Why is maximum work done for reversible paths in isothermal expansion and not minimum work done? [closed]

Work done is given by $$W = -P_{ext}dV$$ according to convention.

Which should mean that in cases of expansion, work done on the system would be negative. If work done is negative, isn't it wrong to say that work done in isothermal reversible process is maximum? Because in irreversible expansions, area covered is smaller but then, if the value is negative, then wouldn't it be that minimum work is done in case of reversible isothermal expansion and maximum for irreversible case?

Image for clarification. This image is for expansion. The dark gray area is for irreversible process and the light gray + dark gray is the work done by reversible process. Clearly reversible process has higher area and hence higher negative value.

• What is the maximum of -5 and -8?
– Karsten
Commented May 16, 2020 at 19:18
• The work done by the system is maximum. Commented May 16, 2020 at 19:59
• -5 is bigger than -8. Magnitude wise, 8 is bigger than 5. Commented May 16, 2020 at 20:00
• @BuckThorn Ahh, is that what they actually mean? Commented May 16, 2020 at 20:00
• Well, in the opposite direction (compression) you would say that the work done by the surroundings is least for a reversible process. Commented May 16, 2020 at 20:04

$$W=-P_{ext}dV$$ ,