How to prove a function has no maximum I have a function:
$$p\cdot(w+6s)^2+(1-p)\cdot(w-s)^2$$
where $p\in(0,1)$, $w>0$ and $s\geq0$ is a choice variable.
I am looking for the maximum of the function with respect to $s$, but can quickly see that the maximum is undefined as the function tends towards infinity as $s$ increases.
The question is how can I proof that in a concise way?
Thanks in advance.
 A: You can expand the squares and write the expression as $f(s)=as^2+bs+c$ with $a \gt 0$.  Then you can say that given $N$, you can find an $s$ such that $f(s) \gt N$ and even exhibit such an $s$ (depending on the other parameters).
A: You have a quadratic in $s$. That means that it either has a single, global minumum, e.g. $y=x^2$, or a single, global maximum, e.g. $y=-x^2$. To find out which, let us expand.
If $\operatorname{f}(s) := p(w+6s)^2+(1-p)(w-s)^2$ then
$$\operatorname{f}(s) = (1+35p)s^2+2w(7p-1)s+w^2 $$
As with all graphs $y=as^2+bs+c$, we have:


*

*$y=\operatorname{f}(s)$ has the shape  $\cap$ if the $s^2$ coefficient is negative, i.e. $1+35p<0$, i.e. $p<-\tfrac{1}{35}$.

*$y=\operatorname{f}(s)$ has the shape  $\cup$ if the $s^2$ coefficient is positive, i.e. $1+35p>0$, i.e. $p>-\tfrac{1}{35}$.


Since $0 < p < 1$ in your case, you have a $\cup$-shaped parabola, which has a single, global minimum.
To prove this using calculus, notice that, since $0 < p < 1$, we have:
$$\frac{\operatorname{df}}{\operatorname{d}\!s} = 0 \iff s=\frac{w(1-7p)}{1+35p}$$
The second derivative distinguishes between maxima and minima:
$$\frac{\operatorname{d^2f}}{\operatorname{d}\!s^2} = 70p+2$$
The turning point above $\left(\frac{\operatorname{df}}{\operatorname{d}\!s} = 0\right)$ is a minimum if, and only if, $70p+2>0$, i.e. $p>-\tfrac{1}{35}$. It follows that for all $0 < p < 1$, the one, and only one, turning point is a minimum.
Hence, the point $(x,y)=\left(\frac{w(1-7p)}{1+35p},\frac{49w^2p(p-1)}{1+35p}\right)$ is a global minimum.
Alternatively, you could complete the square since:
$$\operatorname{f}(s)=(1+35p)s^2+2w(7p-1)s+w^2$$
A: Suppose $s \ge 0$. You have 
$p (w+6s)^2+(1-p)\cdot(w-s)^2 \ge p (w+6s)^2 \ge p s^2$, hence $\sup_s p (w+6s)^2+(1-p)\cdot(w-s)^2 = \infty$.
