Precalculus (6th Edition) Blitzer

The maximum value of the given function is $26$.
Let us consider the inequalities and substitute the equals symbol in place of the inequality to rewrite the equation as given below: Find two solutions of the linear equation to plot the graph of the linear equation $x+y=6$ To find the value of the x-intercept, put y = 0 as given below: \begin{align} & x+y=6 \\ & x+0=6 \\ & x=6 \end{align} To find the value of the y-intercept, substitute x = 0 as follows: \begin{align} & 0+y=6 \\ & y=6 \end{align} Therefore, plot the intercepts $\left( 6,0 \right)\text{ and }\left( 0,6 \right)$ and draw a solid line passing through these points because the inequality contains the $\le$ symbol in which the equality is included. Now, this solid line divides the plane into three regions: the line itself and two half planes. Then, take origin $\left( 0,0 \right)$ as a test point and check the region in the graph to shade: \begin{align} x+y\le 6 & \\ 0+0\overset{?}{\mathop{\le }}\,6 & \\ 0\le 6 & \\ \end{align} Since the test point satisfies the inequality, shade the half plane containing this point, that is, the origin $\left( 0,0 \right)$. Now, to plot the graph of the inequality $x\ge 0$, graph the line $x=0$ which is a vertical line parallel to the y-axis and shade the right part as the inequality contains the $\ge$ symbol. Similarly, to plot the graph of the inequality $y\ge 0$, graph the line $y=0$ which is a horizontal line parallel to the x-axis and shade the upper part as the inequality contains the $\ge$ symbol. Similarly, to plot the graph of $x\ge 2$, graph the line $x=2$ and shade the right part as the inequality contains the $\ge$ symbol. So, the graph of these inequalities is the common shaded region. Substitute the value of $x=2$ in the equation $x+y=6$ to obtain the value of the y-intercept: \begin{align} & 2+y=6 \\ & y=6-2 \\ & y=4 \end{align} Therefore, the intersection point of $x+y=6$ and $x=2$ is $\left( 2,4 \right)$. And the corner points of the given graph are $(2,0),\ (6,0),\ \text{ and }\ \left( 2,4 \right)$. The value of the objective function at each corner point can be computed as given below: \begin{matrix} \text{Corner points} & \begin{align} & \text{Value of} \\ & z=3x+5y \\ \end{align} & {} \\ \left( 2,0 \right) & 3\left( 2 \right)+5\left( 0 \right)=6 & {} \\ \left( 6,0 \right) & 3\left( 6 \right)+5\left( 0 \right)=18 & {} \\ \left( 2,4 \right) & 3\left( 2 \right)+5\left( 4 \right)=26 & \text{Max} \\ \end{matrix} Thus, the maximum value of the objective function is $26$ at the corner point $\left( 2,4 \right)$. 