TITLE: Maximum of $(1-q_1)(1-q_2)\ldots(1-q_n)$
QUESTION [10 upvotes]: 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.

REPLY [3 votes]: ${\bf 1\ }$ 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
$$\max f=\left(1-{1\over\sqrt{2}}\right)^2={1\over2}(3-2\sqrt{2})\doteq0.0858\ .$$