Thomas H. answered • 09/15/24
Calc AB, Pre-Calc, Trig, Alg 2, Alg 1, SAT/ACT Prep, Business Math
Generally, when you're dealing with relations in two variables then the clearest visualization is going to be a regular vertical/horizontal graph like you're used to using with regular x-and-y functions.
In this problem you have two variables. You have C and S.
The only difference between these and the x-and-y is that in this case one variable is not dependent on the other (whereas y is often dependent on x).
Hence you can pick whichever one you want to be on whichever axis and put the other one on the other axis. It doesn't matter as far as solving the problem is concerned. Do what makes sense to you. In fact, I encourage you to do it differently than I did and check that you get the same answer.
I'll assume that you put chairs (C) on the horizontal axis and Sofas (S) on the vertical axis.
I also have to assume that the intent is to maximize profit. So your objective function is $10 for every chair and $30 for every sofa. We can write that, using our C and S variables as 10C+30S. This is what you're trying to maximize. This doesn't have a dollar value yet. The goal of the problem is to find the maximum value given the two constraints.
To use the graphical solution method you need to know what the lines look like on a graph. One way to get a sense of what that movable line (the objective function) is going to look like is to pick an arbitrary value, graph the line, and then understand that it can move. I arbitrarily pick a dollar value of 30. That would give 10C + 30S = 30. If C = 0 then S = 1, and if S = 0 then C = 3. Because C is on the horizontal axis and S is on the vertical axis this means that our objective function can be visualized as a movable line with a slope of -1/3. Note that the final value of your objective function likely won't be 30. We just used that to get an idea of what things look like. You'd get the same "slope" if you used 60 as your arbitrary value (i.e. points 0,2 and 6,0).
Now it's time to graph your constraints. These are inequalities, and you can graph them the same way you've graphed x,y inequalities. Note that in this problem there are an additional two implied constraints. Those are that C >= 0 and S >= 0. This means we only have to work in the first quadrant (upper right) of the graph.
We can use the same method as earlier to find these lines(or you can use a different one if you wish; do what makes sense to you).
The first constraint is that 2C + 3S <= 16. If C = 0 then S = 16/3 (5 and 1/3). If S = 0 then C = 8. So this is a line segment from the point (0,16/3) to (8,0), and all the valid values are below it. You can check this by picking a point on one side and checking whether it satisfies the inequality or not. 0,0 is easy to check. it satisfies the inequality because 2*0 + 3*0 <= 16 is true.
The second constraint is that 1C + 4S <= 16. In the same manner we see that if C = 0 then S = 4, and if S = 0 then C = 16. Draw the line segment and check a point on one side. (0,0) makes the inequality true, so the part we care about is below the graph.
At this point you've defined your feasible region. It's a quadrilateral area that satisfies all four of the constraint inequalities.
Determining the maximum value of your objective function is now simply a matter of figuring out which direction of moving makes it bigger (in this case, as we saw earlier, this is moving the line up or to the right) and moving it as far as it will go without exiting the feasible area.
That single solitary point will be your answer. By moving around your objective function (you are totally allowed to make a physical model if it helps) you'll see that this maximum point is at the intersection of the two non-trivial constraints (the ones that aren't only about zero). So go ahead and use your favorite method to determine the intersection of 2C + 3S = 16 and 1C + 4S = 16. You'll find that C = 16/5 and S = 16/5 is that point.
At this point we run into a problem, though. In the real world we can't have 3.2 Sofas or 3.2 chairs. So we have to start checking the integer points around our final computed answer to determine an answer that will actually work. Our four closest integer points are (3,3), (3,4), (4,3), and (4,4). You can use whatever method you want to determine whether the point is outside the feasible region. Personally, I like plugging real numbers into things, so that's what I'm going to do. (4,4) violates both of the non-trivial constraints because 20 is greater than 16. (3,4) violates them both as well because 18 and 17 are bigger than 16. (4,3) violates the first one by a little bit (17 is bigger than 16), so we can eliminate that one too. That leaves us with (3,3) as the only feasible answer.
C = 3, S = 3, profit = 3*10 + 3*30 = 120 dollars.
