Question 1131669
<br>
let x = number of standard models
let y= number of deluxe models<br>
Constraints:<br>
# of standard models: x >= 6  (the statement of the problem seems to say x must be EQUAL to 6; I assume it is supposed to mean AT LEAST 6)
assembly time: 30x+30y <= 450  -->  x+y <= 15
inspection time: 10x+15y <= 180  -->  2x+3y <= 36<br>
Graph the constraint boundary lines<br>
{{{graph(400,400,-2,20,-2,20,15-x,12-(2/3)x)}}}<br>
The corners of the feasibility region, remembering that x >= 6, are (6,0), (6,8), (9,6), and (15,0).<br>
Evaluate the objective function at each corner:<br>
(6,0): 40(6)+60(0) = 240
(6,8): 40(6)+60(8) = 720
(9,6): 40(9)+60(6) = 720
(15,0): 40(15)+60(0) = 600<br>
The maximum profit is $720, with EITHER 6 standard models and 8 deluxe models, OR 9 standard models and 6 deluxe models.<br>
(If indeed the intent of the problem was supposed to be that EXACTLY 6 standard models are produced, then of course the solution is at (6,8).)<br>
Note as a refinement of this process, you can determine the corner of the feasibility region with the maximum profit without evaluating the objective function at every corner.  The corner with the maximum profit can be found by comparing the slopes of the constraint lines and the objective function.<br>
In this problem, the constraint lines have slopes of -1 and -2/3; the objective function also has a slope of -2/3.  Because the slope of the objective function is equal to the slope of one of the constraint lines, the maximum profit will be anywhere along the boundary of the feasibility region with a slope of -2/3.<br>
In larger problems, this can give several combinations that yield maximum profit; in this problem, where the numbers are small and have to be integers, we find exactly two solutions.<br>
In more complex linear programming problems, where the feasibility region has several corners, this can save a lot of time in finding the answer.