Riemann surface
matlab
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Example 3: Given the objective function $C=12 x+4 y$ and the following feasible set,
A. Find the maximum value.
B. Find the minimum value.
.
Solution: Notice that the feasible set is unbounded. This means that there may or may not be an optimal solution which results in a maximum or minimum function value. The vertices (corner points) of the feasible set are $(0,7),(4,2)$, and $(8,0)$.
We substitute each corner point into the objective function, as shown in the chart below.
Example 4: Use the graphical method to solve the following linear programming problem.
$$
\begin{aligned}
& \text { Maximize } R=4 x+11 y \
& \text { subject to: }\left{\begin{array}{l}
x+y \leq 3 \
2 x+y \leq 4 \
x \geq 0 \
y \geq 0
\end{array}\right.
\end{aligned}
$$
Solution: We need to graph the system of inequalities to produce the feasible set. We will start by rewriting each inequality as an equation, and then number the equation for each line.
$$
\begin{aligned}
x+y & =3 \
2 x+y & =4 \
x & =0 \
y & =0
\end{aligned}
$$
We want to graph each of the lines and determine the proper shading for their respective inequalities.
To graph Line (1), we will find its intercepts.
$$
\begin{array}{ll}
x+y=3 & x+y=3 \
x+0=3 & 0+y=3 \
x=3 & y=3
\end{array}
$$
The $x$ – and $y$-intercepts for Line (1) are $(3,0)$ and $(0,3)$, respectively. Since the inequality $x+y \leq 3$ contains an equal sign, a solid line can be drawn through those two intercepts. We need to choose a test point to substitute into the original inequality to determine which half-plane to shade. We will choose the point $(0,0)$ :
$$
\begin{aligned}
& x+y \leq 3 \
& 0+0 \stackrel{?}{\leq} \
& 0 \leq 3 \
&
\end{aligned}
$$
The point $(0,0)$ satisfies the inequality, so we will shade the half-plane containing $(0,0)$.
To graph Line (2), we will find its intercepts.
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