It's no bother...if it was, then I would have no business trying to offer help!
Let's look at what we have:
f(x,y)=(−0.0005x2+0.04x+0.15)y0.32–6.6y
Computing our partials, and equating to zero, we obtain the system:
fx(x,y)=(−0.001x+0.04)y0.32=0
fy(x,y)=0.32(−0.0005x2+0.04x+0.15)y−0.68–6.6=0
The first equation implies two cases, but the case
y=0 is invalid given the second equation. So we have:
−0.001x+0.04=0∴x=40
Substituting this value for
x into the second equation, we find:
0.32(−0.0005(40)2+0.04(40)+0.15)y−0.68–6.6=0
0.32(0.95)y−0.68=6.6
0.304y−0.68=6.6
y0.68=82538
(y0.68)1725=(82538)1725
y=(82538)1725≈0.010822019346
So, our critical point is:
⎝⎜⎜⎛40,(82538)1725⎠⎟⎟⎞
Now, using the second partials test for relative extrema, we compute:
fxx(x,y)=−0.001y0.32
fyy(x,y)=−0.2176(−0.0005x2+0.04x+0.15)y−1.68
fxy(x,y)=0.32(−0.001x+0.04)y−0.68
Evaluated at the critical point, we find:
fxx⎝⎜⎜⎛40,(82538)1725⎠⎟⎟⎞=−0.001⎝⎜⎜⎛(82538)1725⎠⎟⎟⎞258=−0.001(82538)178
fyy⎝⎜⎜⎛40,(82538)1725⎠⎟⎟⎞=−0.2176(−0.0005(40)2+0.04(40)+0.15)⎝⎜⎜⎛(82538)1725⎠⎟⎟⎞−2542=
−0.2176(0.95)(38825)1742=−0.20672(38825)1742
fxy⎝⎜⎜⎛40,(82538)1725⎠⎟⎟⎞=0.32(−0.001(40)+0.04)⎝⎜⎜⎛(82538)1725⎠⎟⎟⎞−2517=0.32(0)⋅38825=0
And so:
D⎝⎜⎜⎛40,(82538)1725⎠⎟⎟⎞=⎝⎜⎜⎛−0.001(82538)178⎠⎟⎟⎞⎝⎜⎜⎛−0.20672(38825)1742⎠⎟⎟⎞−(0)2>0
Since
fxx⎝⎜⎜⎛40,(82538)1725⎠⎟⎟⎞=−0.001(82538)178<0 then we conclude the critical point is at a relative maximum.