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