Problem 61 - Precalculus

Question

In Exercises \(55-62,\) minimize or maximize each objective function subject to the constraints. Maximize \(z=\frac{1}{4} x+\frac{2}{5} y\) subject to $$\begin{array}{rr}x+y \geq 5 & x+y \leq 7 \\\\-x+y \leq 5 & -x+y \geq 3\end{array}$$

Step-by-Step Solution

Answer

The maximum value is 2.65 at the point (1, 6).

Step 1: Identify the feasible region

First, let's plot the constraints in the xy-plane to identify the feasible region. The constraints are: 1. \(x + y \geq 5\) 2. \(x + y \leq 7\) 3. \(-x + y \leq 5\) 4. \(-x + y \geq 3\).This involves solving for the boundaries:- \(x + y = 5\) and \(x + y = 7\), which are downward-sloping lines.- \(-x + y = 5\) and \(-x + y = 3\) which are upward-sloping lines.The feasible region is the area where all constraints overlap.

Step 2: Finding the vertices of the feasible region

We need to find the points where the lines intersect to determine the vertices of the feasible region. Solve the equations of the lines to find these intersection points.- Intersection of \(x + y = 5\) and \(-x + y = 3\): Solve by adding the equations: \[ 2y = 8 \Rightarrow y = 4 \] Substitute \(y = 4\) into \(x + y = 5\): \[ x + 4 = 5 \Rightarrow x = 1 \]- Intersection of \(x + y = 7\) and \(-x + y = 5\): Solve by adding the equations: \[ 2y = 12 \Rightarrow y = 6 \] Substitute \(y = 6\) into \(x + y = 7\): \[ x + 6 = 7 \Rightarrow x = 1 \]- Intersection of \(x + y = 5\) and \(-x + y = 5\): Solve by adding the equations: \[ 2y = 10 \Rightarrow y = 5 \] Substitute \(y = 5\) into \(x + y = 5\): \[ x + 5 = 5 \Rightarrow x = 0 \]- Intersection of \(x + y = 7\) and \(-x + y = 3\): Solve by adding the equations: \[ 2y = 10 \Rightarrow y = 5 \] Substitute \(y = 5\) into \(x + y = 7\): \[ x + 5 = 7 \Rightarrow x = 2 \]

Step 3: Evaluate the objective function at each vertex

Now evaluate the objective function at each of the intersection points found in Step 2.1. At \((1, 4): z = \frac{1}{4}(1) + \frac{2}{5}(4) = \frac{1}{4} + \frac{8}{5} = \frac{1}{4} + \frac{32}{20} = \frac{37}{20} = 1.85\).2. At \((1, 6): z = \frac{1}{4}(1) + \frac{2}{5}(6) = \frac{1}{4} + \frac{12}{5} = \frac{1}{4} + \frac{48}{20} = \frac{53}{20} = 2.65\).3. At \((0, 5): z = \frac{1}{4}(0) + \frac{2}{5}(5) = 0 + 2 = 2\).4. At \((2, 5): z = \frac{1}{4}(2) + \frac{2}{5}(5) = \frac{1}{2} + 2 = 2.5\).

Step 4: Determine the maximum value of the objective function

The maximum value of the objective function occurs at the point where \(z\) is largest among the values calculated in Step 3.From the evaluations:- \((1, 6)\) gives \(z = 2.65\), which is the largest value.Therefore, the maximum value is 2.65 at the point \((1, 6)\).

Key Concepts

Feasible Region

In linear programming, the feasible region is the area on the graph where all constraints overlap. This is the region where all the conditions given by inequalities are satisfied at the same time. Imagine it like a playground for finding solutions—only combinations of values within this area can work for our problem.

• First, you graph the inequalities.
• For example, if you have constraints like \( x + y \geq 5 \) and \( x + y \leq 7 \), these form lines with shading above or below them depending on the inequality sign.
• You do this for each constraint to outline the edges of your feasible region.

The feasible region is typically a polygon, and in this case, it's a quadrilateral made by intersecting lines. Identifying this region helps in finding the solution's possible values.

Objective Function

The objective function in linear programming is what you aim to maximize or minimize. It is an equation that describes the problem's goal and is usually represented by a linear equation such as \( z = \frac{1}{4}x + \frac{2}{5}y \).

• The coefficients of the variables tell us how much each variable contributes to the function.
• We evaluate this function at each vertex or corner point of our feasible region.

Consider it the "scorecard" for each point within the feasible region—it helps us decide which point provides the best value (either maximum or minimum). By evaluating the objective function at these points, we determine which vertex yields the optimal result.

Maximize

When the task is to maximize, you are looking for the highest possible value for the objective function. After identifying the feasible region, you focus on maximizing the score of your objective function over this region.
Here’s how you go about maximizing:

• Calculate the value of the objective function at each vertex of the feasible region.
• For instance, given vertices like \((1, 4)\), \((1, 6)\), \((0, 5)\), and \((2, 5)\), you'll substitute these into the objective function \( z = \frac{1}{4}x + \frac{2}{5}y \).
• Compare all calculated values.

The point where the function reaches its highest value is your solution. So, maximizing in linear programming is all about finding that sweet spot where the value of your objective function is the greatest possible.