Problem 17 - 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. \(\vee\) Minimize \(\quad c=3 x-3 y\) subject to \(\begin{aligned} \frac{x}{4} & \leq y \\ y & \leq \frac{2 x}{3} \\ x+y & \geq 5 \\ x+2 y & \leq 10 \\ x \geq 0, y & \geq 0 \end{aligned}\)

Step-by-Step Solution

Answer

The feasible region for the given linear programming problem is a quadrilateral formed by the intersection of the lines \(y = \frac{x}{4}\), \(y=\frac{2x}{3}\), \(y = -x + 5\), and \(y = \frac{1}{2}(10-x)\). The corner points are (2, 0.5), (2, 3), (3, 2), and (0, 0). Evaluating the objective function c = 3x - 3y at these corner points, we find that the minimum value is -3 at the point (2, 3). So, the solution is c = -3 at the point (2, 3).

Step 1: Draw the Inequality Constraints on a Graph

First, we will plot each inequality on a graph. To do this, treat each inequality like an equation, and then shade the region satisfying the inequality: 1. \(\frac{x}{4}\leq y\) becomes y = \(\frac{x}{4}\), shade above the line. 2. \(y \leq \frac{2x}{3}\) becomes y = \(\frac{2x}{3}\); shade below the line. 3. \(x+y \geq 5\) becomes y = -x + 5; shade above the line. 4. \(x+2y \leq 10\) becomes y = \(\frac{1}{2}\)(10 - x); shade below the line. 5. \(x \geq 0\), shade right of the y-axis. 6. \(y \geq 0\), shade above the x-axis. Please note that we're working with x and y greater than or equal to 0, so our graph will be in the first quadrant.

Step 2: Identify Feasible Region

The feasible region is the area in which all shaded regions overlap. In this case, it is a quadrilateral formed by the intersection of the lines \(y = \frac{x}{4}\), \(y=\frac{2x}{3}\), \(y = -x + 5\), and \(y = \frac{1}{2}(10-x)\).

Step 3: Identify the Corner Points

The corner points are the vertices of the quadrilateral we identified as the feasible region. These are the points where two constraints intersect. Find these points by solving the corresponding equation pairs: 1. Intersection of \(y = \frac{x}{4}\) and \(y = \frac{1}{2}(10 - x)\): (2, 0.5) 2. Intersection of \(y = -x + 5\) and \(y = \frac{1}{2}(10 - x)\): (2, 3) 3. Intersection of \(y = -x + 5\) and \(y = \frac{2x}{3}\): (3, 2) 4. Intersection of \(y = \frac{x}{4}\) and \(y = \frac{2x}{3}\): (0, 0)

Step 4: Evaluate Objective Function at Corner Points

Plug in the x and y coordinates of each corner point in the objective function c = 3x - 3y, and see which coordinates minimize the objective function: 1. c(2, 0.5) = 3(2) - 3(0.5) = 6 - 1.5 = 4.5 2. c(2, 3) = 3(2) - 3(3) = 6 - 9 = -3 3. c(3, 2) = 3(3) - 3(2) = 9 - 6 = 3 4. c(0, 0) = 3(0) - 3(0) = 0

Step 5: Determine the Minimum Value

The smallest value among the evaluated corner points is when c = -3 at the point (2, 3). So, the optimal solution exists, and the minimum value of the objective function is -3.

Key Concepts

Objective Function Optimization

Objective Function Optimization in linear programming deals with finding the best possible outcome for a linear equation, usually in terms of minimizing or maximizing a particular value. In this context, the objective function is given as \( c = 3x - 3y \). When we say that we want to "minimize" this function, we are looking for the values of \( x \) and \( y \) that make \( c \) as small as possible.

To solve this, we plug in the values of \( x \) and \( y \) from specific points, known as corner points, into the function \( c = 3x - 3y \). These corner points are derived from the feasible region, where all constraints are satisfied. By evaluating each point, we determine which coordinates minimize the function. For instance, by calculating \( c \) at each corner, we can see how the output changes and identify the smallest \( c \).

This procedure helps in determining not just any solution, but the optimal solution that leads to the minimal value of \( c \), which in this case is -3, occurring at the point (2, 3).

Inequality Constraints

Inequality Constraints are a fundamental part of linear programming and serve as the boundaries within which the solution must lie. For this problem, the constraints are a set of inequalities involving \( x \) and \( y \):

• \( \frac{x}{4} \leq y \)
• \( y \leq \frac{2x}{3} \)
• \( x + y \geq 5 \)
• \( x + 2y \leq 10 \)
• \( x \geq 0 \)
• \( y \geq 0 \)

These inequalities describe relationships between \( x \) and \( y \), establishing limits within which solutions can exist. Graphically, each inequality represents a region on the Cartesian plane, and only the area where all these conditions meet is considered feasible.

To interpret these constraints, you graph each inequality and carefully shade the region that satisfies each one. The challenges often involve ensuring this overlap is properly identified, as the final solution is contingent on adhering strictly to these constraints.

Accurate representation of these boundaries ensures that all calculations and outcomes will respect the given limits, with viable solutions situated only within this prescribed space.

Feasible Region Identification

Feasible Region Identification is a key step in solving linear programming problems and refers to the area where all constraint inequalities intersect, the region where a viable solution might exist.

To determine this region, you first draw the lines representing each inequality on a coordinate plane. Then, you identify the intersection or overlap of all these different shaded regions. In our example, the feasible region is a quadrilateral, formed by the lines \( y = \frac{x}{4} \), \( y = \frac{2x}{3} \), \( y = -x + 5 \), and \( y = \frac{1}{2}(10-x) \).

This region must also satisfy all non-negativity constraints (\( x \geq 0 \), \( y \geq 0 \)), meaning it lies entirely within the first quadrant of the graph. By finding where these lines cross, we define the vertices or corner points of the feasible region, which is crucial in determining the optimal solution. Each vertex is a potential candidate for optimizing the objective function.

The feasible region essentially sets the stage for where solutions can exist, guiding the process of finding maximum or minimum values for the objective function based on the given constraints.