Question 1023709
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The general idea is to graph all of the inequalities and the area where they ALL overlap is called the "feasibility area".  Any ordered pair in the feasibility area will satisfy ALL of the constraints, and, if there is an optimum solution, it will be one of the feasibility area polygon vertices (what you called corner points).


The problem with graphing all of the inequalities as they are written is that it gets difficult to tell exactly where they all overlap.  To mitigate this difficulty I graph the inequalities in the OPPOSITE sense which results in the feasibility polygon having no shading; therefore much easier to see.


The constraint inequalities that you have are all correct except for two of them.  Yes, it is indeed true that you are contracting to haul 50 people, but this constraint needs to be couched as an inequality that says you have contracted to haul AT LEAST 50 passengers.  Also, since the MOST drivers you can have is 5, you need *[tex \Large x\ +\ y\ \leq\ 5]


You also need to constrain both *[tex \Large x] and *[tex \Large y] to the positive integers.  You can't have a negative number of either type of bus, nor can you have a fractional number of buses of either type.  Hence, your constraints are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ \leq\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ \leq\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ \in\ \mathbb{Z}\ :\ x\ \geq\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ \in\ \mathbb{Z}\ :\ y\ \geq\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ \leq\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15x\ +\ 10y\ \geq\ 50]


Now, when you go to graph these constraints, just flip the inequality symbol over.  I graphed all but the last constraint in the first diagram and then added the last one in the other figure.


*[illustration LinearOptimization_(3)_1.jpg]


*[illustration LO_3_2]


Note the white quadrilateral that is formed.  This is your feasibility polygon.  Any points wholly within or on the boundaries of this polygon that have INTEGER coefficients are feasible solutions.  Constraining the solutions to the integers changes the rules about where to find the optimum solution.  In this case, you actually have four feasible points.


You did state the Objective Function correctly.  All you need to do is substitute the *[tex \Large x] and *[tex \Large y] values of each of the four points into the objective function and do the arithmetic.  Since your objective is to minimize your objective function value, choose the point that gives you the smallest value of the objective function.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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