Question 453169
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Your objective function is correct, but you are way out to lunch on your constraints.


Your objective function is using *[tex \Large x] and *[tex \Large y] to represent the number of device A and device B units produced, hence for the constraints, *[tex \Large x] and *[tex \Large y] have to mean the same thing.


It takes 3 hours to create each set of components for type A and 7 hours to create the set of components for type B, and you have a total of 182 hours available for component creation, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ +\ 7y\ \leq\ 182]


Likewise, the other two labor hours constraints are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ +\ 4y\ \leq\ 122]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ +\ y\ \leq\ 68]


You should also constrain both *[tex \Large x] and *[tex \Large y] to be non-negative since you cannot manufacture a negative number of things (except maybe if you were using anti-matter -- but that is another problem altogether).


I presume you know how to graph a 2 variable inequality.  The graphs of all of your constraints will create an irregular pentagon region of feasibility.  You will need to solve for the intercepts and the points of intersection.  A fundamental theorem of Operations Research says that a unique optimum solution exists, then it will be one of the vertices of the region of feasibility.


Find the ordered pairs that describe the vertices of your feasible pentagon and test each one in the objective function.  The one that gives the biggest result is your solution.


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