The other answers are also correct but I missed the relation to how efficiency is defined, so here's my additional answer:
For most amplifiers, the total powerconsumption consists of two parts,
a constant part (for biasing etc.) and a variable part which depends on how much power the amplifier is delivering:
(1) \$P_{total} = P_{constant} + P_{out}\$
If we then define the efficiency as the ratio between the delivered power and the total power consumption:
(2) Efficiency = \${P_{out}}/{P_{total}}\$
Filling in (1) into (2):
(3) Efficiency = \${P_{out}}/(P_{constant} + P_{out}) = 1/(\frac{P_{constant}}{P_{out}}+1)\$
From (3) we can now see that the efficiency becomes higher as \$P_{out}\$ increases.
So for maximum efficiency, you need to make \$P_{out}\$ as large as possible.
The largest "usable" power output of an amplifier is just before the point where it starts to saturate. If you go any further (more power) then the amplifier distorts and it doesn't behave as a "proper" amplifier anymore.
For a small output power, efficiency will be low as the \$P_{constant}\$ part is more dominant. Then most of the power is consumed by the amplifier itself instead of appearing as "usable" power at the output.