Let $0 \leq a \leq x \leq 100$ and $f(x)=|x-a|+|x-100|+|x-a-50|$. Then the maximum value of $f(x)$ becomes $100$ when $a$ is equal to

We need to find the value of $a$ such that the maximum value of $f(x) = |x-a| + |x-100| + |x-a-50|$ equals 100, given the constraint $0 \leq a \leq x \leq 100$.

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Since $f(x)$ is a sum of absolute value functions, it's piecewise linear. The maximum value will occur either at the endpoints of our domain $[a, 100]$ or at critical points where the expressions inside the absolute values equal zero.

The critical points are where:

$x = a$

$x = 100$

$x = a + 50$

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Let us substitute $a = 50$ into our function:

$f(x) = |x-50| + |x-100| + |x-50-50|$

$f(x) = |x-50| + |x-100| + |x-100|$

$f(x) = |x-50| + 2|x-100|$

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Since our domain is $x \in [50, 100]$ (because $a \leq x \leq 100$), we need to determine the sign of each absolute value expression:

For $x \in [50, 100]$: $x - 50 \geq 0$, so $|x-50| = x-50$

For $x \in [50, 100]$: $x - 100 \leq 0$, so $|x-100| = 100-x$

Therefore:

$f(x) = (x-50) + 2(100-x)$

$f(x) = x - 50 + 200 - 2x$

$f(x) = 150 - x$

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Since $f(x) = 150 - x$ is a decreasing linear function on the interval $[50, 100]$, the maximum occurs at the left endpoint $x = 50$:

$f(50) = 150 - 50 = 100$

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The maximum value of $f(x)$ becomes 100 when $a = 50$.

When dealing with sums of absolute value functions, always identify the critical points and check the behavior on each interval. The maximum often occurs at domain endpoints for such piecewise linear functions.