It's no bother...if it was, then I would have no business trying to offer help!
Let's look at what we have:
\(\displaystyle f(x,y)=(-0.0005x^2+0.04x+0.15)y^{0.32}–6.6y\)
Computing our partials, and equating to zero, we obtain the system:
\(\displaystyle f_x(x,y)=(-0.001x+0.04)y^{0.32}=0\)
\(\displaystyle f_y(x,y)=0.32(-0.0005x^2+0.04x+0.15)y^{-0.68}–6.6=0\)
The first equation implies two cases, but the case \(\displaystyle y=0\) is invalid given the second equation. So we have:
\(\displaystyle -0.001x+0.04=0\,\therefore\,x=40\)
Substituting this value for
x into the second equation, we find:
\(\displaystyle 0.32(-0.0005(40)^2+0.04(40)+0.15)y^{-0.68}–6.6=0\)
\(\displaystyle 0.32(0.95)y^{-0.68}=6.6\)
\(\displaystyle 0.304y^{-0.68}=6.6\)
\(\displaystyle y^{0.68}=\dfrac{38}{825}\)
\(\displaystyle \left(y^{0.68} \right)^{\dfrac{25}{17}}= \left(\dfrac{38}{825}\right)^{\dfrac{25}{17}}\)
\(\displaystyle y=\left(\dfrac{38}{825}\right)^{\dfrac{25}{17}} \approx0.010822019346\)
So, our critical point is:
\(\displaystyle \left(40,\left(\dfrac{38}{825}\right)^{\dfrac{25}{17}} \right)\)
Now, using the second partials test for relative extrema, we compute:
\(\displaystyle f_{xx}(x,y)=-0.001y^{0.32}\)
\(\displaystyle f_{yy}(x,y)=-0.2176(-0.0005x^2+0.04x+0.15)y^{-1.68}\)
\(\displaystyle f_{xy}(x,y)=0.32(-0.001x+0.04)y^{-0.68}\)
Evaluated at the critical point, we find:
\(\displaystyle f_{xx} \left(40, \left( \dfrac{38}{825} \right)^{ \dfrac{25}{17}} \right)=-0.001 \left( \left( \dfrac{38}{825} \right)^{\dfrac{25}{17}} \right)^{\dfrac{8}{25}}=-0.001 \left( \dfrac{38}{825} \right)^{ \dfrac{8}{17}}\)
\(\displaystyle f_{yy} \left(40, \left( \dfrac{38}{825} \right)^{ \dfrac{25}{17}} \right)=-0.2176(-0.0005(40)^2+0.04(40)+0.15) \left( \left( \dfrac{38}{825} \right)^{ \dfrac{25}{17}} \right)^{- \dfrac{42}{25}}=\)
\(\displaystyle -0.2176(0.95) \left( \dfrac{825}{38} \right)^{ \dfrac{42}{17}}=-0.20672 \left( \dfrac{825}{38} \right)^{ \dfrac{42}{17}}\)
\(\displaystyle f_{xy} \left(40, \left( \dfrac{38}{825} \right)^{ \dfrac{25}{17}} \right)=0.32(-0.001(40)+0.04) \left( \left( \dfrac{38}{825} \right)^{ \dfrac{25}{17}} \right)^{- \dfrac{17}{25}}=0.32(0) \cdot \dfrac{825}{38}=0\)
And so:
\(\displaystyle D\left(40,\left(\dfrac{38}{825}\right)^{\dfrac{25}{17}} \right)= \left(-0.001 \left( \dfrac{38}{825} \right)^{ \dfrac{8}{17}} \right) \left(-0.20672 \left( \dfrac{825}{38} \right)^{ \dfrac{42}{17}} \right)-(0)^2>0\)
Since \(\displaystyle f_{xx} \left(40, \left( \dfrac{38}{825} \right)^{ \dfrac{25}{17}} \right)=-0.001 \left( \dfrac{38}{825} \right)^{ \dfrac{8}{17}}<0\) then we conclude the critical point is at a relative maximum.