Max. value of z, subject to 3x + y <= 24, 6x + 4y <= 6
- Thread starter Navyguy
- Start date
Re: Maximum value of z, need help please
Hello, Navyguy!
This is a Linear Programming problem.
Game plan
. . (1) Graph the region.
. . (2) Test the vertices.
Since \(\displaystyle x\,\geq\,0,\:y\,\geq\,0\), we are in Quadrant I.
To graph \(\displaystyle 3x\,+\,y\:\leq\:24\) . . .
Graph the line: \(\displaystyle \:3x\,+\,y\:=\:24\)
. . It has intercepts \(\displaystyle (8,\,0),\
0,\,24)\)
. . Draw the line, shade the region below the line.
To graph \(\displaystyle 6x\,+\,4y\:\leq\: 66\) . . .
Graph the line: \(\displaystyle \:6x\,+\,4y\:=\:66\)
. . It has intercepts \(\displaystyle (11,\,0),\0,\,16.5)\)
. . Draw the line, shade the region below the line.
The region looks like this:
Three of the vertices are obvious: \(\displaystyle \0,\,0),\8,\,0),\0,\,16.5)\)
The fourth vertex is the intersection of: \(\displaystyle \,\left\{\begin{array}{cc}3x\,+\,y\:=\:24 \\ 6x\,+\,4x\:=\:66\end{array}\)
. . Solve the system and get: \(\displaystyle \,(5,\,9)\)
Test the vertices in the \(\displaystyle z\)-function.
\(\displaystyle \begin{array}{cccc}(0,\,0): & z\:=\:20(0)\,+\,8(0) & = & 0 \\
(8,\,0): & z\:=\:20(8)\,+\,8(0) & = & 160 \\
(0,\,16.5): & \;z\:=\:20(0)\,+\,8(16.5) & \;=\; & 132 \\
(5,\,9): & z\:=\:20(5)\,+\,8(9) & = & \underbrace{172}_{\text{max}} \end{array}\)
Therefore, \(\displaystyle x\,=\,5,\:y\,=\,9\,\) produces the maximum \(\displaystyle z\).
Hello, Navyguy!
This is a Linear Programming problem.
Game plan
. . (1) Graph the region.
. . (2) Test the vertices.
Since \(\displaystyle x\,\geq\,0,\:y\,\geq\,0\), we are in Quadrant I.
To graph \(\displaystyle 3x\,+\,y\:\leq\:24\) . . .
Graph the line: \(\displaystyle \:3x\,+\,y\:=\:24\)
. . It has intercepts \(\displaystyle (8,\,0),\
0,\,24)\). . Draw the line, shade the region below the line.
To graph \(\displaystyle 6x\,+\,4y\:\leq\: 66\) . . .
Graph the line: \(\displaystyle \:6x\,+\,4y\:=\:66\)
. . It has intercepts \(\displaystyle (11,\,0),\
0,\,16.5)\). . Draw the line, shade the region below the line.
The region looks like this:
Code:
|
24*
| *
| *
16.5 o *
|:::* *
|:::::::* * (5,9)
|:::::::::::o
|:::::::::::::* *
|:::::::::::::::* *
|:::::::::::::::::* *
|:::::::::::::::::::* *
|:::::::::::::::::::::* *
--o-----------------------o-----------*----
| 8 11Three of the vertices are obvious: \(\displaystyle \
0,\,0),\8,\,0),\0,\,16.5)\)The fourth vertex is the intersection of: \(\displaystyle \,\left\{\begin{array}{cc}3x\,+\,y\:=\:24 \\ 6x\,+\,4x\:=\:66\end{array}\)
. . Solve the system and get: \(\displaystyle \,(5,\,9)\)
Test the vertices in the \(\displaystyle z\)-function.
\(\displaystyle \begin{array}{cccc}(0,\,0): & z\:=\:20(0)\,+\,8(0) & = & 0 \\
(8,\,0): & z\:=\:20(8)\,+\,8(0) & = & 160 \\
(0,\,16.5): & \;z\:=\:20(0)\,+\,8(16.5) & \;=\; & 132 \\
(5,\,9): & z\:=\:20(5)\,+\,8(9) & = & \underbrace{172}_{\text{max}} \end{array}\)
Therefore, \(\displaystyle x\,=\,5,\:y\,=\,9\,\) produces the maximum \(\displaystyle z\).