Question 1041919
your objective function is to minimize the cost.


set x equal to the number of engines shipped to plant I.
set y equal to the number of engines shipped to plant II.


total cost = 30x + 40y


that's the function you want to minimize.
it costs 30 dollars to ship 1 engine to plant I.
it costs 40 dollars to ship 1 engine to plant II.


set c = total cost.


your objective function is c = 30x + 40y.


your constraints are:


x >= 45
y >= 32


this means you need to ship at least 45 engines to plant I and at least 32 engines to plant II.


x >= 0
y >= 0


this means that the number of engines shipped has to be equal to or greater than 0.


20x + 10y >= 1500


this means that the total rebates provided by the assembly plants has to be greater than or equal to 1500.


x + y < = 120


this means that the total number of engines that can be shipped has to be less than or equal to 120.


summarize your formulas:


objective function:


c = 30x + 40y


constraints:


x + y <= 120
x >= 45
y >= 32
20x + 10y >= 1500



graph the constraint equations and then shade the areas on the graph that satisfy the inequalities.


this is what my graph looks like.


<img src = "http://theo.x10hosting.com/2016/081401.jpg" alt="$$$" </>


i graphed the equations.
i then shaded the areas that satisfied the inequalities.


then you need to find the corner points of the shaded area.


those are:


(45,75)
(45,60)
(59,32)
(88,32)


you then analyze the objective function at those corner points.


objective function is c = 30x + 40y


at (45,75), you get 30*45 + 40*75 = 4350


at (45,60), you get 30*45 + 40*60 = 3750


at (59,32), you get 30*59 + 40*32 = 3050


at (88,32), you get 30*88 + 40*32 = 3920


your minimum cost is when x = 59 and y = 32.


your minimum cost is 3050.


all your constraints need to be satisfied as well.


x + y <= 120 becomes 59 + 32 <= 120 which becomes 91 <= 120 which is true.


x >= 45 becomes 59 >= 45 which is true.


y >= 32 becomes 32 >= 32 which is true.


x and y >= 0 are automatically satisfied when x >= 45 and y >= 32.


20x + 10y >= 1500 becomes 59*20 + 32*10 >= 1500 which becomes 1180 + 320 >= 1500 which becomes 1500 >= 1500 which is true.


all the constraints are satisfied.


your solution is that the minimum cost is 3050.