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.
-3≤ y≤ 7 & (I) 4y≥4x-8 & (II) 6y+3x≤24 & (III)
Inequality I
To graph a , we can separate it into two cases.
Compound Inequality:& -3 ≤ y ≤ 7
Case I:& -3 ≤ y
Case II:& y ≤ 7
To make things a bit more simple, we will draw the graph of each case and then combine them.
Case I
The inequality -3 ≤ y describes all values of y that are greater than or equal to -3. This means that every coordinate pair with an y-value that is greater than or equal to -3 will be included in the shaded region. Notice that the inequality is not strict, so the boundary line will be solid.
Case II
Since the inequality represents the values of y that are less than or equal to 7, every coordinate pair with an y-value that is less than or equal to 7 will be included in the shaded region. The boundary line will be solid because the inequality is non-strict.
Combining the Cases
Let's draw the graphs of both cases on the same coordinate plane and shade the overlapping region. This will give us the graph of the compound inequality.
Inequality II
Let's start by changing the inequality symbol in the second inequality to an equals sign.
Inequality:& 4y≥4x-8
Boundary Line:& 4y=4x-8
The second step will be to rewrite the equation in the .
Now, we can determine its m and b to draw the line.
Slope-Intercept Form:& y= mx+b
Boundary Line:& y= 1x+(-2)
Let's use the slope and y-intercept 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.
4y≥4x-8
4( 0)? ≥4( 0)-8
0? ≥0-8
0≥-8 ✓
Inequality III
Let's again start by changing the inequality symbol in the last inequality to an equals sign.
Inequality:& 6y+3x≤24
Boundary line:& 6y+3x=24
Next we need to rewrite the equation in the .
6y+3x=24
6y=-3x+24
y=-3/6x+4
y=-1/2x+4
6y+3x ≤ 24
6( 0)+3( 0) ? ≤ 24
0 ≤ 24 ✓
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 lines intersect at the points (-1,-3), (4,2), and (-6,7). These are our vertices. Notice that the feasible region is unbounded. This means that it can go on forever in the negative vertical direction and there are no other vertices.
( -1, -3), ( 4, 2), and ( -6, 7)
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 (-9,6).
Let's start with calculating the function value in the vertex ( -1, -3).
f(x,y)=-12x+9y
f( -1, -3)=-12( -1)+9( -3)
f(-1,-3)=12-27
f(-1,-3)=-15
| Vertex or Testing Point
|
f(x,y)=-12x+9y
|
Value
|
| ( -1, -3)
|
f( -1, -3)=-12( -1)+9( -3)
|
f(-1,-3)=-15
|
| ( 4, 2)
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f( 4, 2)=-12( 4)+9( 2)
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f(4,2)=-30
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| ( -6, 7)
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f( -6, 7)=-12( -6)+9( 7)
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f(-6,7)=135
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| ( -9, 6)
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f( -9, 6)=-12( -9)+9( 6)
|
f(-9,6)=162
|
Looking at the table, we can see that the value of the function in (-9,6) is greater than the values in the vertices. That means the maximum value of given function does not exist because the values of this function are increasing as we are moving away from the vertices. However, the minimum value is -30 and it occurs at (4,2).
