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≥|x+1|-2 & (I) 0≤ y≤ 6 & (II) -6≤ x≤ 2 & (III) x+3y≤14 & (IV)
The boundary line of an inequality can be determined by replacing the inequality symbol with an equal sign.
Inequality:& y ≤ |x+1|-2
Boundary Line:& y = |x+1|-2
The graph of this function is the graph of y=|x| after a few transformations. Let's first figure out which transformations were involved so that we can graph it.
In order to decide which part of the plane to shade, we can test a point which is not on the boundary line. Let's test the point ( 0, 0).
If the point satisfies the inequality, we shade the region that contains the point. Otherwise, we shade the region that does not contain the point.
Since the point satisfies the inequality, we will shade the region that contains the point.
Inequality II
Again to graph a compound inequality, we can separate it into two cases.
Compound Inequality:& 0 ≤ y ≤ 6
Case I:& 0 ≤ y
Case II:& y ≤ 6
Let's draw the graph of each case.
Case I
The inequality 0 ≤ y describes all values of y that are greater than or equal to 0. This means that every coordinate pair with an y-value that is greater than or equal to 0 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 6, every coordinate pair with an y-value that is less than or equal to 6 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 III
To graph a compound inequality, we can separate it into two cases.
Compound Inequality:& -6 ≤ x ≤ 2
Case I:& -6 ≤ x
Case II:& x ≤ 2
To make things a bit more simple, we will draw the graph of each case and then combine them.
Case I
The inequality -6 ≤ x describes all values of x that are greater than or equal to -6. This means that every coordinate pair with an x-value that is greater than or equal to -6 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 x that are less than or equal to 2, every coordinate pair with an x-value that is less than or equal to 2 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 IV
Let's again start by changing the inequality symbol in the last inequality to an equals sign.
Inequality:& x+3y≤14
Boundary Line:& x+3y=14
The second step will be to rewrite the equation in the .
x+3y=14
3y=- x+14
y=-1/3x+14/3
x+3y≤14
0+3( 0)? ≤14
0 ≤ 14 ✓
Combining the Inequality Graphs
Let's draw the graphs of the inequalities on the same coordinate plane.
Now that we can see the overlapping region, let's highlight the vertices.
Looking at the graph, we can write down all seven vertices.
( -3, 0), ( -6, 3), ( -6, 6), ( -4, 6),
( 2, 4), ( 2, 1) and ( 1, 0)
Having all the vertices we can find the maximum and minimum of the given function.
Finding the Maximum and Minimum Values
Substituting the vertices into the given function, f(x,y)=5x+4y, we will determine its maximum and minimum values. Let's start with the vertex ( -3, 0).
f(x,y)=5x+4y
f( -3, 0)=5( -3)+4( 0)
f(-3,0)=-15+0
f(-3,0)=-15
| Vertex
|
f(x,y)=5x+4y
|
Value
|
| ( -3, 0)
|
f( -3, 0)=5( -3)+4( 0)
|
f(-3,0)=-15
|
| ( -6, 3)
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f( -6, 3)=5( -6)+4( 3)
|
f(-6,3)=-18
|
| ( -6, 6)
|
f( -6, 6)=5( -6)+4( 6)
|
f(-6,6)=-6
|
| ( -4, 6)
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f( -4, 6)=5( -4)+4( 6)
|
f(-4,6)=4
|
| ( 2, 4)
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f( 2, 4)=5( 2)+4( 4)
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f(2,4)=26
|
| ( 2, 1)
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f( 2, 1)=5( 2)+4( 1)
|
f(2,1)=14
|
| ( 1, 0)
|
f( 1, 0)=5( 1)+4( 0)
|
f(1,0)=5
|
Looking at the table, we can see that the maximum value of given function is 26 and it is reached at (2,4). The minimum value is -18 and it occurs at (-6,3).
