Problem 57 - Precalculus

Question

In Exercises \(55-62,\) minimize or maximize each objective function subject to the constraints. Maximize \(z=4 x+3 y\) subject to $$x \geq 0 \quad y \leq-x+4 \quad y \geq-x$$

Step-by-Step Solution

Answer

Maximize at point (0, 4) with value 12.

Step 1: Identify the Constraints

We have three constraints given:1. \( x \geq 0 \)2. \( y \leq -x + 4 \)3. \( y \geq -x \)These inequalities describe a region in the coordinate plane. To find where to maximize the function, it is helpful to graph these lines and identify the feasible region.

Step 2: Graph the Constraints

First, plot the line \( y = -x + 4 \). This is a line with a y-intercept at 4 and slopes downwards with slope -1. Next, plot the line \( y = -x \). This is a line through the origin with slope -1.The inequality \( x \geq 0 \) ensures we only consider the region on or to the right of the y-axis.The feasible region is bounded by these lines within the first quadrant.

Step 3: Determine the Feasible Region

The feasible region is the area where all constraints overlap. Based on the graph from Step 2:- It will be above the line \( y = -x \) because of the inequality \( y \geq -x \).- Below the line \( y = -x + 4 \) because of the inequality \( y \leq -x + 4 \).- On or to the right of the y-axis because of \( x \geq 0 \).The intersection points of these lines within this region are potential candidates for maximizing the objective function.

Step 4: Identify Intersection Points

Compute the intersection of the lines that form the boundaries of the feasible region:- Intersection of \( y = -x + 4 \) and \( y = -x \) gives the point \( (2, -2) \).- Intersection of \( y = -x + 4 \) and the y-axis (\( x = 0 \)) gives the point \( (0, 4) \).- Intersection of \( y = -x \) and the y-axis (\( x = 0 \)) gives the point \( (0, 0) \).Note: \( (2, -2) \) is not feasible since \(y\) must be positive for feasible points in this context.

Step 5: Evaluate the Objective Function at Key Points

Now, compute \( z = 4x + 3y \) for the feasible points:- At \((0, 4)\), \( z = 4(0) + 3(4) = 12 \).- At \((0, 0)\), \( z = 4(0) + 3(0) = 0 \).The maximum value occurs at \((0, 4)\) with \( z = 12 \).

Step 6: Conclude the Maximum Value

The objective function reaches its maximum value, 12, at the point \((0, 4)\). Hence, the coordinates \((0, 4)\) provide the optimal solution for this problem.

Key Concepts

Objective Function

In linear programming problems, the **objective function** is the equation that needs to be maximized or minimized. It's composed of variables representing quantities we want to optimize under given conditions. In this exercise, the objective function is \( z = 4x + 3y \). Here, \( z \) measures the quantity we want to maximize, which depends on the values of \( x \) and \( y \). The coefficients \( 4 \) and \( 3 \) signify how much each unit of \( x \) and \( y \) respectively contributes to maximizing \( z \). Thus, understanding this function helps us determine which combinations of \( x \) and \( y \) will give the highest value within the allowed constraints.

Constraints

Constraints are the conditions that the solution of an optimization problem must satisfy. They limit the possible values of variables within the objective function. In our problem, the constraints are:

• \( x \geq 0 \)
• \( y \leq -x + 4 \)
• \( y \geq -x \)

These constraints establish boundaries or limits, like a fence around our choices of \( x \) and \( y \). The constraint \( x \geq 0 \) ensures that no negative values for \( x \) are considered, keeping solutions to the right of the y-axis. The other two form diagonal boundaries that must be considered in plotting our feasible region, showing the interplay between \( x \) and \( y \). By graphing the constraints, you visually depict the realm of possibilities where solutions may lie.

Feasible Region

In linear programming, the **feasible region** is the specific area on a graph where all the constraints overlap. It's like the overlapping section of a Venn diagram. This region defines the set of all possible solutions. For this exercise, the feasible region is where:

• The area above our line \( y = -x \)
• Below the line \( y = -x + 4 \)
• To the right of the line \( x = 0 \)

The section where these conditions are all true is the feasible region - and it's only here that the objective function can reach its maximum value. This region can be visualized as a geometric shape on the graph, often forming a polygon. The vertices of this polygon are crucial as they are possible candidates for optimal solutions, acting like potential corners to check for the maximum or minimum values of the objective function.

Intersection Points

Intersection points are where the constraint lines meet. These are key in identifying feasible solutions. In linear programming, these points are typically examined to find the optimal solution, as they frequently represent the maximum or minimum values of an objective function. In this context, the steps involved:

• The intersection of \( y = -x + 4 \) and \( y = -x \) would be at the point \((2, -2)\), but this point is outside our feasible region due to the constraints that preserve positive values for \( y \).
• \( (0, 4) \) where \( y = -x + 4 \) intersects the y-axis is a valid point within the feasible region.
• \( (0, 0) \), the point where \( y = -x \) intersects the y-axis is another valid feasible point.

Checking these intersection points is essential as they help determine where the objective function might achieve maximum or minimum values, depending on the presence and type of optimization, and assessing them gives us the maximum point at \((0, 4)\) with a value of \( z = 12 \).