Maximize
P=60x+50y .
x+y≤80
5x+10y≤560
50x+20y≤1600
x≥0
y≥0
Just need help in finding s=
I wish I knew more about what method you are using, graphical, analytical, or computer. I will explain grahical.
Draw the lines on an x y graph (x horizontal) , scale x from zero to 150, y from zero to 80
Line 1: x=0
line 2: y=0
line 3: y=-x+80
line 4: y=-2.5x+80
line 5: y=-0.5x+56
those are from the constraints given.
Now look: an area is enclosed. Within that area is the "area of acceptable solutions to the constraints". Now we have a very nice theorem that tells us somewhere on the boundary (usually at the corner intersections) will be the maximum or minimum of any profit function. Your function is P=60x+50y
figure that value for each corner. Obviously, the minimum will be at (0,0). Find the coordinates (x,y) for the maximum.
To find the maximum value of P, we need to determine the values of x and y that satisfy the given constraints and maximize the objective function P.
First, let's graph the feasible region represented by the constraints:
1. Graph the equation x + y = 80:
- Plot the points (80, 0) and (0, 80).
- Draw a straight line through these points.
- Shade the region below or on the line to represent the inequality x + y ≤ 80.
2. Graph the equation 5x + 10y = 560:
- Divide both sides of the equation by 10 to simplify: x + 2y = 112.
- Plot the points (112, 0) and (0, 56).
- Draw a straight line through these points.
- Shade the region below or on the line to represent the inequality 5x + 10y ≤ 560.
3. Graph the equation 50x + 20y = 1600:
- Divide both sides of the equation by 20 to simplify: 5x + 2y = 80.
- Plot the points (80, 0) and (0, 40).
- Draw a straight line through these points.
- Shade the region below or on the line to represent the inequality 50x + 20y ≤ 1600.
The feasible region is the overlapping shaded area resulting from the three constraints.
To find the vertices of the feasible region, we identify the points where the lines intersect. In this case, the feasible region has four vertices.
Now, substitute the coordinates of each vertex into the objective function P = 60x + 50y to find the value of P at each vertex.
s1 = P(0, 0)
s2 = P(Vertex 1)
s3 = P(Vertex 2)
s4 = P(Vertex 3)
s5 = P(Vertex 4)
Compare the values of P at each vertex to determine the maximum value.
The vertex that gives the maximum value of P is the optimal solution.