SOLUTION: If x+y+z=16 then find the maximum value of (x-3)(y-5)(z-2). Given that (x-3)>0, (y-5)>0, (z-2)>0

We wish to maximize 

 

So we solve x+y+z=16 for z and substitute:



 



Next we find both partial derivatives and set them equal to zero:







Setting that equal to zero gives:

y-5=0;  -2x-y+17=0

Since y-5 > 0 we ignore the first

-2x-y+17=0









x-3=0;  -2y-x+19=0

Since x-3 > 0 we ignore the first.

-2y-x+19=0

So we solve this system:



That has solution (x,y)=(5,7)

Substituting in





So the maximum value of (x-3)(y-5)(z-2) is 8.

Edwin

By analogy with the well known AM-GM inequality ("Arithmetic Mean - Geometric Mean inequality") for two variables "a" and "b"


    ab <= ,         (1)


there is AM-GM inequality for three variables "a", "b" and "c"


    abc <= .      (2)


Inequalities (1) and (2) are valid for any two and three variables, respectively, that are real non-negative numbers.


Apply inequality (2), taking  


    a = x-3,  b = y-5,  c = z-2.


You will get


    (x-3)*(y-5)*(z-2) <=  =  =  =  =  = 8.


Thus for any 3 values of x, y and z, restricted by the equality


    x + y + z = 16   and  inequalities  x >= 3, y >= 5  and  z >= 2,


the inequality  


    (x-3)*(y-5)*(z-2) <= 8 


is held.


From the other side, at x= 5, y= 7  and z= 4 we have


    (x-3)*(y-5)*(z-2) = (5-3)*(7-5)*(4-2) = 2*2*2 = 8.


and the values of x, y and z satisfy all needed restrictions.



Thus the maximum value of (x-3)*(y-5)*(z-2), where  x, y and z are restricted by 


    x + y + z = 16,    x >= 3, y >= 5  and  z >= 2


is 8.        ANSWER