lagrange multipliermultivariable-calculusoptimizationpartial derivative
$f(x, y) = xy^2 + 4$, $ D$ = {$(x, y)$ | $x ≥ 0, y ≥ 0, x^2 + y^2 ≤ 3$}
I'm lost on this one. I've tried plugging in critical points but my $D$ ends up being 0, meaning I can conclude nothing about extrema. I've tried Lagrange multipliers and I've gotten nowhere with that. If I were to guess, I'd say the minimum value is 4, but I can't figure out what to try to prove this right.
Lagrange multipliers do work for your problem, but that involves solving three simultaneous non-linear equations in three unknowns. There is another way that uses only one variable: parameterization.
Get a parameterization that describes the given curve in terms of only one variable. In the case of your ellipse you can use
$$x=\cos t, \quad y=\frac 17\sin t, \quad 0\le t\le 2\pi$$
Now you want to want to find the extrema of
$$\hat f(t)=4x+y=4\cos t+\frac 17\sin t$$
There are several ways to find the extrema of that, using calculus or just trigonometry. Here is a calculus way:
$$\hat f'(t)=-4\sin t+\frac 17\cos t=0$$ $$28\sin t=\cos t$$ $$\tan t=\frac 1{28}$$ $$\cos t=\sqrt{\frac 1{\tan^2 t+1}}=\pm\frac{28}{\sqrt{785}}$$ $$\sin t=\sqrt{1-\cos^2 t}=\pm\frac{1}{\sqrt{785}}$$ $$x=\cos t=\pm\frac{28}{\sqrt{785}}$$ $$y=\frac 17\sin t=\pm\frac{1}{7\sqrt{785}}$$
where $x$ and $y$ have the same sign. This means the maximum of $f$ is
$$4x+y=4\frac{28}{\sqrt{785}}+\frac{1}{7\sqrt{785}}=\frac{785}{7\sqrt{785}}=\frac{\sqrt{785}}{7}\approx 4.00255$$
and the minimum is the negative of that. This graph confirms the maximum.
Taking from where you left off. By equating coordinates of both sides, you have: $2z = 4\lambda z\implies z = 0$ or $\lambda = \dfrac{1}{2}$. If $\lambda = \dfrac{1}{2} \implies x = 8$ which is not possible since $x^2 \le 6$. So what is left is $z = 0$ or $\lambda = 0$. Also $\lambda \neq 1$ for otherwise $2x-8 = 2x$ which is not possible. $\lambda \neq 0$, for if it were $0$, then you have: $2x-8 = 0 \implies x = 4$, and this is not possible since $x^2 \le 6$. Thus $2x-8 = 2\lambda x, 2y-4 = 2\lambda y\implies x = \dfrac{4}{1-\lambda}, y = \dfrac{2}{1-\lambda}\implies \dfrac{16}{(1-\lambda)^2}+\dfrac{4}{(1-\lambda)^2}=6\implies (1-\lambda)^2 = \dfrac{20}{6} = \dfrac{10}{3}\implies \lambda = 1\pm \dfrac{\sqrt{30}}{3}$. Can you take it from here ?. It looks as if one of the lambdas will yield a min and the other a max.
Best Answer
Note that $$f(x,y)=xy^2+4\le x(3-x^2)+4=-x^3+3x+4$$ Now for $0\le x\le \sqrt{3}$ you have by differentiating $-x^3+3x+4$ $$-3x^2+3=0 \implies x^2=1 \implies x=1$$ with $(-3x^2+3)'_{x=1}=-6<0$ that it is maximized at $x=1$. So the max is $$x=1, y^2=3-1^2 \implies y=\sqrt{2}$$
For the min: $xy^2\ge 0$ in the given domain and $0$ is attained so this is indeed the minimum. Note however that all the points $x=0, y$ are in $argmin$.