Our first step in finding the maximum and minimum values of the given function is to graph the system and determine the vertices of the overlapping region. Substituting these vertices into the equation, we will find these values.
y≤2x+6 & (I) y≥2x-8 & (II) y≥-2x-18 & (III)
Inequality I
The equation of the boundary line can be written by changing the inequality symbol in the inequality to an equals sign.
Inequality:& y≤2x+6
Boundary Line:& y=2x+6
Since this equation is in , we can determine its m and y-intercept b to draw the line.
Slope-Intercept Form:& y= mx+b
Boundary Line:& y= 2x+6
Now that we know the slope and y-intercept, let's use these to draw the boundary line. Notice that the inequality is non-strict, which means that the line will be solid.
To complete the graph, we will test a point that is not on the line and decide which region we should shade. Let's test the point ( 0, 0). If the point satisfies the inequality, we will shade the region that contains the point. Otherwise, we will shade the other region.
y ≤ 2x+6
0 ? ≤ 2( 0)+6
0 ≤ 6 ✓
Since the point satisfies the inequality, we will shade the region that contains the point.
Inequality II
Let's again start by changing the inequality symbol in the second inequality to an equals sign.
Inequality:& y≥2x-8
Boundary Line:& y=2x-8
Next, we can determine its m and y-intercept b to draw the line.
Slope-Intercept Form:& y= mx+b
Boundary Line:& y= 2x+(-8)
Now that we know the slope and y-intercept, let's use these to draw the . Notice that the inequality is non-strict, which means that the line will be solid.
To complete the graph, we will test a point that is not on the line and decide which region we should shade. Let's test the point ( 0, 0). If the point satisfies the inequality, we will shade the region that contains the point. Otherwise, we will shade the other region.
y≥2x-8
0? ≥2( 0)-8
0? ≥0-8
0≥-8 ✓
Inequality III
For the last time let's start by changing the inequality symbol in the last inequality to an equals sign.
Inequality:& y≥-2x-18
Boundary Line:& y=-2x-18
Next, we will determine its m and y-intercept b to draw the line.
Slope-Intercept Form:& y= mx+b
Boundary Line:& y= -2x+(-18)
Now that we know the slope and y-intercept, let's use these to draw the boundary line. Notice that the inequality is non-strict, which means that the line will be solid.
To complete the graph, we will test a point that is not on the line and decide which region we should shade. Let's test the point ( 0, 0). If the point satisfies the inequality, we will shade the region that contains the point. Otherwise, we will shade the other region.
y≥-2x-18
0? ≥-2( 0)-18
0? ≥0-18
0≥-18 ✓
Combining the Inequality Graphs
Let's draw the graphs of the inequalities on the same .
Now that we can see the overlapping region, let's highlight the vertices.
Looking at the graph, we can see that the line y=-2x-18 intersects the other lines at the points (-6,-6) and (-2.5,-13). These are our vertices. Notice that the feasible region is unbounded. This means that it can go on forever in the positive vertical direction and there are no other vertices.
( -6, -6), and ( -2.5, -13)
Having all the vertices we can find the maximum and minimum of the given function.
Finding the Maximum and Minimum Values
Since our region is unbounded in one direction, it will have either a maximum or a minimum, but not both. To check which of these it has, we need to calculate the values of the given function at the vertices and at a test point that lies inside the feasible region. Let's use (0,4).
Let's start with calculating the function value in the vertex ( -6, -6).
f(x,y)=5x-4y
f( -6, -6)=5( -6)-4( -6)
f(-6,-6)=-30+24
f(-6,-6)=-6
| Vertex or Testing Point
|
f(x,y)=5x-4y
|
Value
|
| ( -6, -6)
|
f( -6, -6)=5( -6)-4( -6)
|
f(-6,-6)=-6
|
| ( -2.5, -13)
|
f( -2.5, -13)=5( -2.5)-4( -13)
|
f(-2.5,-13)=39.5
|
| ( 0, 4)
|
f( 0, 4)=5( 0)-4( 4)
|
f(0,4)=-16
|
Looking at the table, we can see that the value of the function in (0,4) is less than the values in the vertices. That means the minimum value of given function does not exist because the values of this function are decreasing as we are moving away from the vertices. However, the maximum value is 39.5 and it occurs at (-2.5,-13).
