Exercise 24 Page 162

We have been given a feasible region with an objective function $P=x+2y.$

We will substitute the vertices into the objective function and determine the maximum value of $P.$ Let's begin by the point $B(10,0).$ The value of $P$ at $B(10,0)$ is $10.$ We can find the value of $P$ for the other vertices in the same way.

Vertex	Objective Function	$P$
$B(10,0)$	$P=10+2(0)$	$10$
$C(10,3)$	$P=10+2(3)$	$16$
$D(4,6)$	$P=4+2(6)$	$16$
$E(0,6)$	$P=0+2(6)$	$12$

At the points $C$ and $D,$ the value of $P$ is $16$ which is maximum. This means that the objective function passes through the points $C$ and $D.$ Therefore, the points on the segment $\overline{CD},$ yield the maximum profit. Let's choose two more point on the segment.

Let's check the points!

Vertex	Objective Function	$P$
$F(8,4)$	$P=8+2(4)$	$16$
$G(6,5)$	$P=6+2(5)$	$16$

The points $F$ and $G$ also yield the maximum profit. As a result, we have four points $C,D,F,G$ that yield the maximum profit.