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 $2401$
Let $$f(x,y,z,w) = (x-1)(y+3)(z-1)(w-2)$$ and $$g(x,y,z,w) = x + y + w - 29$$
We want to $$\max\{f(x,y,x,w)\}$$ 
subject to:
$$g(x,y,z,w) = 0, \ \ \ x,y,z &gt; 2$$
Let
\begin{align*}
\mathcal{L}(x,y,z,\lambda) &amp;= f(x,y,z,w) + \lambda g(x,y,z,w)\\ &amp;= (x-1)(y+3)(z-1)(w-2) + \lambda(x + y + w - 29)
\end{align*}
Then 
$$\nabla \mathcal{L}(x,y,z,\lambda) = 0$$ yields $4$ equations:

\begin{equation}{\tag{1}}
(y+3)(z-1)(w-2) + \lambda = 0
\end{equation}

\begin{equation}{\tag{2}}
(x-1)(z-1)(w-2) + \lambda = 0
\end{equation}

\begin{equation}{\tag{3}}
(x-1)(y+3)(w-2) + \lambda = 0
\end{equation}

\begin{equation}{\tag{4}}
(x-1)(y+3)(z-1) + \lambda = 0
\end{equation}

Now when you set each one equal to each other and find another four equations where $y$ will be a linear combination of $x,y,w$ we find another $4$ equations:

\begin{equation}{\tag{5}}
y = x - 4
\end{equation}

\begin{equation}{\tag{6}}
y = z - 4
\end{equation}

\begin{equation}{\tag{7}}
y = y
\end{equation}

\begin{equation}{\tag{8}}
y = w - 5
\end{equation}

Now, we see that 
$$(y+4) + y + (y + 4) + (y+5) = 29 \Rightarrow y = 4$$

Then it is trivial to find $x,z,w$ by plugging in $y=4$ to the equations above. Hope that helps!
$x=8, y=4, z=8, w=9$  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