SOLUTION: A bed manufacturer makes two types of bed: standard and luxury. The cost of manufacturing each type of bed is 2300$ for a standard model and 3700$ for a luxury model. It costs 300

Let X = # of standard beds;
    Y = # of luxury   beds.


The objective (profit) function is F(X,Y) = 6X + 8Y, in thousand dollars.


The constraints are

2.3X + 3.7Y <= 851      (1)    (maximum weekly cost, in thousand dollars)

0.3X + 0.4Y <= 120      (2)    (maximum weekly shipping, in thousand dollars)

X + Y <= 300            (3)

X >= 0,  Y >= 0.        (4)    (non-negativity)


The problem is to maximize the objective function under the given restrictions.


The feasible domain is shown in the Figure below.

It is a quadrilateral in QI under the red, green and blue lines - factually, under the red and blue lines.






Plot  2.3X + 3.7Y = 851 (red),  0.3X + 0.4Y = 120 (green)  and  X + Y <= 300 (blue)


The maximum (the solution to the problem) is achieved in one of the three corner points:

    P1 = (0,230)      (red line Y-intercept)

    P2 = (185,115)    (red line and blue line intersection point)

    P3 = (300,0)      (blue line X-intercept)


Now, calculate the value of the objective function at each of this three corner points

    at P1:  F(0,230)   = 6*0 + 8*230   = 1840 thousand dollars;

    at P2:  F(185,115) = 6*185 + 8*115 = 2030 thousand dollars;   and  

    at P3:  F(300,0)   = 6*300 + 8*0   = 1800 thousand dollars.


The maximum is achieved at P2, and this point gives the solution.


ANSWER.  The maximum profit is achieved when  185 standard beds and 115 luxury beds are produced per week.

         The maximum profit then is 2030 thousand dollars per week.