3
$\begingroup$

I am having troubles with one part of this homework problem. Hopefully somebody can help me out:

Find the minimum and maximum values of the function $f(x,y)=x^2+y^2$ subject to the given constraint $x^4+y^4=18$.

Using Lagrange multipliers, I can easily solve for the maximum:

$f_x(x,y)=2x$ and $f_y(x,y)=2y$.

If we call the second equation $g$, then: $g_x=4x^3$ and $g_y=4y^3$.

Then we apply the Lagrange multiplier:

$2x=\lambda 4x^3$ and $2y = \lambda 4y^3$

By solving for $x$ and $y$ and plugging in to $g$, we get $\lambda=1/\sqrt{36}=\pm1/6$.

To find the maximum, I will use the positive $1/6$, and solve for $x^2=3$ and $y^2=3$, resulting in a maximum of $6$, which the system spits out as correct.

For a minimum, I originally thought $0$ because $x^2$ and $y^2$ must be positive numbers, but that is not correct. Then I noticed that since $x^2=1/(2\lambda)$, and $y$ also, then when $\lambda$ is negative that would be the minimum, resulting in $-6$, which is also not correct.

So long story short, how do I find the minimum value in this case?

Thanks!

$\endgroup$
2
$\begingroup$

When you solve for $\lambda$ using the systems:

$\begin{cases} 2x = \lambda 4x^3 & (1) \\ 2y = \lambda 4y^3 & (2) \\ x^4 + y^4 = 18 & (3) \end{cases}$

You cancel out $x$ in (1) to get $x^2 = 1/(2\lambda)$, you missed a case that $x=0$.

If $x=0$, then $y^4 = 18$, yields $y = \pm {18}^{\frac{1}{4}}$.

The minimum value of $f(x,y)$ would be $\sqrt{18}$, occurs at $(0,\pm {18}^{\frac{1}{4}})$ or $(\pm {18}^{\frac{1}{4}},0)$.

$\endgroup$
2
$\begingroup$

$$\mathcal{L}(x,y,\lambda)=x^2+y^2-\lambda(x^4+y^4-18)$$ We have $$\nabla \mathcal{L}(x,y,\lambda) = \left(\begin{array}{c}2x-4\lambda x^3\\2y-4\lambda y^3 \\ x^4+y^4-18 \end{array}\right).$$ Setting this equal to zero gives us $$\frac{1}{2\lambda}=x^2 \;\text{ -or- }\; x=0$$ and $$\frac{1}{2\lambda}=y^2 \;\text{ -or- }\; y=0$$ Clearly both $x$ and $y$ cannot be zero at the same time. So we have three cases:

Case 1) $$x^2=y^2=\frac{1}{2\lambda} \;\Longrightarrow\; \frac{1}{4\lambda^2}+\frac{1}{4\lambda^2}=18 \;\Longrightarrow\; \lambda = \pm \frac{1}{6}.$$ Since the function is of $x^2$ and $y^2$ we need only consider one of the two cases; let's pick $\lambda=\frac{1}{6}$ which results in $$f(x,y)=x^2+y^2=\frac{1}{2\lambda}+\frac{1}{2\lambda}=6.$$

Case 2) $$x=0, \qquad y^2=\frac{1}{2\lambda}$$ Putting this into the constraint gives us $$0+y^4=18\;\Longrightarrow\; \frac{1}{4\lambda^2}=18 \;\Longrightarrow\; \frac{1}{72}=\lambda^2 \;\Longrightarrow\; \lambda = \pm \frac{1}{6\sqrt{2}}.$$ This results in $$f(x,y)=x^2+y^2=0+\frac{1}{2\lambda}=\frac{6\sqrt{2}}{2}=3\sqrt{2}.$$

Case 3) is symmetric with case (2) since the function and constraint is symmetric w.r.t. $x$ and $y$.

We have enumerated all possible values of $f$, so you can easily see which are maxima and minima. You will have to use some care if you need to enumerate all minimizers and maximizers, as there are a few cases and some symmetry to consider.

P.S., the accepted capitalization of Joseph-Louis Lagrange's surname is with lower-case `g's. This is different from some other similar words, e.g., LaGrange County, LaGrange College, etc. I cannot recommend strongly enough sticking with "Lagrange" for capitalization.

$\endgroup$
  • $\begingroup$ +1 Great answer, i appreciate the detailed explanation. I accepted wonghang's answer simply for the fact that it succinctly hit my question and pointed out my error directly. $\endgroup$ – Riley Feb 26 '14 at 5:58
1
$\begingroup$

Note that the function $ \ f \ $ is the "distance-squared" function, measured from the origin. Your "Lagrange equations" give $ \ \lambda = \frac{1}{2x^2} = \frac{1}{2y^2} \ \Rightarrow \ y^2 = x^2 \ . $ If we insert this into the constraint equation, we have $ \ x^4 = 9 \ \Rightarrow \ x, y = \pm \sqrt{3} , $ which leads to the results you already found.

The equation that ought to be written is $ \ 2x - 4 \lambda x^3 = 0 , $ for which $ \ x = 0 \ $ is also a solution. (The other Lagrange equation leads to $ \ y = 0 \ $ .) There are four points $ \ (0, \pm 18^{1/4} ) \ \text{and} \ ( \pm 18^{1/4} , 0 ) , $ which are the points at the minimum distance from the origin.

The constraint equation describes a "superellipse", for which a graph is shown below. What you found are the points on the diagonals, $ \ y = x \ \text{and} \ y = -x \ , \ (\pm \sqrt{3}, \pm \sqrt{3} ) \ . $

enter image description here

[It seems like there's a bit of an optical illusion here, as the axis intercepts are actually closer to the origin than the "diagonal corner" points.]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.