Problem 7 - Finite Mathematics and Applied Calculus

Question

Solve the LP problems. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. Maximize \(\quad p=3 x+2 y\) subject to \(\begin{aligned} 0.2 x+0.1 y & \leq 1 \\ 0.15 x+0.3 y & \leq 1.5 \\\ 10 x+10 y & \leq 60 \\ x \geq 0, y \geq 0 & \end{aligned}\)

Step-by-Step Solution

Answer

The feasible region for the given constraints is a polygon with vertices A(4, 2), B(6, 0), C(0, 5), and D(0, 0). By testing each vertex, the maximum value of the objective function is found at vertex B(6, 0) with an optimal value of \(p = 18\).

Step 1: Simplify Constraints

First, let's simplify the given constraints. We can simplify the third constraint by dividing both sides by 10. We now have the following constraints: \(\quad 0.2 x + 0.1 y \leq 1\) \(\quad 0.15 x + 0.3 y \leq 1.5\) \(\quad x + y \leq 6\) \(\quad x \geq 0, \quad y \geq 0\)

Step 2: Determine the Feasible Region

Graph the constraints on a coordinate plane: 1. The constraint \(0.2x + 0.1y \leq 1\) can be rewritten as \(y \leq 10 - 2x\). It's a line with negative slope and intercept (0,10), so the feasible region will be below the line. 2. The constraint \(0.15x + 0.3y \leq 1.5\) can be rewritten as \(y \leq 5 - 0.5x\). It's a line with negative slope and intercept (0,5), so the feasible region will be below the line. 3. The constraint \(x + y \leq 6\) is a line with negative slope and intercept (0,6), so the feasible region will be below the line. 4. The constraints \(x \geq 0\) and \(y \geq 0\) means that the feasible region will be in the first quadrant (upper right corner) of the graph. Sketching these lines on a coordinate plane, we can find the feasible region as a small polygon with vertices A(4, 2), B(6, 0), C(0, 5), and D(0, 0).

Step 3: Find the Maximum Value of the Objective Function

We want to maximize \(p = 3x + 2y\), so we need to test each of the vertices of the feasible region. 1. For vertex A(4, 2), \(p_A = 3(4) + 2(2) = 12 + 4 = 16\) 2. For vertex B(6, 0), \(p_B = 3(6) + 2(0) = 18 + 0 = 18\) 3. For vertex C(0, 5), \(p_C = 3(0) + 2(5) = 0 + 10 = 10\) 4. For vertex D(0, 0), \(p_D = 3(0) + 2(0) = 0\) Since the maximum value among \(p_A\), \(p_B\), \(p_C\), and \(p_D\) is 18, the optimal solution is at vertex B(6, 0) with an optimal value of p=18.

Key Concepts

LP Problem Solving

Linear Programming (LP) is an essential method used to determine the best outcome in a mathematical model whose requirements are represented by linear relationships. This mathematical optimization technique is widely applied in various fields, such as economics, business, engineering, and military applications.

The process of solving an LP problem typically involves the following steps: defining the variables, formulating the objective function, establishing the constraints, graphing the feasible region, and then finding the optimal value of the objective function within this region. In the exercise provided, the objective is to maximize the function 'p' with respect to the given constraints.

It is crucial to first simplify the constraints to make the subsequent steps more straightforward. As seen in the step-by-step solution, each constraint is converted into a simpler form to facilitate the graphical approach. The variables 'x' and 'y' are non-negative, highlighting that the feasible region will be in the first quadrant of the coordinate plane. Through simplification and careful graphing, the LP problem becomes more manageable and easier to analyze.

Feasible Region Determination

The feasible region is the set of all possible points that satisfy all the constraints of a linear programming problem. In graphical terms, it is the area where the constraint lines or planes intersect in a way that adheres to all imposed conditions. Identifying the feasible region is a crucial step in visualizing the constraints and envisioning where the optimal solution could lie.

In the context of the given problem, the constraints are graphed on the coordinate plane, with particular attention paid to their intercepts and slopes. These constraints form boundaries of the feasible region. The region is then located in the first quadrant due to the non-negativity constraint of 'x' and 'y'. As with the solution provided, it is advisable to label the vertices of the feasible region clearly. The outcome is a polygonal shape—the region within which potential solutions exist. Assessing this visually, helps students understand the limits within which they can maximize or minimize their objective functions.

Objective Function Maximization

The objective function in a linear programming problem represents the quantity that needs to be maximized or minimized. Its optimization is the goal of LP problems. Maximization involves finding the highest value of the objective function, while minimization seeks the lowest value, given the problem's constraints.

After determining the feasible region, the next step is to find the point within this region that yields the highest (or lowest) value of the objective function. For maximization problems like in the exercise, this is typically achieved through the method of corners, which states that the optimal point must lie at one of the vertices of the feasible region. As confirmed by the solution, each vertex is evaluated under the objective function. The vertex that gives the highest function value is where the maximum lies. In this case, vertex B(6, 0) is where 'p' reaches its maximum value, making it the optimal solution for the given LP problem. This practical approach allows students to easily pinpoint the best solution.