This one was actually kind of tough - I checked out the
examiner's report and the
itute solutions and neither of them really gave a clear, step by step explanation on how to do the question - although itute was a bit better than the VCAA.
Because Z is maximised at the points M and N, we can say that Z(m) and Z(n) are equal. For that to make much sense though, we need to know the coordinates of M and N. They're pretty clear on the graph as long as you work out the the interval is 5 (and maybe have a ruler to double check your points line up) but if you really wanted to you could set up some simultaneous equations to find them, too.
In any case, the coordinates of M and N are (20,50) and (45,25) respectively. We can set up two equations/functions for Z with this information:
Z = 20a + 50b (M)
Z = 45a + 25b (N)
Because we know Z has to be equal for both M and N to find a and b, we can quickly throw out some of the options that would give us something ridiculous - this means we don't have to worry about either of the options with negative values (A, E) and can probably get away without checking that D fits our rules because the numbers are pretty different to A, B and C. (Note: With multiple choice questions, you can usually eliminate two options after reading the question or minimal working out: in this case it was D and E)
So that leaves us with option B and option C to check, like so:
B. Z = 45*15 + 25*15 =1050
B. Z = 20*15 + 50*15=1050
We can already see that these are equal (and therefore, B is the answer), but let's check C to be sure we haven't messed up somewhere:
C. Z = 45*15 + 25*25 =1300
C. Z = 20*15 + 50*25 =1550
So B is our answer - of course some of the constraints I set were maybe a little too general (option A could've probably been a reasonable choice) but in the end, there's only five equations to check so realistically you probably wouldn't need to set any at all.