<s>[INST] If $x+y+z+w=29$ where x, y and z are real numbers greater than 2, then find the maximum possible value of $(x-1)(y+3)(z-1)(w-2)$ 
If $x+y+z+w=29$ where x, y and z are real numbers greater than 2, then find the maximum possible value of $(x-1)(y+3)(z-1)(w-2)$. 

$(x-1)+(y+3)+(z-1)+(w-2)=x+y+z+w-1=28$
Now $x-1=y+3=z-1=w-2=7$ since product is maximum when numbers are equal
My answer came out to be as $6*10*6*5=1800$ but the answer is $2401$. What am I doing wrong? And also, how we will get the answer $2401$
 [/INST] $x=8, y=4, z=8, w=9$  is the solution.
edit:
your solution is also correct. I don't know why are you multiplying $6∗10∗6∗5$?
$x−1=y+3=z−1=w−2=7$  means 7*7*7*7=2401    
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