Maximum of $(1-q_1)(1-q_2)\ldots(1-q_n)$ I'm trying to find the maximum of $(1-q_1)(1-q_2)\ldots(1-q_n)$ where $n\ge 2$, on a the set $\{(q_1,\ldots , q_n) :q_1^2+q_2^2+\ldots+q_n^2=1 \ q_i\ge 0 \}$ (With the condition $q_i\ge0$ this is just the upper half of the sphere). This appeared to be a simple Lagrange multipliers question but calculating the derivative is becoming a problem. How is this done?
I was thinking of trying the fact that the product is the volume of a box having $(1,1\ldots 1)$ and $(q_1,\ldots ,q_n)$ at its diagonal but I can't figure out how to continue.(I'm guessing this reduces the problem to something geometrical)
Updates: Empirical evidence suggests that the maximum is when the $q_i$ are equal as pointed out by @Sabyasachi , I let WolframAlpha solve the Laplace multiplier equations for $n=2,3,4$ and got the same result(that they are equal). One solution which does not turn up in WA's solutions is that $q_3=0$ and $q_1=q_2=\frac {1} {\sqrt{2}}$ which is on the sphere and returns a larger value. 

Addendum: I've included the full set of solutions Mathematica finds for the case $n=3$ at the end of this answer.

The comments and answers so far don't show how Lagrange multipliers can solve the problem (because of the non-negativity condition). Here's how that technique can be used.
To maximize a function $f(q_1,\dots, q_n)$ with the requirement that $q_i\ge0$ and an additional constraint $c(q_1,\dots, q_n)=0$, it suffices to maximize the function $f(s_1^2,\dots, s_n^2)$ subject to the constraint $c(s_1^2,\dots, s_n^2)=0$, then choose those solutions with all $s_i$ real that give the largest value of $f$. For these, $(s_1^2,\dots, s_n^2)$ is a non-negative solution to the original constrained maximization question. 
For this particular question, then, maximize $\prod(1-s_i^2)$ subject to $\sum s_i^4=1$.
Here’s the three-variable case, where we need to maximize $(1-a)(1-b)(1-c)$ for $a,b,c\ge0$, subject to $a^2+b^2+c^2=1$.
Maximize $(1-x^2)(1-y^2)(1-z^2)$ subject to $x^4+y^4+z^4=1$. Using Lagrange multipliers, let $g(x,y,z,\lambda)=(1-x^2)(1-y^2)(1-z^2)-\lambda(x^4+y^4+z^4-1)$, and solve the following system of equations over the real numbers:
$$\begin{align}
0 = \frac{\partial f}{\partial x} = & -2x(1-y^2)(1-z^2)+4x^3\lambda\\
0 = \frac{\partial f}{\partial y} = & -2y(1-x^2)(1-z^2)+4y^3\lambda\\
0 = \frac{\partial f}{\partial z} = & -2z(1-x^2)(1-y^2)+4z^3\lambda\\
0 = \frac{\partial f}{\partial \lambda} = & x^4+y^4+z^4-1\\
\end{align}$$
When a solution $(x,y,z,\lambda)$ of this system gives a maximum among all real solutions, $(x^2,y^2, z^2)$ is a solution to the original question.
The system is no fun to solve, and I didn't try to generalize to see if it gives a nice answer the original question in general. In theory it will, but whether it's any simpler than the other answers, I don't know.

Here are all of Mathematica’s real solutions to the system above, written in terms of $a=x^2,b=y^2,c=z^2$.
$
\begin{array}{llll}
 (a,b,c) & (1-a)(1-b)(1-c)\\ \hline\\
\left(\frac{1}{\sqrt{2}},  \frac{1}{\sqrt{2}},  0\right) & \left(1-\frac{1}{\sqrt{2}}\right)^2 \\
 \left(\frac{1}{\sqrt{2}} , 0 , \frac{1}{\sqrt{2}}\right) & \left(1-\frac{1}{\sqrt{2}}\right)^2 \\
 \left(0 , \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right)& \left(1-\frac{1}{\sqrt{2}}\right)^2 \\
 \left(\frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} \right)& \left(1-\frac{1}{\sqrt{3}}\right) ^3 \\
 \left(\frac{2}{3}, \frac{2}{3} , \frac{1}{3} \right)& \frac{2}{27} \\
 \left(\frac{2}{3} , \frac{1}{3} , \frac{2}{3} \right)& \frac{2}{27} \\
 \left(\frac{1}{3} , \frac{2}{3} , \frac{2}{3} \right)& \frac{2}{27} \\
 (1 , 0 , 0) & 0 \\
 (0 , 1 , 0) & 0 \\
 (0 , 0 , 1) & 0 \\
\end{array}
$
