Question

Solve the following linear programming problem by the graphical method, under the following constraints: \[ x + 3y \leq 60, \quad x + y \geq 10, \quad x \leq y, \quad x \geq 0, \quad y \geq 0. \] Find the minimum and maximum values of \[ Z = 3x + 9y. \]

Hint:

In linear programming problems, the maximum and minimum values of the objective function are found at the vertices of the feasible region.

Answer 1

Step 1: Write the Constraints as Equations.
We start by converting each inequality into an equation in order to graph them: 1. \( x + 3y = 60 \) 2. \( x + y = 10 \) 3. \( x = y \) 4. \( x = 0 \) and \( y = 0 \) are just the axes of the coordinate plane. Now we plot the lines corresponding to these equations.
Step 2: Find the Intersection Points.
To identify the feasible region, we solve for the intersection points of the lines by solving the corresponding systems of equations. Intersection of \( x + 3y = 60 \) and \( x + y = 10 \):
Solving the system of equations: \[ x + 3y = 60 \quad \text{and} \quad x + y = 10, \] From the second equation, we get \( x = 10 - y \). Substitute this into the first equation: \[ (10 - y) + 3y = 60, \] \[ 10 - y + 3y = 60 \quad \Rightarrow \quad 10 + 2y = 60 \quad \Rightarrow \quad 2y = 50 \quad \Rightarrow \quad y = 25. \] Substitute \( y = 25 \) into \( x + y = 10 \): \[ x + 25 = 10 \quad \Rightarrow \quad x = -15. \] However, since \( x \geq 0 \), this intersection does not lie within the feasible region. Intersection of \( x + 3y = 60 \) and \( x = y \):
Substitute \( x = y \) into the first equation: \[ x + 3x = 60 \quad \Rightarrow \quad 4x = 60 \quad \Rightarrow \quad x = 15. \] Thus, the intersection point is \( (15, 15) \). Intersection of \( x + y = 10 \) and \( x = y \):
Substitute \( x = y \) into \( x + y = 10 \): \[ x + x = 10 \quad \Rightarrow \quad 2x = 10 \quad \Rightarrow \quad x = 5. \] Thus, the intersection point is \( (5, 5) \).
Step 3: Plot the Feasible Region.
Plot the points of intersection \( (15, 15) \) and \( (5, 5) \), and draw the lines \( x + 3y = 60 \), \( x + y = 10 \), and \( x = y \). The feasible region is bounded by these lines.
Step 4: Evaluate the Objective Function \( Z = 3x + 9y \) at the Corner Points.
We now evaluate the objective function at the corner points of the feasible region, which are \( (0, 0) \), \( (5, 5) \), and \( (15, 15) \). At \( (0, 0) \): \[ Z = 3(0) + 9(0) = 0. \] At \( (5, 5) \): \[ Z = 3(5) + 9(5) = 15 + 45 = 60. \] At \( (15, 15) \): \[ Z = 3(15) + 9(15) = 45 + 135 = 180. \]
Step 5: Conclusion.
Thus, the minimum value of \( Z \) is \( 0 \) at \( (0, 0) \), and the maximum value of \( Z \) is \( 180 \) at \( (15, 15) \).