Welcome!
In this tutorial, we will graphically solve this minimization problem using the Objective
Function Line Approach.
Note that the use of the graphical approach is limited to 2 decision variables.
We start by finding points for the constraint lines.
Let’s label them C1 to C3 for reference purposes.
For Constraint 1) when X = 0, Y = 2 And when Y = 0, X = 6.
For Constraint 2) when X = 0, Y = 5.
And when Y = 0, X = 2.
For Constraint 3) Y is just always less than or equal to 4 irrespective of the value of
X.
In drawing the graph, note here that the highest value of X is 6 and the highest value of Y
is 5.
And that is what tells us how far we need to go on the X and Y axes.
You can see that I went a little bit higher than 6 on the X axis, and a little more than
5 in the Y axis because I have the space.
So drawing the line for the first constraint, we have the points (0, 2) and (6, 0)
which we join to form the constraint line.
For the second constraint, we have (0, 5) and (2, 0)
which we also join to form the constraint line.
For Y ≤ 4, the line is simply at Y = 4.
Now since the first 2 constraints are “greater than” constraints,
they will be satisfied in the region above the lines away from the origin.
And Y ≤ 4 is simply satisfied below the line.
The directions of the arrows show that the region satisfying all the constraints at the
same time (called the feasible region) is this area shaded in blue.
Now the optimal solution to this problem will occur at an extreme or corner point of this
feasible region.
Let’s label them A, B, and C for reference purposes.
Now recall that the objective function is to Minimize 5X + 7Y.
So the corner point that gives us the minimum value of the objective function is the optimal
point.
Instead of testing every single point, we will use the objective function line approach.
We begin by setting the objective function to any number of our choice.
To make the process easy, we can simply multiply the coefficients of X and Y.
In this example we have 5 times 7 which equals 35.
35 seems to be appropriate for this problem.
Note that if this number is too small or too large, you can always multiply or divide it
by an appropriate value.
Next we find points to draw the objective function line.
When X = 0, Y = 5 And when Y = 0, X = 7.
So the objective function line is this blue dotted line here.
Recall that is this a minimization problem.
So to use this line to determine the best extreme point, we slowly move it down towards
the origin parallel to itself.
If this were a maximization problem, we would move the line upwards away from the origin.
The last extreme point the line touches before leaving the feasible region is the optimal
solution point.
So let’s slowly move the objective function line parallel to itself.
You can see here that it passes points A and C, so it is obvious that Point B is the optimal
solution point.
We now need to solve the two lines that intersect at point B simultaneously to determine the
coordinates at that point.
Here are the equations.
Let’s use the substitution method.
From C1 we can write X in terms of Y to give X = 6 – 3Y.
We can then substitute that expression into C2.
And that simplifies to 30 - 15Y + 2Y = 10, which gives -13Y = -20
And that gives Y = 1.54.
Substituting that Y value in the X equation we have X = 6 - 3(1.54).
That is, X = 1.38.
So the coordinates at point B are X = 1.38 and Y = 1.54.
Let’s now plug the point into the objective function.
We have 5(1.38) + 7(1.54) which gives 17.7.
So the optimal solution occurs at point B and the corresponding minimum value of the
objective function is 17.7.
See you in the next video.
And thanks for watching.