Riemann surface
matlab

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.

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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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