# Find the max/min values of a function when (x,y) is a point in the set...

• Oct 22nd 2010, 11:47 AM
Runty
Find the max/min values of a function when (x,y) is a point in the set...
This normally wouldn't be so hard for me, except for the awkward constraint we've been given.

Find the maximum and minimum values of the function
$\displaystyle f(x,y)=2x^4-4x^2y^2+5y^4$
when $\displaystyle (x,y)$ is a point in the set
$\displaystyle \{(x,y):x^4+y^4\leq 80\}$.

I've been told that using Lagrange Multipliers is the proper method to solve this, but I'm not so sure. I'm new to using Lagrange Multipliers, and I haven't seen a question for which the constraint is a set that doesn't have a definite value.
• Oct 22nd 2010, 02:11 PM
Krizalid
you just apply the method for $\displaystyle x^4+y^4=80,$ once got the max. or min. points, keep those which satisfy $\displaystyle x^4+y^4\le80.$
• Oct 23rd 2010, 07:43 AM
HallsofIvy
I'm not sure what Krizalid is saying. If you "apply the method" (Lagrange multipliers) for $\displaystyle x^4+ y^4= 80$, any points you get will satisfy $\displaystyle x^4+ y^4= 80$ and so $\displaystyle x^4+ y^4\le 80$.

Here's what I would do. Find all x and y such that $\displaystyle \nabla f= 0$, that is, such that $\displaystyle \frac{\partial f}{\partial x}= 0$ and $\displaystyle \frac{\partial f}{\partial y}= 0$. Note those (x, y) that satisfy $\displaystyle x^4+ y^4< 80$.

Then use Lagrange multipliers to find those (x, y) that maximize or minimize f with the constraint $\displaystyle x^4+ y^4= 80$.

Finally, check the value of f at each of those (x, y) points, both inside and on $\displaystyle x^4+ y^4= 80$, to determine the maximum and minimum values of f.
• Oct 23rd 2010, 10:58 AM
Runty
HallsofIvy, either I did your method wrong, or something else happened that I'm not familiar with. I've not seen the method you're using in our course material yet.

When I went for $\displaystyle \frac{\partial f}{\partial x}=0$ and $\displaystyle \frac{\partial f}{\partial y}=0$, I end up with $\displaystyle \frac{\partial f}{\partial x}=8x(x^2-y^2)$ and $\displaystyle \frac{\partial f}{\partial y}=4y(5y^2-2x^2)$, which leads to $\displaystyle x=0,y=0$ in both cases.

Did I do this incorrectly somehow, or this some sort of a "trick question" thing?

EDIT: Okay, using the Lagrange method of $\displaystyle u=f(x,y)+\lambda g(x,y)$ seems to either be malfunctioning on me, or I'm doing it wrong. Again, our source textbook is next-to-worthless.

Using your constraint $\displaystyle x^4+ y^4= 80$, I have $\displaystyle u=2x^4-4x^2y^2+5y^2+\lambda (x^4+y^4)$. Taking partial derivatives gets me $\displaystyle \frac{\partial u}{\partial x} =8x^3-8xy^2+4\lambda x^3$ and $\displaystyle \frac{\partial u}{\partial x} =20y^3-8x^2y+4\lambda y^3$. Finding $\displaystyle \lambda$, however, gets me $\displaystyle \lambda =\frac{2x^4-4x^2y^2+5y^2}{80}$, which I'm guessing is a mistake of some sort.

Is this method incorrect?
• Oct 23rd 2010, 07:39 PM
wopashui
for fx=fy=0, when x= -1/2 and y=-1/5 is also sasitfy the equation other x=y=0, so will that be 4 critipal points?
• Oct 24th 2010, 04:38 AM
HallsofIvy
$\displaystyle f(x,y)=2x^4-4x^2y^2+5y^4$ so $\displaystyle \nabla f= (8x^3- 8xy^2)\vec{i}+ (-8x^2y+ 20y^3)\vec{j}$. At any max or min in the interior of the set we must have $\displaystyle 8x^2- 8xy^2= 8x(x^2- y^2)= 8x(x- y)(x+ y)= 0$ so one of x= 0, x=y, x=-y must be true. Also $\displaystyle -8x^2y+ 20y^3= 4y(5y^2- 2x^2)= 4y(\sqrt{5}y- \sqrt{2}x)(\sqrt{5}y+ \sqrt{2}x)= 0$ so one of y= 0, $\displaystyle \sqrt{5}y= \sqrt{2}x$, or $\displaystyle \sqrt{5}y= -\sqrt{2}x$ must be true.

If x= 0, only y= 0 satisfies the other equations. If y= x again we have only x= y= 0. I don't see how x= -1/2, y= -1/5 satisfy those equations. I find only x= y= 0.

On the boundary, $\displaystyle g(x, y)= x^4+ y^4= 80$, $\displaystyle \nabla g= 4x^3\vec{i}+ 4y^3\vec{j}$ so we must have $\displaystyle (8x^3- 8xy^2)\vec{i}+ (-8x^2y+ 20y^3)\vec{j}= \lambda(4x^3\vec{i}+ 4y^3\vec{j}$ or $\displaystyle 8x^3- 8xy^2= \lambda(4x^3)$ and $\displaystyle -8x^2y+ 20y^3= \lambda(4y^3)$.

It's never necessary to find $\displaystyle \lambda$ since that is not relevant to the solution. Instead, eliminate $\displaystyle \lambda$ by dividing one equation by the other:
$\displaystyle \frac{8x^3- 8xy^2}{-8x^2y+ 20y^3}= \frac{4x^3}{4y^3}$
$\displaystyle \frac{2(x- y)(x+ y)}{(\sqrt{5}y- \sqrt{2}x)(\sqrt{5}y- \sqrt{2}x)}= \frac{x^2}{y^2}$.
That, together with $\displaystyle x^4+ y^4= 80$ gives two equations to solve for x and y.
• Oct 24th 2010, 03:30 PM
Runty
Quote:

Originally Posted by HallsofIvy
$\displaystyle f(x,y)=2x^4-4x^2y^2+5y^4$ so $\displaystyle \nabla f= (8x^3- 8xy^2)\vec{i}+ (-8x^2y+ 20y^3)\vec{j}$. At any max or min in the interior of the set we must have $\displaystyle 8x^2- 8xy^2= 8x(x^2- y^2)= 8x(x- y)(x+ y)= 0$ so one of x= 0, x=y, x=-y must be true. Also $\displaystyle -8x^2y+ 20y^3= 4y(5y^2- 2x^2)= 4y(\sqrt{5}y- \sqrt{2}x)(\sqrt{5}y+ \sqrt{2}x)= 0$ so one of y= 0, $\displaystyle \sqrt{5}y= \sqrt{2}x$, or $\displaystyle \sqrt{5}y= -\sqrt{2}x$ must be true.

If x= 0, only y= 0 satisfies the other equations. If y= x again we have only x= y= 0. I don't see how x= -1/2, y= -1/5 satisfy those equations. I find only x= y= 0.

On the boundary, $\displaystyle g(x, y)= x^4+ y^4= 80$, $\displaystyle \nabla g= 4x^3\vec{i}+ 4y^3\vec{j}$ so we must have $\displaystyle (8x^3- 8xy^2)\vec{i}+ (-8x^2y+ 20y^3)\vec{j}= \lambda(4x^3\vec{i}+ 4y^3\vec{j}$ or $\displaystyle 8x^3- 8xy^2= \lambda(4x^3)$ and $\displaystyle -8x^2y+ 20y^3= \lambda(4y^3)$.

It's never necessary to find $\displaystyle \lambda$ since that is not relevant to the solution. Instead, eliminate $\displaystyle \lambda$ by dividing one equation by the other:
$\displaystyle \frac{8x^3- 8xy^2}{-8x^2y+ 20y^3}= \frac{4x^3}{4y^3}$
$\displaystyle \frac{2(x- y)(x+ y)}{(\sqrt{5}y- \sqrt{2}x)(\sqrt{5}y- \sqrt{2}x)}= \frac{x^2}{y^2}$.
That, together with $\displaystyle x^4+ y^4= 80$ gives two equations to solve for x and y.

Alright, so using your method, I ended up with the following roots:
$\displaystyle x=0,y=0$
$\displaystyle x=-2\sqrt{2},y=-2$
$\displaystyle x=-2\sqrt{2},y=2$
$\displaystyle x=2\sqrt{2},y=-2$
$\displaystyle x=-2\sqrt{2},y=2$

According to this result I got off WolframAlpha, however, I have the following max/min values:
Max: $\displaystyle f(x,y)=400$ at $\displaystyle (x,y)=(0,-2\sqrt[4]{5})$
Min: $\displaystyle f(x,y)=0$ at $\displaystyle (x,y)=(0,0)$
WolframAlpha isn't 100% reliable at times, but I calculated $\displaystyle x^4+y^4$ using the max roots and it worked out to equal 80. However, I'm wondering if the program misinterpreted what I put in.