Example 3.2
Graphical Solution Method
Background Information

To illustrate the graphical approach, we will use a
slightly different scaled down version of Monet’s
product mix problem.

Now there are only two frame types,1 and 2, and
only two scarce resources, labor hours and metal.

The algebraic model is given below:
max 2.25x1 + 2.60x2 (profit objective)
subject to 2x1 + x2  4000 (labor constraint)
x1 + 2x2  5000 (metal constraint)
x1, x2  0 (nonnegativity constraint)
3.1 | 3.1a | 3.3
Background Information –
continued

The objective implies that each type 1 frame
contributes a profit of $2.25, whereas each type 2
frame contributes a profit of $2.60.

The first constraint is a labor hour constraint. There
are 4000 hours available. Each type 1 frame requires
2 labor hours, and each type 2 frame requires 1 labor
hour.

Similarly, the second constraint is a metal constraint.
There are 5000 ounces of metal available. Each type
1 frame requires 1 ounce of metal and each type 2
frame requires 2 ounces of metal.

Find the optimal product mix graphically.
3.1 | 3.1a | 3.3
Solution

The idea is to graph the constraints on a twodimensional graph to see which points (x1, x2) satisfy
all of the constraints. This set of points is labeled the
feasible region. Then we see which point in the
feasible region provides the largest profit.

The graphical solution appears on the next slide.
3.1 | 3.1a | 3.3
Graphical Solution
3.1 | 3.1a | 3.3
Solution – continued

To produce the graph, we first locate the lines where
the constraints hold as equalities.

For example, the line for labor is 2x1 + x2 = 4000. The
easiest way to graph this is to find the two points
where it crosses the axes.

Joining the points (0,4000) and (2000,0), we get the
line where the labor constraint is satisfied exactly,
that is, as an equality.

All points below and to the left of this line are also
feasible; there are these are the points where less
than the maximum number of 4000 labor hours are
used.
3.1 | 3.1a | 3.3
Solution – continued

We indicate the feasible side of the line by the short
arrows pointing down to the left from the labor
constraint line.

Similarly the metal constraint line crosses the axes at
the points (0,2500) and (5000,0), so we join these
two points to find the line where all 5000 ounces of
metal are used.

Finally the points on or below both of these lines
constitute the feasible region. These are the point
below the heavy lines.
3.1 | 3.1a | 3.3
Solution – continued

You can think of the feasible region as all points on or
inside the figure formed by four points: (0,0),
(0,2500), (2000,0), and the point where labor hour
and metal constraint lines intersect.

The next step is to bring profit into the picture. We do
this by constructing “isoprofit” lines – that is lines
where total profit is a constant. Any such line can be
written as 2.25x1 + 2.60x2 = P where P is a constant
profit level. Solving for x2, we can put this equation in
slope-intercept form: x2 = P/2.60 – (2.25/2.60)x1
3.1 | 3.1a | 3.3
Solution – continued

This shows that any isoprofit line has slope –
2.25/2.60, and it crosses the vertical axis at the value
P/2.60. Three of these isoprofit lines appear in the
chart as dotted lines.

Therefore, to maximize profit, we want to move the
dotted line up and to the right until it just barely
touches the feasible region.

Graphically, we can see that the last feasible point it
will touch is the point indicated in the figure, where
the labor hour and metal constraint lines cross.
3.1 | 3.1a | 3.3
Solution – continued

We can then solve two equations in two unknowns to
find the coordinates of this point. They are x1 = 1000
and x2 = 2000, with a corresponding profit of P =
$7450.

Note that if the slope of the isoprofit lines were much
steeper, the the optimal point would be (2000,0). On
the other hand,m if the slope were mush less steep,
the optimal point would be (0,2500).These
statements make intuitive sense.

If the isoprofit lines are steep, this is because the unit
profit from frame type 1 is large relative to the unit
profit from frame type 2.
3.1 | 3.1a | 3.3
Solution – continued

The crucial point, however, is that only three points
can be optimal: (2000,0), (0, 2500), or (1000, 2000),
the three “corner” points [other than (0,0)] in the
feasible region.

The best of these depends on the relative slopes of
the constraint lines and isoprofit lines in the graph.
3.1 | 3.1a | 3.3