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

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) .

P = x + 2 y

SubstituteII

$x = 10$ , $y = 0$

P = 10 + 2 (0)

ZeroPropMult

Zero Property of Multiplication

P = 10 + 0

AddTerms

Add terms

P = 10

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 $\overline{C D},$ 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.

Exercise