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?



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 $(\pm {18}^{\frac{1}{4}},0)$.


$$\mathcal{L}(x,y,\lambda)=x^2+y^2-\lambda(x^4+y^4-18)$$ 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.

  • $\begingroup$ +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

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.]


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