Function 2 variables - Maximum

Olelarsen

New member
Joined
Dec 1, 2012
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8
Hey i need help with this function of 2 variables.
F(x, y) = (-0,0005x^2 + 0,04x + 0.15) y^0,32 – 6,6y
i have found the partial derivatives and set them = 0.
f1' = (-0.001x + 0.04)y^0.32 = 0
f2' = (-0.0005x^2 + 0.04x) * 0.32y^-0.68 – 6.6 = 0
and i get the point ---> (40, 0.00841) . I just need to know what i have to do to like, find out if this is a maximum or minimum? . I would be so grateful if someone could help me! . :(
 
Now you want to use the second partials test for relative extrema:

Let \(\displaystyle (a,b)\) be a critical point of \(\displaystyle z=f(x,y)\) and suppose \(\displaystyle f_{xx},\,f_{yy}\) and \(\displaystyle f_{xy}\) are continuous in a rectangular region containing \(\displaystyle (a,b)\).

Let \(\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2\).

i) If \(\displaystyle D(a,b)>0\) and \(\displaystyle f_{xx}(a,b)>0\), then \(\displaystyle f(a,b)\) is a relative minimum.

ii) If \(\displaystyle D(a,b)>0\) and \(\displaystyle f_{xx}(a,b)<0\), then \(\displaystyle f(a,b)\) is a relative maximum.

iii) If \(\displaystyle D(a,b)<0\), then \(\displaystyle f(a,b)\) is not an extremum.

iv) If \(\displaystyle D(a,b)=0\), then no conclusion can be drawn concerning a relative extremum.
 
Thank you SO much for answering my question so fast! what can I say, i suck at math ^o^ haha.. Oh i see. so hmm.. I just wanna be totally sure of what im doing.. That means i will end up with these 3 results?? right?
f11’’ = (-0,001) * y0,32
f12’’ = (-0,001x + 0,04) * 0,32y-0,68
f22’’= (-0,0005x2 + 0,04x) * - 0,21y

Am i doing this the right way?? or?? .. and like, if i try to say: f11’’(x, y) f22’’(x, y) – (f12’’(x, y))2

I end up getting like, a totally huge solution:
(1.05*10^7 * (x^4 – 80x^3 + 3047,62x + 121905)) / x^0.68) ?? .
 
You have incorrectly calculated the second partial with respect to y, \(\displaystyle f_{yy}(x,y)\).
 
You have incorrectly calculated the second partial with respect to y, \(\displaystyle f_{yy}(x,y)\).



Oh hmm.. you mean this one right ? --> f22’’= (-0,0005x2 + 0,04x) * - 0,21y ... should it be without the Y in the end? like this-->
f22’’= (-0,0005x2 + 0,04x) * - 0,21 ??? or? .. :(:(:confused::confused:
 
edit: Actually you incorrectly calculated the first partial with respect to y.

You should find

\(\displaystyle f_y(x,y)=(-0.0005x^2+0.04x+0.15)(0.32y^{-0.68}) – 6.6=0.32(-0.0005x^2+0.04x+0.15)y^{-0.68} – 6.6\)

Now, if we take the partial of this with respect to y, we obtain:

\(\displaystyle f_{yy}(x,y)=(-0.68\cdot0.32)(-0.0005x^2+0.04x+0.15)y^{-1.68}=-0.2176(-0.0005x^2+0.04x+0.15)y^{-1.68}\)

Do you see why?
 
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edit: Actually you incorrectly calculated the first partial with respect to y.

You should find

\(\displaystyle f_y(x,y)=(-0.0005x^2+0.04x+0.15)(0.32y^{-0.68}) – 6.6=0.32(-0.0005x^2+0.04x+0.15)y^{-0.68} – 6.6\)

Now, if we take the partial of this with respect to y, we obtain:

\(\displaystyle f_{yy}(x,y)=(-0.68\cdot0.32)(-0.0005x^2+0.04x+0.15)y^{-1.68}=-0.2176(-0.0005x^2+0.04x+0.15)y^{-1.68}\)

Do you see why?

Ahhhhhh i think i got it :), when you move down the -0,68 you also say y^-0,68-1 ? Right? .. Oh so i guess the other two are correct? .. and like, as you said earlier i'll just plot in my (a,b) which is (40, 0.00841) in my case into f11’’(x, y) f22’’(x, y) – (f12’’(x, y))2
to see if its <0 and then plot my (A,b) into Fxx as well to see if its <0 . right ? . :)
 
...when you move down the -0,68 you also say y^-0,68-1 ? Right?...

Yes, exactly! :mrgreen:

...Oh so i guess the other two are correct?

Yes, both partials with respect to x are correct.

...i'll just plot in my (a,b) which is (40, 0.00841) in my case into f11’’(x, y) f22’’(x, y) – (f12’’(x, y))2
to see if its <0 and then plot my (A,b) into Fxx as well to see if its <0 . right ? . :smile:

You need to recalculate your critical point(s) using the revised partial with respect to y.
 
Yes, exactly! :mrgreen:



Yes, both partials with respect to x are correct.



You need to recalculate your critical point(s) using the revised partial with respect to y.

Thank you so much :) !!!!! omg!!! :) :) :)
 
Glad to help and welcome to the forum! :D
 
Glad to help and welcome to the forum! :D


Omg im sorry to bother you once again, but like, i get D = (-0,000003*(x^4-15,959x^2-3,28125x+51,6953) / y^1,36 ) and when i plot in my (a, b) it gives -3404,97 =S =S ! and im like pretty sure my critical point, which is now (40, 0.010) should be a maximum =S =S !!! . grr what am i doing wrong :( :( ?
 
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.
 
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Omg you have no idea how much this has helped me.. THANK YOU SOOOOOOOOO MUCH!!!!! :), omg, i finally understand everything perfectly now!. thank you once again =], really! .. I hope you have a nice day AND a nice upcoming christmas + new year :)
 
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