Hi John, Thank you for the lecture. I have a few questions please – 1) What was the purpose of drawing the dotted line for contribution? We already had point B from the intersection of the Material and Labour lines. 2) If the maximum demand for E was 4 chairs, then the demand line would intersect with the labor line somewhere where S would be 34ish. What would be the optimum point in that case? B will fall outside the feasible area if E was less than or equal to 4? Apologies if the misunderstood something here. Thanks in advance Regards

The optimal mix could be at any of the corners, and is not necessarily going to be at the intersection of the materials and labour lines. I do explain this in the lecture. If the maximum demand for E was 4 chairs then the feasible region and the optimum point would be different.

Dear John Moffat, I can’t imagine my Studies without your help, We can’t reward you easily, thank you so much. I will try my best to donate to this website. Much love and respect for you my mentor.

What if the maximum demand for executive chairs is 4, would the optimum production plan still be to produce 5 units of executive chairs and 30 standard chairs? If not what would be the optimum production plan?

No it would not – there would be no point in producing 5 chairs if they can only sell a maximum of 4 chairs!!!!

You would need to draw the executive chair constraint as 4 instead of 10, identify the new feasible area, and find the corner furthest away from the origin when moving out the contribution line.

Since it is less likely that we would be required to draw a graph in the exam, is it possible to work that out arithmetically. I understand that the simultaneous equations will give you a point where the two lines meet, how do we incorporate the demand line (constraint) in the calculations.

You cannot be asked to draw a graph in the exam, but you could be given a graph and expect to be able to interpret it. You would draw the new demand line on the graph provided, identify the corner furthest away from the origin and then solve the relevant two equations together in the same way as I do in the exam. If one of the equations is E = 4, then it is ridiculously easy to solve the two equations together ðŸ™‚

I wanted to ask if we have 3 constraints equations, namely material,labour, and demand, and in one situation the constraint for demand has been changed, can you please tell me which other constraint, apart from demand, needs to be considered to calculate maximum contribution?

That depends on the angle of the lines and which of the intersections the contribution line meets last when moving away from the origin. Obviously you cannot be asked to draw the graph, but you can be asked to interpret it.

Hi john. great thanks for amazing lecture. i was asking how will you move the line practically in exams(CBE scenario)? or is there any laid mathematical relationship of the slope and finding the maximum point?

You cannot be asked to draw the graph in the exam. If the question gives you the graph then it will be pretty clear from the drawing of it which corner is the optimum. If not then you can do as I state in the lecture and check the contribution at each corner – the optimum is the one giving the highest contribution.

You can calculate the angle of the line and use that by comparing it will the angle of the other lines, but this is likely to take longer and is not necessary.

Hi! I solved this question using excel’s solver, and it is the exact units for contribution. But on the $90 part (which is a “pretend contribution” ) when maximizing contribution, I manually solved it with another “pretend contribution” of $100. When I see the results, it was different, assuming it’ll stay the same. Why?

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 ðŸ™‚ )

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 ðŸ™‚ )

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 â€˜Sâ€™ or â€˜Eâ€™, 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

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)

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.

#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..

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?

sir, I have a bit confused about the optimum point. From the graph, B is the furthest point from the origin and was proved to presume contribution = $90, then E will be equal to 10 and S will be equal to 15, move further away, then B is the optimum point. When can we assume that the optimum point is C? Should we need to plot another contribution line (with a different slope and angle), then move further to see if it touches point C?

From this part, we need to understand the feasible area where constraints are in place and what is the optimal production plan for maximizing the contribution. Correct?

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.

shopnipung@gmail.com says

Hi John, Thank you for the lecture. I have a few questions please –

1) What was the purpose of drawing the dotted line for contribution? We already had point B from the intersection of the Material and Labour lines.

2) If the maximum demand for E was 4 chairs, then the demand line would intersect with the labor line somewhere where S would be 34ish. What would be the optimum point in that case? B will fall outside the feasible area if E was less than or equal to 4?

Apologies if the misunderstood something here.

Thanks in advance

Regards

John Moffat says

The optimal mix could be at any of the corners, and is not necessarily going to be at the intersection of the materials and labour lines. I do explain this in the lecture.

If the maximum demand for E was 4 chairs then the feasible region and the optimum point would be different.

mashcal5 says

Dear John Moffat, I can’t imagine my Studies without your help, We can’t reward you easily, thank you so much. I will try my best to donate to this website. Much love and respect for you my mentor.

John Moffat says

Thank you for your comment ðŸ™‚

alin.sivi says

Thank you sir for making this topic easy to understand ðŸ™‚

John Moffat says

Thank you for your comment ðŸ™‚

2dop says

Hi John,

What if the maximum demand for executive chairs is 4, would the optimum production plan still be to produce 5 units of executive chairs and 30 standard chairs? If not what would be the optimum production plan?

Thank you.

John Moffat says

No it would not – there would be no point in producing 5 chairs if they can only sell a maximum of 4 chairs!!!!

You would need to draw the executive chair constraint as 4 instead of 10, identify the new feasible area, and find the corner furthest away from the origin when moving out the contribution line.

2dop says

Thank you for the prompt reply Sir.

Since it is less likely that we would be required to draw a graph in the exam, is it possible to work that out arithmetically. I understand that the simultaneous equations will give you a point where the two lines meet, how do we incorporate the demand line (constraint) in the calculations.

Thank you.

John Moffat says

You cannot be asked to draw a graph in the exam, but you could be given a graph and expect to be able to interpret it.

You would draw the new demand line on the graph provided, identify the corner furthest away from the origin and then solve the relevant two equations together in the same way as I do in the exam. If one of the equations is E = 4, then it is ridiculously easy to solve the two equations together ðŸ™‚

2dop says

Thank you John, I have understood. You are awesome.

John Moffat says

Thank you for your comment ðŸ™‚

licelleanessa says

Thank you for this lecture, explanation simple and easy to follow

John Moffat says

Thank you for the comment ðŸ™‚

jatingupta@2097 says

Hello John,

Hope you are safe and well!

I wanted to ask if we have 3 constraints equations, namely material,labour, and demand, and in one situation the constraint for demand has been changed, can you please tell me which other constraint, apart from demand, needs to be considered to calculate maximum contribution?

Thanks in advance,

Jatin

John Moffat says

That depends on the angle of the lines and which of the intersections the contribution line meets last when moving away from the origin. Obviously you cannot be asked to draw the graph, but you can be asked to interpret it.

Ermali says

Sir, isn’t the point A the furtherest from origin? clearly i am missing smthng but i do understood the whole flow very well.

John Moffat says

We need the point at which the contribution line is furthest away from the origin, not when the point itself is necessarily furthest away.

shakir7385 says

Dear John,

Why there was even a need to draw contribution line since where the material and labour line intersect is proved to be the optimum point.

Regards

John Moffat says

Because the intersection of the material and labour lines would not always be the optimum point.

ABDULLAHI312 says

Hi john. great thanks for amazing lecture. i was asking how will you move the line practically in exams(CBE scenario)? or is there any laid mathematical relationship of the slope and finding the maximum point?

John Moffat says

You cannot be asked to draw the graph in the exam.

If the question gives you the graph then it will be pretty clear from the drawing of it which corner is the optimum.

If not then you can do as I state in the lecture and check the contribution at each corner – the optimum is the one giving the highest contribution.

You can calculate the angle of the line and use that by comparing it will the angle of the other lines, but this is likely to take longer and is not necessary.

ictiancris says

Hi! I solved this question using excel’s solver, and it is the exact units for contribution. But on the $90 part (which is a “pretend contribution” ) when maximizing contribution, I manually solved it with another “pretend contribution” of $100. When I see the results, it was different, assuming it’ll stay the same. Why?

John Moffat says

The $90 was only to get the angle of the line. Using $100 will give a different line, but the angle will be the same which is all that matters.

(Appreciate that you will not have Excel in the exam. The spreadsheet in the exam is not Excel and does not have built-in formulae.)

poddar8697 says

Sir, what if ‘C’ is optimum point then what will be the solution or equation from and respectively E and S values.

John Moffat says

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 ðŸ™‚ )

ketra1 says

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

John Moffat says

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 ðŸ™‚ )

AverageAndy says

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.

robelandat says

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 â€˜Sâ€™ or â€˜Eâ€™, which 2 equations out of the 3 would we pick?

I hope my question makes sense.

Thanks in advance.

AverageAndy says

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.

7fsa says

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.

John Moffat says

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.

kapa says

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.

John Moffat says

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)

aleonard013 says

Hi Sir,

Thanks a lot for this explanation. It was really helpful !

John Moffat says

You are welcome ?

jelderfieldscott says

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

Shivangi says

#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)

mami9561 says

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..

leobenny says

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?

John Moffat says

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!

lwhnatalie says

sir,

I have a bit confused about the optimum point.

From the graph, B is the furthest point from the origin and was proved to presume contribution = $90, then E will be equal to 10 and S will be equal to 15, move further away, then B is the optimum point. When can we assume that the optimum point is C? Should we need to plot another contribution line (with a different slope and angle), then move further to see if it touches point C?

From this part, we need to understand the feasible area where constraints are in place and what is the optimal production plan for maximizing the contribution. Correct?

fgnhgn says

kindly mention the reason to use 2.5 for equation 3 @ 09:20

John Moffat says

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.