Maximize p = 6x + 8y + 4z subject to
3x + y + z ≤ 15
x + 2y + z ≤ 15
x + y + z ≤ 12
x ≥ 0, y ≥ 0, z ≥ 0.

Separate the variables that you know, 6+8+4=18. Now look at your answer, we know that 18 is higher than 15, so one of the variables x, y, or z has to be a zero. 6+8=14 or 6+4=10, so the 8(y)=0 and y=0. Now the 6 has to have the y=1 and the 4 and be multiplied by 2 or 1. Whit this there are only three answers, the first three. If 6(1)+4(1)=10 then x+y+z≤12 could be our answer. If 6(1)+4(2)=14 then we are left with 2 answers, 3x+y+z≤15 or x+2y+z=15. When you only need one answer you will go with x+y+z≤12 

It’s d for sure i just did also that’s why