: Why can't wearing both ear muffs (whether passive or active) and earplugs simultaneously yield more than 36 dB noise reduction ratio (NRR)? I read on Wikipedia: Using both ear muffs (whether passive or active) and earplugs

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In addition to what @Tom_C said, another reason is that both earplugs and earmuffs are very good at blocking high-frequency sounds, but very poor at blocking low frequency ones.
Presumably the "30dB reduction" assumes the sound is equally loud across some standard range of frequencies, but while wearing both pieces of gear, the "input" to the earplugs will consist largely of the low frequencies that the earmuffs weren't able to block.

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Seems like a point of confusion on this question is actually the physics, not the math.
A simple experiment you can try at home that explains why earplugs + earmuffs doesn't double the dB of attenuation:
You need two powered speakers and a dB meter. Create a mono track in your favorite DAW of some sound source, pink noise would be perfect. Or if you have a noise generator, use that. Play the noise mixed equally through both powered speakers. Turn off one speaker and adjust the other one so that you have a clear reading in the dB meter - something like 70 dB would be good. Then turn off the speaker you just adjusted, turn the other one on, and adjust it to match the first (it helps to have a tripod to mount the dB meter so that it doesn't move between readings).
Now you should have two speakers each calibrated to play correlated white noise at 70 dB. When you turn on both speakers, the resulting SPL is not 140 dB, which is a damn good thing because that would damage your hearing. The SPL meter should show something close to 76 dB. In practice, it will probably be lower than 76 dB because even though the signal being fed to both speakers is correlated, the resulting sound arriving at the dB meter from each speaker is no longer correlated because of different path lengths and reflections from room surfaces.
If you want to explore further, calibrate both speakers to play the noise at 60 dB (instead of 70). Now when they are both on, you should get a level of approximately 66 dB. Doubling the amount of sound is an increase of 6 dB. When you turn off one of the speakers, the level drops by 6 dB. Halving the amount of sound is a drop of 6 dB.
It's not a perfect analogy, but the physics and math match up with NRRs. If you have 30 NRR earplugs and you add 30 NRR earmuffs, you get an extra 6 dB of reduction for a total of 36. If you instead had 20 NRR earplugs and added 20 NRR earmuffs, you would end up with an NRR of 26. If the two numbers don't match, then it gets a little messier with the math. The point is, you can't add NRRs, because that is not acoustically analogous to a 30 dB reduction followed by another 30 dB reduction. The same way adding a second speaker playing the same sound is not acoustically adding 70 dB of gain to another 70 dB of gain.

A simpler explanation:
Your intuition that starting with 30 dB of reduction and providing an additional 30 dB of reduction is doubling the reduction. That is correct. What's confusing is that when you double the reduction, you increase that reduction by 6 dB. The reason why 2x = 6 dB is what I wrote below:

This cuts right to heart of what decibels actually are. One way to think about them is as ratios. Another way, which helps us understand why they don't add or subtract linearly, is to think of them as exponents.
For example, 22 + 22 does not equal 24, it equals 23. So if we ignore the common base of all those exponents, it appears that 2 + 2 = 3! Also note that:

23 + 23 = 24
24 + 24 = 25
25 + 25 = 26
etc.

A decibel is an exponent with the base removed, so combining two different decibel values does not result in a new value that is the sum of the two values. When we add an exponent of 2 to itself, the value we get is the exponent plus 1 of 2. So in "exponent of 2" math:

x (+) x := x + 1

Where (+) is the special "adding exponents of 2" operation and := is a special kind of "equals" for the operation.
Just like we can make that special "exponents of 2" math, there is special math for "adding" decibel values. In decibel math:

x (+) x := x + 6

(Experts in decibels will want to point that that this is not always true - there are actually two different kinds of decibels that have slightly different math for each of them. For this answer, let's just focus on the kind of decibel that obeys the above "formula".)
In other words, if we have a 30 dB sound and we add another 30 dB sound to it, the resulting sound level will be 36 dB, not 60 dB.
The same relationship holds in reverse for negative decibels. If we have 30 dB of sound reduction, and we add another 30 dB of sound reduction, the total reduction is 36 dB.
The reason why 36 dB is a soft limit on portable sound reduction is because you can't keep adding on more and bigger earmuffs.
You might be thinking, "Hey Todd, you said 30 dB plus 30 dB equals 36 dB, but what about 30 dB plus 33 db?" The answer is that those 3 extra dB aren't that much when combined with the other 30 dB, and we can easily round the total down a little to 36 dB.

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Small disclaimer about decibels
For sound pressures, decibels are defined as follow:
XdB = 20 log (p1/p0)
with p1 being the amplitude of the pressure field of the sound, and p0 a reference (20microPascal of pressure).
This translates in terms of power/loudness as:
Pow_dB = 10 log(P1/P0)
Because of this definition, you cannot simply add sound powers:
10 log((P1+P2)/P0) different from 10 log (P1/P0) + 10 log (P2/P0)

However thanks to the first formula here they will add up when using decibels for gain or attenuation: let's start with a sound pressure P1, and let's apply a two times gain. The resulting pressure is 2*P1. In dB:
X = 10 log(2*P1/P0) = 10 log(2) + 10 log(P1/P0) approx 10 log(P1/P0) + 3dB

It we multiply this last sound level by 2 another time, it will finally ends up that we added 3dB again, resulting in a final value of
X = 10 log(4*P1/P0) = 10 log(4) + 10 log(P1/P0) approx 10 log(P1/P0) + 3dB + 3dB = 10 log(P1/P0) + 6dB

Obviously this does work the other way around. Let's divide our sound by two:
X = 10 log(P1/(2*P0) = 10 log(1/2) + 10 log(P1/P0) approx 10 log(P1/P0) - 3dB

Dividing the power is equivalent to subtracting dB.
That is neat, because, it is much simpler for us human behind a mixer, to think in terms of addition or subtraction when behind a console.
About your problem
According to what I said, actually, you should be able to add up the dB reduction of protection…
But: these protection are protecting your ears, and, unfortunately, sound also travels into flesh. Let's have a look at the transmission coefficient of sound:
T = 2 Z2 / (Z1+Z2)

Where Z are the acoustic impedance of our two medias (let's say, air and water, mainly like flesh).
One can calculate these two impedances:
Zwater = 1.5 10^6 Pa s/m
Zair = 408 Pa s/m

Resulting in:
T = 2Zair/(Zair + Zwater) = 0.000544

Let's transform that in terms of dB attenuation:
10 log (2*Z1/(Z1+Z2)) = -32.64 dB
You will note that this is not far for -36dB, especially considering that one's head is only composed of water! Things will be a bit different when considering the skull and so, but the general picture is here.
This is why an ear protection cannot reduce more than that: it does not prevent the sound to reach your inner ear from the flesh, or your skull…
Q.E.D.