Maximum value of $k$ so that the inequality $\sqrt{y-5}+\sqrt{8-y} \ge k$ should have answer 
Suppose $k$ be a real number so that the inequality
$\sqrt{y-5}+\sqrt{8-y} \ge k$ should have answer. what is the maximum
possible value of $k$?
$1)\sqrt{6}-1\quad\quad2)\frac12+\sqrt{3}\quad\quad3)\sqrt6\quad\quad4)2\sqrt2\quad\quad5)\sqrt6+1$

So I translated the question into finding the minimum value of $f(y)$ where $f(y)=\sqrt{y-5}+\sqrt{8-y}\quad$.(because for $k=\min(f(y))$ the inequality certainly has answer.) therefore we should solve  $f'(y)=0$:
$$f'(y)=\frac{1}{\sqrt{y-5}}-\frac{1}{\sqrt{8-y}}=0$$
$$y-5=8-y\rightarrow\quad y=6.5$$
And $f(6.5)=2\sqrt{\frac32}=\sqrt{6}$. so the answer is third choice.
Is my answer right? I know to find maximum or minimum of a function we take derivative of it and equate it to $0$. so here I don't know why $f(6.5)$ is minimum. can you please explain it to me?
EDIT:
I just noticed that $\sqrt6$ is not the minimum because $f(5)=f(8)=\sqrt3<\sqrt6$ maybe it is maximum! so I conclude the answer to this problem should be $\sqrt3$ but I don't have this option in the choices
 A: There is already an answer that clears your question
We can find the best minimum without using derivatives $${(\sqrt{y-5}+\sqrt{8-y})}^2=3+2\sqrt{8-y}\sqrt{y-5}\ge 3$$  because  $\sqrt{8-y}\sqrt{5-y}\ge 0$ $$\implies \sqrt{y-5}+\sqrt{8-y}\ge \sqrt{3}$$
A: Let's start with a theorem:

Suppose $f : [a, b] \to \Bbb{R}$ is continuous and differentiable on $(a, b)$. Further, let $c, d \in [a, b]$ such that $f(c) = \min f$ and $f(d) = \max f$. Then $c = a$, $c = b$, or $f'(c) = 0$ (and similarly for $d$).

In other words, the global maximum/minimum must either occur at a stationary point, or at an endpoint of the interval.
In this case, we have a function defined on the interval $[5, 8]$. You have already computed the stationary point $x = 6.5$ (beware, your derivative is missing factors of $1/2$, but they would have cancelled out when solving $f'(x) = 0$ anyway). So, our maximum and minimum must occur at $5$, at $8$, or at $6.5$. So, we just need to figure out which has the largest function value, and which has the least function value.
We have $f(6.5) = \sqrt{6}$ as you said, and $f(5) = f(8) = \sqrt{3}$. So, we have an absolute minimum of $\sqrt{3}$ and an absolute maximum of $\sqrt{6}$.
That said, even though you found the maximum instead of the minimum, $\sqrt{6}$ is still correct. It means that there is one and only one value of $y$ such that $\sqrt{y - 5} + \sqrt{8 - y} \ge \sqrt{6}$, but one and only one value of $y$ is still a solution! Anything larger than $\sqrt{6}$, and there'd be no solution.
The theorem applies to maximising/minimising a function on a closed and bounded interval. If you have open ends, or infinite ends, you can compute the limit(s) of the function as it approaches these end(s). So, instead of comparing the function value at the endpoint, compare the limit(s) at these end(s) to the function value at the stationary points.
(Note that I used neither the first nor second derivative tests to determine global maxima/minima. These tests are for testing local maxima/minima. While every global maximum/minimum is also local, these tests are useless for determining global maximma/minima.)
A: You found the correct derivative of $f(y)$, but you missed a couple cases.  Also, you are looking for the maximum of $f$, not the minimum, because you only need to have any solutions.  When $k$ is the maximum, only maximizers will solve the inequality.
When solving a problem like this, you should always graph it if you can, or try to visualize it if you are confident.  The graph has a domain of $[5,8]$ and is convex from above so the point $6.5$ is actually a maximum.
When testing for a maximum or minimum, you have to check not only the zeros of the derivative but points where it is undefined and boundaries of the search region as well.  In this case, $5$ and $8$ are both points where the derivative is undefined and boundaries of the search region (because the search region is the whole domain of $f(y)$).
Finally, you should always check if a zero of the derivative is a maximum or minimum.  This can be done in several ways: the second derivative test, comparison to another point which cannot be extremal, or you can simply calculate the function at all zeros/undefined points of the derivative and boundaries and find the largest/smallest value.
In this case, both $5$ and $8$ produce the smallest value of $\sqrt{3}$, so the value of $\sqrt{6}$ you found is indeed the answer, but because it is the maximum, not the minimum.
Edit: Also, you can solve this without calculus.  We want $$\sqrt{y-5}+\sqrt{8-y}\ge k$$ which implies (by squaring both sides) $$3+2\sqrt{-y^2+13y-40}\ge k^2.$$
The maximum of the lhs is $3+2\sqrt{\max(-y^2+13y-40)}$, but that polynomial is a parabola that is convex from above so its vertex is its maximum.  A parabola's vertex is at $\frac{-b}{2a}$, which here is $\frac{13}{2}$.  The value of the parabola at the vertex is thus $\frac{9}{4}$, which we can plug in to get the a maximum of 6 for $k^2$ and thus a maximum of $\sqrt{6}$ for $k$.
A: $k$ is the maximum of the function $\sqrt{y-5}+\sqrt{8-y}$ on $[5,8]$.
Using Cauchy-Schwarz you get immediately
$$\sqrt{y-5}+\sqrt{8-y} \leq \sqrt 2 \sqrt{y-5+8-y} = \sqrt 6$$
Equality is reached for $y-5 = 8-y \Leftrightarrow y =\frac{13}{2}$.
So, $\boxed{k=\sqrt 6}$.
A: Since $\sqrt x$ is concave, via Jensen's inequality we have
$$k \le \sqrt{y-5}+\sqrt{8-y} \le 2 \sqrt{\frac{(y-5)+(8-y)}{2}} = \sqrt 6$$
A: We need $y-5\ge0\iff y\ge5$ and $8-y\ge0\iff y\le8$
$$5\le y\le8\iff5-\dfrac{8+5}2\le y-\dfrac{8+5}2\le8-\dfrac{8+5}2$$
WLOG $y-6.5=1.5\cos2t$  where  $0\le2t\le\pi$
$$\sqrt{y-5}+\sqrt{8-y}$$
$$=\sqrt{6.5+1.5\cos2t-5}+\sqrt{8-(6.5+1.5\cos2t)}$$
$$=\sqrt3(\cos t+\sin t)$$
Now as $\cos t,\sin t\ge0;$
$$\cos t+\sin t=\sqrt2\cos\left(t-\dfrac\pi4\right)\le\sqrt2$$
