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Sir, what if ‘C’ is optimum point then what will be the solution or equation from and respectively E and S values.
Point C is where the for E = 10 crosses the line 2S + 4E = 80.
So we solve the two equations together (which is rather simple since we know that E = 10 🙂 )
Hello John
Thank you for the clear explanation but i’d like to ask how did you know that 2.5 was going to give you 5 in the 3rd equation
There is 2S in one equation and 5S in the other.
5 divided by 2 is 2.5.
(As I say in the lecture, you will no doubt have been taught how to solve simultaneous equations in school algebra. There is more than one way of solving so if you were taught a different way then by all means use it 🙂 )
Sir, firstly many thanks for the lucid explanation of such a complex topic.
Actually, I wanted to know what if there’s a third product also? Because then we will have 3 variables in one constraint equation, which I guess would get very elusive to solve.
Hello John,
Great lecture, quite easy to follow.
Although I have a question which leans more to math.
What if we had more than 3 constraints, say a third line? How would we solve the equation?
In order to find ‘S’ or ‘E’, which 2 equations out of the 3 would we pick?
I hope my question makes sense.
Thanks in advance.
Even if we would have had 3 constraints, i.e. 3rd line, we would still choose the point that is furthest away from the iso-contribution line(keeping in mind the slope of the iso-contrn line). And consequently find the intersection of the two lines upon which the furthest point lie. Then finally substitute the obtained values of variables in the contrn equation.
P.S.- In a circumstance where at the furthest point, 3 lines intersect at the same time, then we can choose any two lines out of the two. At the end of the day, they all meet at the same coordinates, so it doesnt matter which two lines we choose.
Hello John sir,
How are you,
I hope you are great,
Thank you for appreciated time,
I have a simple question
How can i make sure that this is the optimum combination after getting the values of S and E and after trying all crosses how to make sure that this is the optimum combination of Sand E that achieve the maximum contribution.
Thank you in advance.
By putting the values of S and E at all the intersections into the equation for the contribution.
However, this takes time, and given that in the exam the graph will be given you (you cannot be expected to draw it yourself) it should be very clear by moving the contibution line which intersection is the optimum.
Hi
Thank you for the lecture, I have question though
How important it is to label your axes properly? In your case vertical is S and horizontal is E. When you were calculation maximum contribution the demand line didnt cross point B. However I tried to do it different way (E was vertical and S was horizontal) and I got totally different result. My equations are based on material and demand: E=10 and 2S+4E=80
I am confused.
As I say in the lecture, you cannot be asked to draw the graph yourself. The graph will be drawn for you and so it will not be up to you to label the axes.
(Even if you were having to draw it yourself, the solution would be the same however you chose the axes even though the lines would obviously be in different places)
Hi Sir,
Thanks a lot for this explanation. It was really helpful !
You are welcome ?
Thank you for the informative lectures. I can see you’ve answered a similar question from another student. The 2.5 figure used to multiply in equation 3 @9.20- I understand the reasons why you’ve done it. Can any figure be used to multiply to get the same number of ‘S’s’ in an equation. For example;
Mats: 20S + 6E = 100
Lab: 9S + 2E= 50
Would the multiple figure in this scenario be 2.22?
I’d like to continue using this method and my only confusion at the moment is the above. Is the multiple figure adapted to the scenario depending on the figures or is it only roung figures that coincidentally fit. Thank you in advance.
Jess
#Look up Simultaneous Equations for how to solve them.
Whatever coefficient you choose to make equal- use that multiple figure throughout the entire equation so you can get a zero. (like having two 6E’s cancel each other out or having two 20S’s cancel each other out -you could even flip the equations to keep smaller values and have two 2E’s or two 9S’s). But it will Not work if you have eg: a 5.999E and a 6E because then the answer would not be zero but something like 0.001E and you would still have two unsolved coefficients (S and E) in your equations.
So in your example, to make your life easier, make the E coefficient equal – where the multiple is a simple 3 instead of using 20/9 and ending up with equation 3 being 20S+40/9E=1000/9
(Also, when you have a never-ending decimal number, its better to stick with the fraction version instead of rounding off)
Thanks alot for your amazing explaining.. Before it was so difficult to sort out the issue BUT NOW it’s so easy to understand and solve it after watched your lecture. Absolutely recommended watching your lectures..
Thanks again and it’s much appreciated Sir..
Sir since we took point B
We assumed the linear equations to be based on materials and labour..
But what if it was point C?
Would we take the linear equation based on demand and materials?
Would that give the same result?
If it was point C, then yes you would solve the equations for demand and for materials.
No – it would not give the same result. Just look at the graph and you can see immediately that the values of E and S will be completely different!
kindly mention the reason to use 2.5 for equation 3 @ 09:20
As I do explain in the lectures, it is so as to end up with the same number of S’s in both equations.
However, as I also explain, there are many ways of solving simultaneous equations (all, obviously, giving the same end result – it doesn’t matter which way you do it in the exam, and so if the way you were taught at school was different then by all means do it that way.