Model Solution
The Graphical Method
—Determining the Optimal Solution—
Part 3. Objective function
The final part of the model to be graphed is the objective function. Objective functions are always equations, so the plot will be a line, not a half-plane. In the Lego problem, the objective function is:
Max 16x + 10y.
Now, the algebraic portion of this expression (omitting for now the logical operator, "Max") generates profits, and can be restated as:
16x + 10y = Z
where Z is a variable dependent on the values assumed by x and y. This profit equation conforms to the standard linear equation except for the fact that the third parameter, c , is not a constant but a variable, Z. We need a numeric constant in order to be able to plot the line.
Clearly, the value of Z varies according to the values assumed by x and y. (Or to put it in economic terms, the profit varies according to the number of tables and chairs produced.) But regardless of the profit level Z, the slope of the profit equation remains constant:
m = -a / b = -16 / 10 = -1.6.
In effect, the profit equation stands for an entire family of lines, all of which share the same slope. The only impact different values of Z have on this family of lines is to determine different y -intercepts, and hence different particular lines, since B = c / b = Z / 10.
Given this state of affairs, we would like to plot one line of this family —any line— to see how its slope relates to the feasible region. Our concern is not to determine the optimal line at the outset but to get a qualitative appreciation of how an arbitrary profit function interacts with the set of feasible solutions. The beauty of coordinate geometry is that once we see the general pattern existing between profit lines and feasible region, the optimal solution can be visually ascertained. And since coordinate geometry links the graph to specific algebraic equations, the optimal solution will be readily obtained.
Now, to plot an arbitrary profit line we need to assign an arbitrary value to Z. Any number will do. This being the case, the wise analyst chooses the easiest option. The easiest option is to let Z = ab, the product of the first two parameters, in our case: 16 × 10. The reason for choosing this particular number is that in order to plot the profit line, we need two points. Watch this. We have the following profit equation:
16x + 10y = (16)(10).
Let’s first determine the y -intercept. This means that x will be equal to zero. But if x = 0, the equation simplifies to:
10y = (16)(10).
Since 10 appears on both sides of the equation, it is canceled out, leaving:
y = 16.
Thus, the coordinates for the y -intercept are (0, 16). By the same token, the coordinates for the x -intercept are (10, 0). The upshot is that if we let Z = ab, the two intercepts will have coordinates (b, 0) and (0, a ). No calculation required. Sure beats work.
The “switcheroo” technique can sometimes yield intercepts that are too large or too small in relation to the size of the graph. Never fear. If the points are off scale, divide the coordinates by an appropriate multiple of ten (ten itself included) to rescale them. (If they are too small, multiply them by a multiple of ten. Other multiple values are acceptable as well.) In our case, we handily come up with the rescaled coordinates (0, 1.6) and (1, 0), which place the (red) profit line nicely within the feasible region:
The red line is called an isoprofit line because all of its points have coordinate values that when plugged into the objective function yield the same amount of profit, in this case, Z = 16. However, it is clear that $16 is not the maximum possible profit for our problem. All of the feasible points to the right of this isoprofit line yield a higher profit because their x and y coordinates have greater values — meaning, greater production of tables and chairs. Now consider, for illustration purposes, doubling production. Profits would then double to Z = 32:
Thus, increasing production (greater x and y values) shifts the isoprofit line in a northeasterly direction. The new isoprofit line Z = 32 is still not optimal because there remain feasible points lying to its right. Clearly, to reach optimality the line would have to be shifted until it was tangent to the edge of the feasible region. This is the insight provided by coordinate geometry: once an arbitrary isoprofit line is plotted, all one has to do to determine the optimal solution is to slide the line in the direction of optimality to the very edge of the feasible region. The last feasible point touched by the shifted red line (a boundary point) is the optimal solution.
Optimal solutions must lie on the boundary of feasible regions. An interior point cannot be optimal because some neighboring points will be feasible and improve the value of the objective function. The only way to avoid having neighboring feasible points in the direction of value improvement is to reach the last feasible point on the boundary of the feasible region.
We have visually ascertained the optimal point. To determine its exact coordinates and therefore the solution to the problem, we solve the system of simultaneous equations corresponding to the constraint lines that intersect and give rise to the optimal point. In our case, the system is:
2x + y = 6
2x + 2y = 8
whose solution is x = 2, y = 2, as depicted in the illustration below. Substituting these values in the objective function yields the value of the problem, Z = $52, the maximum possible profit given the resource constraints.
Math First Aid
MathTools provides links to sites with online software for solving systems of simultaneous equations and many other math problems.
CowPi features a small system solver. They also have Edwin A. Abbott's classic, Flatland: A Romance of Many Dimensions, which is required reading for every educated person.
Mark Chamness has a pretty cool 3 x 3 Linear Equation Solver.
Math-n-stuff's page, Solving Systems of Equations is very informative. They also have a bunch of other stuff.
RIOT provides an online LP Solver for small-sized problems (max 10 variables, 10 constraints). No graphics.
Université de Nice Sophia Antipolis offers a free-form linear solver, among other interesting things.
Here's a tutorial for solving 2 x 2 linear systems.
And Ben Langton's QuickMath (click the nifty logo below) does just about anything. It is powered by Mathematica.
Linear System of Equations
Simultaneous Equations
System of Equations
Simultaneous Equations
System of Linear Equations
