I haven't done this for a long time so i'm bit stuck on this
Max. for with constraints
I found:
Think this is correct step up to this point, but when I try to rearrange above equation to find I couldn't do it.
Please help me, thank you.
I haven't done this for a long time so i'm bit stuck on this
Max. for with constraints
I found:
Think this is correct step up to this point, but when I try to rearrange above equation to find I couldn't do it.
Please help me, thank you.
Putting these equal to 0, we have
..... (1)
..... (2)
..... (3)
Subtract (2) from (1): . Therefore either or . Similarly
either or ; and
either or .
From those equations it clearly follows that x, y and z cannot all be different. On the other hand, they cannot all be the same, because then the equation x+y+z=3 would mean that they are all equal to 1, which would contradict the equation yz+zx+xy=1.
Thus two of x,y and z are equal, and the third is different, say z=y≠x. Then the equations yz+zx+xy=1 and x+y+z=3 become , . Substitute the value of x from the second of those equations into the first to get a quadratic for y. The numbers don't come out particularly pleasantly, but I get the maximum value of V as .