 20*15 = 900;
at P4: P(X,Y) = P(0,0) = 0*25 + 0*15 = 0.
The maximum value is 1000 at P2; so, X = 40, Y = 0 is the solution to the given problem.
Answer. 40 bikes of the model A and 0 (zero, ZERO) bikes of the model B satisfy the constraints
and provide the maximum profit of 1000 dollars.