Quadratic equation to find max and min

[Mark44]

x = 1 -> y=+-$\sqrt2$
y = 0 -> x=+-$\sqrt2$

OK, those look fine.

 

[Pi-is-3]

$2x^2 + y^2 = 4$
$4x + y^2$

I know you are done with the question, but I wanted to say that lagrange multipliers is not the best for this problem.

Set $x=\sqrt{2}cos(\theta), y=2sin(\theta)$

So now we need to optimize $4(\sqrt{2}cos(\theta)+sin^2(\theta) )$

Spoiler: Solution, since you are done with problem

So now we need to optimize $\sqrt{2}cos(\theta)+sin^2(\theta)$Differentiate it w.r.t $\theta$ we get $-\sqrt{2}sin(\theta)+2sin(\theta)cos(\theta)=0$Implies $sin(\theta)=0$ or $cos(\theta)=\frac{1}{\sqrt{2}}$

So max when x=1 and $y=\sqrt{2}$ (so max is 6) and minimum when y=0 and $x=-\sqrt{2}$ (so min is $-4\sqrt{2}$).

 

[Lifeforbetter]

$2x^2 + y^2 = 4$
$4x + y^2$

I know you are done with the question, but I wanted to say that lagrange multipliers is not the best for this problem.

Set $x=\sqrt{2}cos(\theta), y=2sin(\theta)$

So now we need to optimize $4(\sqrt{2}cos(\theta)+sin^2(\theta) )$

Spoiler: Solution, since you are done with problem

So now we need to optimize $\sqrt{2}cos(\theta)+sin^2(\theta)$Differentiate it w.r.t we get $-\sqrt{2}sin(\theta)+2sin(\theta)cos(\theta)=0$Implies $sin(\theta)=0$ or $cos(\theta)=\frac{1}{\sqrt{2}}$

So max when x=1 and $y=\sqrt{2}$ (so max is 6) and minimum when y=0 and $x=-\sqrt{2}$ (so min is $-4\sqrt{2}$).

Ok. Thanks. But where can i know to set $x=\sqrt{2}cos(\theta), y=2sin(\theta)$

 

[Pi-is-3]

Ok. Thanks. But where can i know to set $x=\sqrt{2}cos(\theta), y=2sin(\theta)$

It is very simple. Let us consider $2x^{2} +y^{2}=4$

Then $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1$

Here we make the substitution $\frac{x^2}{2}=cos^2(\theta)$ and $\frac{y^2}{4}=sin^2(\theta)$

And that is how we get it.

 

[Pi-is-3]

Also, if you ever need, $- \sqrt{a^2+b^2} \leq acos(\theta)+bsin(\theta) \leq \sqrt{a^2+b^2}$

 

[Lifeforbetter]

Also, if you ever need, $- \sqrt{a^2+b^2} \leq acos(\theta)+bsin(\theta) \leq \sqrt{a^2+b^2}$

How about this?

 

[Pi-is-3]

How about this?

Do you mean the proof? You can prove it through Lagrange multipliers, or their is a trigonometrical proof too. For the Lagrange multipliers proof, your optimization function is $acos(\theta)+bsin(\theta)$ and restrain function is $cos^2(\theta)+sin^2(\theta)=1$.