Question 250218
<font face="Garamond" size="+2">


You got the correct answer, but I think you got there by chance.  If you graph the  four constraint inequalities, you will see that you have an area of feasibility described by the quadrilateral with vertices (0,0), (0,90), (20,60), and (50,0).


An elementary Linear Optimization theorem says that the optimum values are at a vertex of the feasibility polygon.  Because of the particular profit values given, the actual optimum for this problem occurs at the intersection of the two resource constraint boundary lines.  But if the profit numbers change, you could have a different result.  Completing this problem requires that you check the value of the Profit function for EACH of the feasibility vertices to verify that indeed you have chosen the one that optimizes profit.


As I mentioned, your answer is correct because:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 108(20)\ +\ 60(60)\ =\ 5760]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 108(0)\ +\ 60(90)\ =\ 5400]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 108(50)\ +\ 60(0)\ =\ 5400]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 108(0)\ +\ 60(0)\ =\ 0]


However, if the profit numbers change you can get a different result.  Let the profit on suits remain at 108, but the profit on dresses increase to 75.  Then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 108(20)\ +\ 75(60)\ =\ 6660]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 108(0)\ +\ 75(90)\ =\ 6750]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 108(50)\ +\ 75(0)\ =\ 5400]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 108(0)\ +\ 60(0)\ =\ 0]


And you would want to manufacture zero suits and 90 dresses.


On the other hand, if the profit on suits were to increase to 130 leaving the profit on dresses at 60, then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 130(20)\ +\ 60(60)\ =\ 6200]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 130(0)\ +\ 60(90)\ =\ 5400]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 130(50)\ +\ 60(0)\ =\ 6500]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 108(0)\ +\ 60(0)\ =\ 0]


So you would want to manufacture 50 suits and zero dresses.


Lesson:  Check all your possible solutions before making a judgement.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>