Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
4. Linear Programming
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Exercise 23 Page 162

We have been given four constrains and an objective function.
Let's begin by graphing Constraint III and IV. We should first determine their boundary lines by replacing the inequality sign with the equals sign.
The boundary lines of the first two constraints are the axes. Since both and are greater than or equal to we will shade Quadrant I. Notice that the inequalities are non-strict, so the boundary lines will be solid.
Next, we will graph Constraint I and II by finding their boundary lines.
To graph their boundary lines, we will find the and intercepts of the lines. We will first substitute for into Boundary Line I and solve it for to find its intercept.
The intercept of Boundary Line I is the point Proceeding in the same way, we can find the intercepts of both lines.
Constraint Boundary Line intercept intercept
I
II

Now we will plot the intercepts and connect them with line segments. The lines will be bound by the axes because of Constraint III and IV. Also, notice that the constraints are non-strict. Therefore, the lines will be solid.

Next, we will test the point to determine which region we should shade. Let's start with Constraint I.
For Constraint I, we will shade the region that does not contain the point.
Now we will test the point for Constraint IV.
Therefore, we will again shade the region that does not contain the test point.

The overlapping section of the graph above is the feasible region. Let's remove the unnecessary parts and indicate the vertices of the feasible region.

The coordinates of two of the three vertices are determined. The third vertex is the point of intersection of Constraint I and II. To determine it we will first write a system of equations.
To solve the system, we will equate the coefficients of and use the Elimination Method.
The coordinate of the point is By substituting for into the equation, we can find the coordinate.

Calculate quotient

Then, the third vertex is We can write this point as a mixed number
The coordinates of the point can be rounded to and Let's begin by substituting the point into the objective function and determine at which point it is minimum.
The value of at is We can find the value of for the other rounded coordinates in the same way.
Vertex Objective Function

At the point has the minimum value of