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

Question

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 $2401$

Answer 1

$x=8, y=4, z=8, w=9$ 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

Answer 2

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: $$g(x,y,z,w) = 0, \ \ \ x,y,z > 2$$ Let \begin{align*} \mathcal{L}(x,y,z,\lambda) &= f(x,y,z,w) + \lambda g(x,y,z,w)\\ &= (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!