Problem 27 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Show that the following linear programming problem $$ \begin{array}{ll} \text { Maximize } & P=2 x+2 y-4 z \\ \text { subject to } & 3 x+3 y-2 z \leq 100 \\ & 5 x+5 y+3 z \leq 150 \\ & x \geq 0, y \geq 0, z \geq 0 \end{array} $$ has optimal solutions \(x=30, y=0, z=0, P=60\) and \(x=0, y=30, z=0, P=60\).

Step-by-Step Solution

Answer

The optimal solutions for the given linear programming problem are \(x = 30, y = 0, z = 0, P = 60\) and \(x = 0, y = 30, z = 0, P = 60\), as both of these points provide the maximum value of the objective function \(P = 60\).

Step 1: Identify the constraints and the objective function

The given linear programming problem has the following constraints: 1. \(3x + 3y - 2z \leq 100\) 2. \(5x + 5y + 3z \leq 150\) 3. \(x \geq 0\) 4. \(y \geq 0\) 5. \(z \geq 0\) The objective function to be maximized is \(P = 2x + 2y - 4z\).

Step 2: Graph the feasible region

Graphing the feasible region in 3-dimensional space defined by the constraints can be quite complex. However, we can start by graphing the constraints in two dimensions (i.e., on the xy-plane) and considering the z-values separately. The given inequalities on the xy-plane are: 1. \(3x + 3y \leq 100\) 2. \(5x + 5y \leq 150\) 3. \(x \geq 0\) 4. \(y \geq 0\) The z-values can be plugged into the inequalities to find respective boundaries for z.

Step 3: Identify the corner points of the feasible region

By solving the given inequalities for x and y (ignoring z for now), we can find the corner points of the feasible region: 1. Intersection of constraints 1, 3, and 4: \((0, 0)\) 2. Intersection of constraints 2, 3, and 4: \((30, 0)\) 3. Intersection of constraints 1, 2, and 4: \((0, 30)\) Now, we need to find the z-values corresponding to each corner point by plugging the x and y values into the inequalities that include z: 1. For point \((0, 0)\): \(-2z \leq 100\) and \(3z \leq 150\) Solving these inequalities, we get \(z \geq -50\) and \(z \leq 50\). Since z must be non-negative, the only valid solution is \(z = 0\). 2. For point \((30, 0)\): \(-2z \leq 100 - 90\) and \(3z \leq 150\) Solving these inequalities, we get \(z \geq -5\) and \(z \leq 50\). Again, since z must be non-negative, the only valid solution is \(z = 0\). 3. For point \((0, 30)\): \(-2z \leq 100 - 90\) and \(3z \leq 150 - 150\) Solving these inequalities, we get \(z \geq -5\) and \(z \leq 0\). Since z must be non-negative, the only valid solution is \(z = 0\). Thus, our corner points in 3D space are \((0, 0, 0)\), \((30, 0, 0)\), and \((0, 30, 0)\).

Step 4: Evaluate the objective function at each corner point

Now we will evaluate the objective function \(P = 2x + 2y - 4z\) at each corner point: 1. \(P(0, 0, 0) = 0\) 2. \(P(30, 0, 0) = 60\) 3. \(P(0, 30, 0) = 60\)

Step 5: Determine the optimal solution(s)

Based on the evaluations of the objective function at each corner point, we can see that the maximum value of P is 60, which occurs at the corner points \((30, 0, 0)\) and \((0, 30, 0)\). Therefore, the optimal solutions are \(x = 30, y = 0, z = 0, P = 60\) and \(x = 0, y = 30, z = 0, P = 60\).

Key Concepts

Feasible Region

In linear programming, the feasible region is the area in which all constraints of a problem are satisfied. Think of it as a playground where all possible solutions can "play" without breaking any rules. For this example, the constraints are given as inequalities:

• \(3x + 3y - 2z \leq 100\)
• \(5x + 5y + 3z \leq 150\)
• \(x \geq 0\), \(y \geq 0\), \(z \geq 0\)

These inequalities define limits on the values of \(x\), \(y\), and \(z\). The feasible region is typically visualized using graphs. For 3D problems like this one, visualizing can be tricky, so we often consider simpler projections, like the xy-plane, before incorporating \(z\). Remember, any point within this region is a potential solution but only certain points (like the corners) will give us our best answers.

Objective Function

The objective function is the mathematical expression you want to maximize or minimize. It's the goal of your linear programming problem. In this exercise, the objective function is:\[ P = 2x + 2y - 4z \]Your task is to find values of \(x\), \(y\), and \(z\) that maximize \(P\). The coefficients (like 2 for \(x\) and \(-4\) for \(z\)) in the function indicate how much each variable contributes to the objective. Here, both \(x\) and \(y\) add to \(P\), while \(z\) subtracts.
To find the maximum \(P\), you'll evaluate this function at different possible solutions within the feasible region, focusing primarily on those important corner points, as they often lead to the optimal solution.

Constraints

Constraints are the boundaries that define the feasible region. They limit the possible values of the variables based on specific conditions:

• Inequalities like \(3x + 3y - 2z \leq 100\) and \(5x + 5y + 3z \leq 150\) frame the shape of our feasible region.
• Simple conditions \(x \geq 0\), \(y \geq 0\), and \(z \geq 0\) ensure no negative values since we can't have negative quantities.

Constraints are crucial as they restrict where you can look for solutions. If you imagine the feasible region as a cake, the constraints are the knife cuts that shape it. Solving the problem means finding points that sit within this shaped cake where the objective function reaches its peak value.

Corner Points

Corner points, also known as vertices, are where the constraints' boundaries meet. These are special because, in linear programming, they often have the best solutions.

• For this problem, corner points were calculated by finding intersections of the constraints within the feasible region: \((0, 0, 0)\), \((30, 0, 0)\), \((0, 30, 0)\).
• Evaluations at these points found that \(P\) reaches its maximum value (60) at \((30, 0, 0)\) and \((0, 30, 0)\).

Why the corner points? Linear objective functions reach extreme values (either minimum or maximum) at these intersections. Thus, solving involves evaluating the objective function at these points until you spot which yields the desired outcome.