Answer:
letter A: B = 400; C = 1000; Max Z = $380
Step-by-step explanation:
bear claws (B) need 6 oz of flour (f), 1 oz of yeast (y), 2 ts of paste (p)
croissants (C) need 3 oz of flour (f), 1 oz of yeast (y), 4 ts of paste (p)
putting that information in equations, we have:
B = 6f + y + 2p
C = 3f + y + 4p
The total number of resources (R) are:
R = 6600f + 1400y + 4800p
Let's call M our total profit, "k1" the number of B produced and "k2" the number of C produced.
So, we can state that:
M = k1*0.2 + k2*0.3
The number of resources R2 will demand is calculated like this:
R2 = k1*B + k2*C = (6k1+3k2)f + (k1+k2)y + (2k1+4k2)p
using R2 and R, we can make some inequations:
6k1+3k2 <= 6600 -> 2k1+k2 <= 2200
k1+k2 <= 1400
2k1+4k2 <= 4800 -> k1+2k2 <= 2400
if we try to maximize k2 (as it worths more), we will have k1 = 0 and k2 = 1200 (limited by p), but looking at the resources R, we will still have resources to use (f and y). Looking at B and C expressions, we see that removing one C gives enough 'p' to make 2 B, which is a good trade (as 2B worths 0.4 cents, and 1C worths 0.3 cents). we have 200 'y' remaining, so doing this 200 times give us k1 = 400 and k2 = 1000, and the only resource remaining will be some of 'f'.
calculating the profit M, we have:
M = 400*0.2 + 1000*0.3 = 380$
the right answer is letter A.
