Forums › Ask ACCA Tutor Forums › Ask the Tutor ACCA PM Exams › Chapter 4 – Planning with limiting factors Question help
- This topic has 3 replies, 2 voices, and was last updated 7 years ago by John Moffat.
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- June 2, 2017 at 12:19 pm #389698
Please see question below, at work typing this out, spent half an hour on it last night and could not figure it out :S;
A: X=0 y=3,333.3
Z=10x + 5y= 10(0) + 5(3333.3) = $16,666.50B: Solving 0.18x+0.03y=100, and 0.05x +0.02y =50
Gives x=238.1 and y=1904.8
Z=10(238.1) + 5(1,904.8) = $11,905C: Solving 0.05x + 0.02y= 50 and 0.02x + 0.05y=40
Gives x=809.5 and y=476.2
Z=10(809.5)=5(476.2)=$10476D: X=2,000 y=0
Z=10(2000) + 5 (0) = $20,000I get part A of this and that’s about it. Part B is where I’m completely lost how X=238.1 and Y=1904.8 is worked out?
Kaplan text book doesn’t really have a breakdown, just assumes you should know…
June 2, 2017 at 4:22 pm #389747There are several ways of solving simultaneous equations. I will work through it the way I show in my free lectures on Linear Programming.
For B:
0.18x + 0.03y = 100 (1)
0.05x + 0.02y = 50 (2)If you multiply equation (2) by 1.5 all the way through, you get:
0.075x + 0.03y = 75 (3)
(the reason I did this was to get the same number of y’s in the equation as in the other equation)Now subtract each term in (3) from each term in (1)
0.105x + 0 = 25
x = 25/0.105 = 238.1Now put x = 238.1 in either for the first 2 equations.
If you put it in equation (1), then:
(0.18 x 238.1) + 0.03y = 100
0.03y = 100 – 42.858 = 57.142y = 57.142 / 0.03 = 1904.8
Do watch the free lecture because I explain the same idea with different equations.
(The free lectures are a complete free course for Paper F5 and cover everything needed to be able to pass the exam well)
June 8, 2017 at 6:16 pm #391954Thank you so much John, appreciate it. Will have a look at this tomorrow after work.
June 9, 2017 at 7:46 am #392090You are welcome 🙂
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