We have been given four constrains and an objective function.

C = 6 x + 9 y ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + 2 y \geq 50 2 x + y \geq 60 x \geq 0, y \geq 0

Let's begin by graphing Constraint III and IV. We should first determine their boundary lines by replacing the inequality sign with the equals sign.

III : IV : \underline{Constraint} x \geq 0 y \geq 0 \underline{Boundary Line} x = 0 y = 0

The boundary lines of the first two constraints are the axes. Since both

x

and

y

are greater than or equal to

0,

we will shade Quadrant I. Notice that the inequalities are non-strict, so the boundary lines will be solid.

Next, we will graph Constraint I and II by finding their boundary lines.

I : II : \underline{Constraint} x + 2 y \geq 50 2 x + y \geq 60 \underline{Boundary Line} x + 2 y = 50 2 x + y = 60

To graph their boundary lines, we will find the

x -

and

y -

intercepts of the lines. We will first substitute

0

for

x

into Boundary Line I and solve it for

y

to find its

y -

intercept.

x + 2 y = 50

Substitute

$x = 0$

0 + 2 y = 50

AddTerms

Add terms

2 y = 50

DivEqn

$LHS / 2 = RHS / 2$

y = 25

The

y -

intercept of Boundary Line I is the point

(0, 25) .

Proceeding in the same way, we can find the intercepts of both lines.

Constraint	Boundary Line	$x = 0$	$y -$ intercept	$y = 0$	$x -$ intercept
I	$x + 2 y = 50$	$0 + 2 y = 50$	$(0, 25)$	$x + 2 (0) = 50$	$(50, 0)$
II	$2 x + y = 60$	$2 (0) + y = 60$	$(0, 60)$	$2 x + 0 = 60$	$(30, 0)$

Now we will plot the intercepts and connect them with line segments. The lines will be bound by the axes because of Constraint III and IV. Also, notice that the constraints are non-strict. Therefore, the lines will be solid.

Next, we will test the point

(0, 0)

to determine which region we should shade. Let's start with Constraint I.

x + 2 y \geq 50

SubstituteII

$x = 0$ , $y = 0$

0 + 2 (0) \geq ? 50

ZeroPropMult

Zero Property of Multiplication

0 ≱ 50

For Constraint I, we will shade the region that does not contain the point.

Now we will test the point for Constraint IV.

2 x + y \geq 60

SubstituteII

$x = 0$ , $y = 0$

2 (0) + 0 \geq ? 60

ZeroPropMult

Zero Property of Multiplication

0 ≱ 60

Therefore, we will again shade the region that does not contain the test point.

The overlapping section of the graph above is the feasible region. Let's remove the unnecessary parts and indicate the vertices of the feasible region.

The coordinates of two of the three vertices are determined. The third vertex is the point of intersection of Constraint I and II. To determine it we will first write a system of equations.

Exercise