Exercise 15 Page 161

Let's begin by indicating the constraints on the graph. The error is that the original solver is not considering the constraint $y\le x+3.$ Let's determine the boundary line of the missing constraint to graph it. Since the boundary line is in slope-intercept form, we will plot its $y\text{-}$intercept and draw the line. Remember that we have two constraints $y\ge$ and $x\ge 0.$ Therefore, the boundary line will be bound by the axes. Next, we will test the point $(0,0)$ to decide which region we should shade. Since the point satisfies the constraint, we will shade the region that contains the point.

The overlapping section of the constraints will be the feasible region. Let's remove the unnecessary parts and indicate the vertices of the feasible region.

To find the maximum value of $P,$ we will substitute the vertices into the objective function. Let's begin with the vertex $(0,0).$ For the vertex $(0,0),$ the value of $P$ will be $0.$ We can find the value of $P$ absorption for the other vertices proceeding in the same way.

Vertex	Objective Function	Absorption (lb/yr)
$(0,0)$	$P=\text{-}(0)+3(0)$	$0$
$(0,3)$	$P=\text{-}(0)+3(3)$	$9$
$(1,4)$	$P=\text{-}(1)+3(4)$	$11$
$(3,0)$	$P=\text{-}(3)+4(0)$	$\text{-}3$

Therefore, the maximum value of $P$ is $11$ at $(1,4).$