You will not these days be asked to draw a graph (because of computer based exams).

However you must be able to understand the graph because it is very likely that the examiner will give you a graph and expect you to calculate things from it.

(Solving without a graph will not work – just solving individual pairs of equations together will likely give solutions that break other constraints, and without the graph this will not be clear.)

Thanks John for the presentation. Solving for the optimum solution graphically involves plotting all of the constraints coordinates arrived at in formulating the problem.

Sir,
I do checking for point A,C,D contribution. Is it correct as below? I understood that is no need to calculate in the exam. Thanks a lot.
Point A,(E=0, S=36)
Max C= (6×36)+0 = $216<=
Point C, Mats; 2S+4E=80
Demds; E=10
Substitute: 2S+40=80
S=20
Max C=(6×20)+(9×10)= $210 <=
Point D; (S=0,E=10)
Max C=0+90 = $90 <=
Therefore, The optimum point is B ($225).

You will not be expected to check the other corners in the exam – you are expected to understand the idea of using the iso-profit line to find the ‘best’ corner.

yugunman says

I put E on the y axis and S x axis and 2x+4y=80

5x+6y=180

Does it matter the exam which way I do it

John Moffat says

No, it doesnt matter (but appreciate that you cannot be asked to draw the graph in the exam).

Dee says

Hi John, I wanted to ask how you got the Material and labour calculation.

Material: If S is 0, E= 20, If E= 0, S= 40

Labour If S =0, E= 30 & If E= 0, S =36.

I know it is simple calculation but didn’t get this.

Thank you so much.

John Moffat says

For materials, 2S + 4E = 80

If you put S equal to 0, then 0 + 4E = 80

so 4E = 80. So E = 80/4 = 20

If you put E equal to 0, then 2S + 0 = 80

So 2S = 80. So S = 80/2 = 40

Do exactly the same for the labour equation 🙂

Dee says

Thank you so much! 🙂

John Moffat says

You are welcome 🙂

Krupali says

Dear sir,

is it ok if we dont draw the graph in the exam ?

and just slove it with the help of equation ?

thanks

Krupali

John Moffat says

You will not these days be asked to draw a graph (because of computer based exams).

However you must be able to understand the graph because it is very likely that the examiner will give you a graph and expect you to calculate things from it.

(Solving without a graph will not work – just solving individual pairs of equations together will likely give solutions that break other constraints, and without the graph this will not be clear.)

Samuel Koroma says

Thanks John for the presentation. Solving for the optimum solution graphically involves plotting all of the constraints coordinates arrived at in formulating the problem.

John Moffat says

Yes – and the graphical approach is the only approach in the syllabus for F5.

Samuel Koroma says

Noted sir

May says

Sir,

I do checking for point A,C,D contribution. Is it correct as below? I understood that is no need to calculate in the exam. Thanks a lot.

Point A,(E=0, S=36)

Max C= (6×36)+0 = $216<=

Point C, Mats; 2S+4E=80

Demds; E=10

Substitute: 2S+40=80

S=20

Max C=(6×20)+(9×10)= $210 <=

Point D; (S=0,E=10)

Max C=0+90 = $90 <=

Therefore, The optimum point is B ($225).

May says

Sorry Sir. I am sleepy. This checking should be next video 😛

John Moffat says

You will not be expected to check the other corners in the exam – you are expected to understand the idea of using the iso-profit line to find the ‘best’ corner.