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Next, assign zero values to SB and SA to reach solution values for point M. Substituting zero values for SA and SB in Equations 9.7 and 9.8 results in two equations with two unknowns:

Multiplying Equation 9.8 by two and subtracting this result from Equation 9.7 provides the solution value for Qx:

Then, substituting 6 for Qx in Equation 9.8 finds QY = 4. Total profit contribution in this case is $108 [= ($12 X 6) + ($9 X 4)].

Similar algebraic analysis provides the solution for each remaining corner of the feasible space. However, rather than work through those corner solutions, the results are shown in Table 9.2. It is apparent, just as illustrated in the earlier graphic analysis, that the optimal solution occurs at point M, where 6 units of x and 4 units of Y are produced. Total profit is $108, which exceeds the profit at any other corner of the feasible space.

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