Graph the given system and determine the vertices of the overlapping region. Substituting these vertices into the equation will help you find the maximum value.
P=12 is maximized at (0,4)
Practice makes perfect
Our first step in finding the maximum value of the given equation is to graph the system and determine the vertices of the overlapping region. Substituting these vertices into the equation, we will find the maximum value.
x+y≤ 5 & (I) x+2y ≤ 8 & (II) x≥ 0,y≥ 0 & (III)
We can write the equation of the boundary line by changing the inequality symbol to an equals sign.
Inequality:& y≤- x+5
Boundary Line:& y=- x+5
Since this equation is in slope-intercept form, we can determine its slope m and $y\text{-}$intercept b to draw the line.
Slope-Intercept Form:& y= mx+ b
Boundary Line:& y= -1x+ 5
Now that we know the slope and the y-intercept, let's use them 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.
We can write the equation of the boundary line by changing the inequality symbol to an equals sign.
\begin{aligned}
\textbf{Inequality: }& \ y\leq\N\dfrac{1}{2}x+4 \\
\textbf{Boundary Line: }& \ y=\N\dfrac{1}{2}x+4
\end{aligned}
Since this equation is in slope-intercept form, we can determine its slope $\colIV{m}$ and $y\text{-}$intercept $\colVI{b}$ to draw the line.
\begin{aligned}
\textbf{Slope-Intercept Form: }& \,\ y= \colIV{m}x+\colVI{b}\\
\textbf{Boundary Line: }& \ y= \colIV{\N\dfrac{1}{2}}x+\colVI{4}
\end{aligned}
Now that we know the slope and $y\text{-}$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 $(\col{0},\colII{0}).$ If the point satisfies the inequality, we will shade the region that contains the point. Otherwise, we will shade the other region.
Since the point satisfies the inequality, we will shade the region that contains the point.
Inequality III
Notice that for the third inequality the boundary lines are $x=0$ and $y=0.$ Therefore, they overlap with the $x\text{-}$ and $y\text{-}$axis.
Our inequalities are $x\geq 0$ and $y\geq 0.$ This means that every coordinate pair with $x\text{-}$ and $y\text{-}$value greater than or equal to $0$ needs to be included in the shaded region. This indicates that we should shade the first quadrant.
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 see that the line $y=0$ intersects with the lines $x=0$ and $x=5$ at the points $(0,0)$ and $(5,0),$ respectively. These are two of the vertices.
\begin{gathered}
(\col{0},\colII{0}) \text{ and } (\col{5},\colII{0})
\end{gathered}
We know the $x\text{-}$coordinates of the other vertices are $\col{0}$ and $\col{2}.$ We will find their $y\text{-}$coordinates by substituting them into the equation $y=\N\frac{1}{2}x+4.$ Let's do it!
For the vertex $(\col{0},\colII{0}),$ the value of the function is $0.$ We can determine the values of the function for the other vertices in the same way.
Vertex
$P=x+3y$
Value
$(\col{0},\colII{0})$
$P=\col{0}+3(\colII{0})$
$P=0$
$(\col{5},\colII{0})$
$P=\col{5}+3(\colII{0})$
$P=5$
$(\col{0},\colII{4})$
$P=\col{0}+3(\colII{4})$
$\colIII{P=12}$
$(\col{2},\colII{3})$
$P=\col{2}+3(\colII{3})$
$P=11$
Looking at the table, we can see that the maximum value of the equation is $P=12.$
