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) = $16,666.50
B: 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) = $11,905
C: 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)=$10476
D: X=2,000 y=0
Z=10(2000) + 5 (0) = $20,000
I 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…
Ask the Tutor ACCA PM
Chapter 4 – Planning with limiting factors Question help
9y ago#1There 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 = 238.1
Now 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 = 57.142
y = 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)
Thank you so much John, appreciate it. Will have a look at this tomorrow after work.
9y ago#3You are welcome :-)
This topic is locked — no new replies.