I'm not too sure about it... but that's what I would have done:
For maximising profit, I take that MacGregor needs to make maximum use of all his given time.
Total time for woodworking is:
600 = 5b + 10d
where b and d are the number of bookcases and desks respectively.
Total time for finishing is:
250 = 4b + 3d
By solving these two simultaneous equations, I find that:
b = 28
d = 46
This makes a profit of $ 4710
Now, if he wants to make more desks, he'll need to make 47 desks and 26 bookcases so that the time constraints are still respected.
5(26) + 10(47) = 600 (good)
4(26) + 3(47) = 245 (good, still less than 250)
Profit = $ 4565 (ok, it's less, so, my answer is still the more profitable)
Now, if he wants to make more bookcases, he'll need to make 44 desks and 29 bookcases so that the time constraints are still respected.
5(29) + 10(44) = 585 (good)
4(29) + 3(44) = 248 (good)
Profit = $ 4460 (ok, it's less, so, my answer is still the more profitable)
My conclusion, the number of bookcases and desks to make maximum profit is 26 and 46 respectively