Question 532189
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First let's get your terminology straight.  *[tex \Large Z\ =\ 5x\ +\ 6y] is NOT the maximum profit.  It is the profit when *[tex \Large x] items having a profit of 5 each and *[tex \Large y] items having a profit of 6 each are sold.  Your goal is to maximize this profit amount.


Second, don't mix capital letters and lower case letters for your variables (*[tex \Large x] and *[tex \Large X] for example).  They are different variables and stand for different quantities.  And I'm sure that is NOT what you meant when you wrote out the problem.


Part a).  Graph all four of your constraint inequalities: *[tex \Large 2x\ +\ y\ \leq\ 120], *[tex \Large 2x\ +\ 3y\ \leq\ 120], *[tex \Large x\ \geq\ 0], and *[tex \Large y\ \geq\ 0].  Hint:  Ordinarily when you graph an inequality you shade in the half-plane that represents the solution set.  If you do that here, your area of feasibility will be that area where the four shaded areas (one each from each of your constraint inequalities) overlap.  However, if you do the shading using the opposite sense, that is shade in that part of the plane NOT in the solution set, then the area of feasibility ends up being totally unshaded and therefore much easier for visual comprehension.


Either way, the optimum solution, since none of the slopes of the constraint boundaries are equal to the slope of the function you are trying to optimize, you know that the optimum solution will be one of the four vertices of the feasibility quadrilateral.


So, find the four ordered pairs that represent the four vertices of the feasibility quadrilateral and evaluate the profit function for the values of the coordinates of each of those points.  The optimum solution will make the function have the larger value.


Part b) and c) Change your profit function to *[tex \Large Z_b\ =\ 8x\ +\ 6y] and *[tex \Large Z_c\ =\ 3x\ +\ 6y].  Then evaluate each of these new profit functions using the same potential optimum points as you used for part a.  See if the change in the profit function causes the optimum point to change.


Write back and we can negotiate a price for a full graphical solution if you are interested.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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