We have been given four constrains and an objective function.
C=6x+9y⎩⎪⎪⎨⎪⎪⎧x+2y≥502x+y≥60x≥0, y≥0
Let's begin by graphing Constraint III and IV. We should first determine their by replacing the sign with the equals sign.
III: IV: Constraintx≥0y≥0Boundary Linex=0y=0
The boundary lines of the first two constraints are the axes. Since both
x and
y are
greater than or equal to 0, we will shade Quadrant I. Notice that the inequalities are , so the boundary lines will be
solid.
Next, we will graph Constraint I and II by finding their boundary lines.
I:II:Constraintx+2y≥502x+y≥60Boundary Linex+2y=502x+y=60
To graph their boundary lines, we will find the
x- and
y-intercepts of the lines. We will first substitute
0 for
x into Boundary Line I and solve it for
y to find its .
The
y-intercept of Boundary Line I is the point
(0,25). Proceeding in the same way, we can find the intercepts of both lines.
Constraint
|
Boundary Line
|
x=0
|
y-intercept
|
y=0
|
x-intercept
|
I
|
x+2y=50
|
0+2y=50
|
(0,25)
|
x+2(0)=50
|
(50,0)
|
II
|
2x+y=60
|
2(0)+y=60
|
(0,60)
|
2x+0=60
|
(30,0)
|
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
(0,0) 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 of Constraint I and II. To determine it we will first write a .
{2x+y=60x+2y=50
To solve the system, we will equate the coefficients of
x and use the .
{2x+y=60x+2y=50
{4x+2y=120x+2y=50
{4x+2y−(x+2y)=120−50x+2y=50
{4x+2y−x−2y=120−50x+2y=50
{3x=70x+2y=50
⎩⎪⎨⎪⎧x=370x+2y=50
The
x-coordinate of the point is
370. By substituting
370 for
x into the equation, we can find the
y-coordinate.
⎩⎪⎨⎪⎧x=370x+2y=50
⎩⎪⎪⎨⎪⎪⎧x=370370+2y=50
{x=3702y=380
{x=370y=680
(I), (II): Calculate quotient
{x=23.33y=13.33
Then, the third vertex is
(23.33,13.33). We can write this point as a mixed number
(2331,1331).
The coordinates of the point
((2331,1331) can be rounded to
(24,14),(24,13) and
(23,14). Let's begin by substituting the point
(24,14) into the objective function and determine at which point it is minimum.
C=6x+9y
C=6(24)+9(14)
C=144+126
C=270
The value of
C at
(24,14) is
270. We can find the value of
C for the other rounded coordinates in the same way.
Vertex
|
Objective Function
|
C
|
(24,14)
|
C=6(24)+9(14)
|
270
|
(24,13)
|
C=6(24)+9(13)
|
261
|
(23,14)
|
C=6(23)+9(14)
|
264
|
At the point (24,13), C has the minimum value of 261.