finding minimum of function Can you please give me some hints finding minimum of this function:
$ (r-1)^2 + (\frac{s}{r} -1)^2 + (\frac{t}{s}-1)^2 + (\frac{4}{t}-1)^2$  
where $ 1 \le r \le s \le t \le 4 $,
$r,s,t \in \Bbb R $
 A: Let us write $a = \frac{r}{1}$, $b = \frac{s}{r}$, $c = \frac{t}{s}$, and $d = \frac{4}{t}$. Then the function you want to minimise is
$$f(a,b,c,d) = (a-1)^2 + (b-1)^2 + (c-1)^2 + (d-1)^2,$$
subject to the constraints $a,b,c,d \geqslant 1$ and $p(a,b,c,d) = a\cdot b\cdot c\cdot d = 4$.
That looks like using the method of Lagrange multipliers may help a lot. So considering $\dfrac{\partial f}{\partial a}(a,b,c,d) = 2(a-1)$ and similar for $b,c,d$, and $\dfrac{\partial p}{\partial a}(a,b,c,d) = bcd = \dfrac{p(a,b,c,d)}{a} = \frac{4}{a}$ on the surface $abcd = 4$, again similar for $b,c,d$. So to have $\nabla f = \lambda\cdot \nabla p$, we must have
$$\lambda = \frac{\partial f/\partial a}{\partial p/\partial a} = \frac{a(a-1)}{2} = \frac{b(b-1)}{2} = \frac{c(c-1)}{2} = \frac{d(d-1)}{2}.$$
Thus the only critical point of $f$ on $\{abcd = 4\}$ (with positive $a,b,c,d$) is $a=b=c=d=\sqrt{2}$. It is readily checked that that point is a local minimum, and then one verifies that it is the global minimum under the constraints.
Thus the function is minimised for $r = \sqrt{2}, s = 2, t = 2\sqrt{2}$, with the minimal value $4(3-2\sqrt{2})$.
A: Another way...  As Daniel Fischer has done, define $a=r, b=\frac{s}r, c=\frac{t}s, d=\frac4t$.  Then we have $a, b, c, d \ge 1$ and $abcd = 4$.  
Now using the QM- AM-GM inequalities, we have
$$\sqrt{\frac{(a-1)^2 + (b-1)^2 + (c-1)^2+(d-1)^2}{4}}\ge \frac{a+b+c+d - 4}4 \ge \sqrt[4]{abcd}-1 = \sqrt{2}-1$$
with equality (i.e. minimum) iff $a=b=c=d=\sqrt2$.
A: if it is a function of one variable,derivative with respect to that variable and set to $0$,if it is function of two variable,then do it for both variable   and again set it to $0$ and use this  condition
http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/min_max/min_max.html
EDITED: 
as i see it is related to global  minimum of given function on some bound interval,that why when you derivative and find critical points,take each one so that it fits  it's  given bound interval
