First, given the cost function, we can write average cost function: ¯C(x)=C(x)x=10e0.04xxC¯(x)=C(x)x=10e0.04xx, with the help of the first derivative, we determine the critical points:

¯C(x)=10e0.04xx¯C′(x)=10e0.04x⋅0.04x−10e0.04x⋅1x2¯C′(x)=10e0.04x⋅(0.04x−1)x2=00.04x−1=0x=10.04=25unitsC¯(x)=10e0.04xxC¯′(x)=10e0.04x⋅0.04x−10e0.04x⋅1x2C¯′(x)=10e0.04x⋅(0.04x−1)x2=00.04x−1=0x=10.04=25units

We can prove, we have a maximum:

{x<25→¯C′(x)>0x>25→¯C′(x)<0→maximum{x<25→C¯′(x)>0x>25→C¯′(x)<0→maximum

So, the correct answer is c.