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. 
 A: ${\bf 1\ }$ We begin with the following two-dimensional problem: Maximize
$$f(q_1,q_2):=(1-q_1)(1-q_2)$$
under the constraints 
$$q_1^2+q_2^2=r^2, \quad q_1\geq0, \ q_2\geq0\  .$$
Here $r$ is a parameter, $0<r\leq1$. Analyzing the graphs of the functions
$$g_r(t):=(1-r\cos t)(1-r\sin t)\qquad(0\leq t\leq{\pi\over2})$$
we find the following (see the figure):

(a) When $r<\rho:=2(\sqrt{2}-1)$ then
$$q_1q_2>0\quad\Rightarrow\quad f(q_1,q_2)<f(r,0)\ .$$
(b)  When $r>\rho$ then
$$q_1\ne q_2\quad\Rightarrow\quad f(q_1,q_2)<f\left({r\over\sqrt{2}},{r\over\sqrt{2}}\right)\ .$$
${\bf 2\ }$ Now let $n\geq2$ be arbitrary and assume that $q=(q_1,q_2,\ldots,q_n)$ is an admissible point where $f$ assumes its maximum. Put $$\tau:={1\over2}\rho^2\doteq0.343>{1\over3}\ .$$
In the light of ${\bf 1}$ there can be at most one $i$ with $0<q_i^2<\tau$, and as $\tau>{1\over3}$ there can be at most two $i$ with $q_i^2\geq\tau$. All other $q_i$ have to be zero.
${\bf 3\ }$ When $0<q_1^2<\tau\leq q_2^2\leq q_3^2$ then
$$q_1^2+q_2^2=1-q_3^2\leq 1-\tau<2\tau=\rho^2\ .$$
From ${\bf 1}$(a) we then conclude that $q$ cannot be a maximum point.
${\bf 4\ }$ When only $q_1^2$ and $q_2^2$ are $>0$ then ${\bf 1}$(b) with $r=1$ tells us that we have $q_1=q_2={1\over\sqrt{2}}$, and this leads to the maximal value
$$\max f=\left(1-{1\over\sqrt{2}}\right)^2={1\over2}(3-2\sqrt{2})\doteq0.0858\ .$$
A: 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}
$
A: If any of the $q_i$ are 1, then the product is zero.
Otherwise, take the logarithm, and try to maximize
$\log(1-q_1)+\log(1-q_2)+\ldots+\log(1-q_n)$
