Minimum of maximum

For the sake of easy typing, we shall denote $f_a(x)$ as $f(x)$ and $N_a$ as $N$ . Remark that (when we take the boundary conditions):

$f(-16) = a - 16$

$f(10) = a + 10$

Using the generalised triangle inequality, we can get:

$26 = f(10) - f(-16) ≤ |f(10)| + |f(-16)|$

Therefore it is not possible for each of $|f(10)|, |f(-16)|$ to be less than 13. Whence, $N ≥ 13$

On the other hand, by choosing $f(x) = x - 3$ , we can see that $-13 ≤ f(x) ≤ 13$ for $x ϵ [-16, 10]$ . Hence $N ≤ 13$ .

Therefore, N = 13 .

Since $f_a(x)$ is linear, The extreme cases (where $x=-16,10$ ) should be equal. This means $|-16+a|=|10+a|$ Which solves as $a=-3$ . Therefore, $\displaystyle\max_{x\in [-16,10]}|f_a(x)|=|-16+3|=|10+3|=\boxed{13}$

For a better explanation, I will show two graphs.

Graph of $|f_7(x)|=|x+7|$ ,

Graph 1

Graph of $|f_{-5}(x)|=|x-5|$ ,

Graph 2

Image Credit for Wolfram|Alpha

From the two graphs above, we can see the maximum value of $f_a(x)$ is always in $x=-16$ or $x=10$ .

If $a$ increases, the value of $f_a(10)$ increase while $f_a(-16)$ decreases.

If $a$ decreases, the value of $f_a(10)$ decrease while $f_a(-16)$ increases.

So, we must get an $a$ to let both of them, $f_a(10)$ and $f_a(-16)$ be equal, then the maximum of the function will be minimized. If $a$ goes slightly larger or smaller, one of the value $f_a(10)$ and $f_a(-16)$ will be different and one will get higher than the usual position.

Hence, note

$|f_a(10)|=|f_a(-16)|$

$|10+a|=|-16+a|$

We let $-16+a<0$ ,

$10+a=16-a$

$a=3$

Substitute $a=3$ , we get

$N_3$

$=\displaystyle \max_{x\in[-16,10]}|f_3(x)|$

$=13$

Hence, the answer is $\boxed{13}$ .

To find the minimum value for the maximum value for the absolute value of $f_a(x) = x + a$ , we need to "balance" between the two edge, i.e. positive value and the negative value for $f_a(x)$ in the given range of $-16\leq x\leq10$ . Those in between will always be smaller since this is a straight line equation with gradient 1. Therefore, $|-16+a|=|10+a|$ $-16+a=-10-a$ (taking + - because we are equating the minimum negative and maximum positive value) $2a=6$ , $a=3$ So, the minimum value for maximum of $|f_a(x)|=|-16+3| or |10+3|$ which is 13

Geometrically, $mod(x+a)$ are those V-shaped curves shifted left or right by $a$ . In the domain of $[-16,10]$ , these V-shaped curves will give a maxima either at $x=-16$ or at $x=10$ depending on $a$ . For a certain value of $a$ , both the values at $x=-16$ and at $x=10$ will be equal and that is what we are asked to find out. This curve will have its tip (read: the minima) lying exactly at the center of $-16$ and $10$ , i.e, at $-3$ . Now, if we consider only one side of the V-curve, say the one with the positive slope, i.e, $y=x+a$ this straight line should pass through $(-3,0)$ which implies $a=3$ . Now we can find the value of this curve at $x=10$ , which is $y=10+3=13$

Eh, this problem is easy as soon as you understand it. What it's asking is "what is the minimum value that this function can be if you're adding $a$ to some number between -16 and 10 and taking the largest one?". Gah, it sounds so obnoxiously complex, but for it means that for any given $a$ , there is a maximum value that $|x+a|$ can have if $x$ is in the domain $[-16, 10]$ . Once armed with this information, we can try different values of $a$ and see what the maximum value is for $|f_a(x)|$ . For $a=2$ , the maximum would be 14 ( $|-16+2|$ ). For any $a<3$ , the maximum is simply $|-16+a|$ . For $a>3$ , the maximum is $10+a$ . Thus, the parametric equation defined reaches a minimum at $a=3$ , for which the minimum value is $\boxed{13}$ .

In the coordinate $x-y$ axis, we can draw a series of parallel colored lines at $45^o$ angles bouncing off the x-axis and hitting walls at x=-16 and x=10. For each light beam (i.e. different values of a), we find the maximum of the height that it hits either of the walls. We are interested in the value of $a$ at which the maximum height is the mimimum for all $a$ . The lines bounce off the x-axis at x=-a. Since we can smoothly vary the value of a, we can find an $a$ where the height of the light beam at the two walls is the same. For all other $a$ 's on either side of this $a$ , either the right wall will have higher height or the left wall will have higher height. The height on the right wall, i.e. x=10, is a+10. The height on the left wall, i.e. x=-16, is -(-16+a). Equating the two, we get $a+10=16-a \implies a=3$ . For this value of $a$ , the $N_a$ will be minimum. For this value of $a$ , $N_a=13$ .

The graph of $f_a(x) = |x - a|$ made of two rays at the positive $y$ half-plane that meet at $(-a,0)$ . Now it's clear that to minimize $N_a$ we need to put $a$ in between the endpoints of domain of $x$ . That is $a = -3$ , hence $\min (N_a) = |10 - (-3)| = \boxed{13}$

we see that F(x)=x+a takes different values for different values of x in the range -16 to 10 . F(x) also takes different values for different values of 'a' . now we have to make the maximum value of F(x) minimum. the implication is that we have to find such 'a' such that when we plug the value of 'a' in F(x) ,|F(x)| becomes minimum. if we plug in a=3 we see that |-16+3| = 10+3=13 which is the maximum value of f(x) for given 'a'=3 for any x in the range -16 to 10. for any other value of 'a' the maximum value of f(x) exceeds 13. thus minimum of the maximum value of f(x) is 13.