Find Maximum of $3x-2y$ Given $x^2+4y^2=4$

[Albert1]

$x,y\in R$ and $x^2+4y^2=4$

please find $max(3x-2y)$

 

[MarkFL]

My solution:

Spoiler

We are given the objective function:

\(\displaystyle f(x,y)=3x-2y\)

Subject to the constraint:

\(\displaystyle g(x,y)=x^2+4y^2-4=0\)

Using Lagrange multipliers, we obtain:

\(\displaystyle 3=\lambda(2x)\)

\(\displaystyle -2=\lambda(8y)\)

And this implies:

\(\displaystyle x=-6y\)

Substituting into the constraint, we find:

\(\displaystyle y=\pm\frac{1}{\sqrt{10}}\implies x=\mp\frac{6}{\sqrt{10}}\)

We then find:

\(\displaystyle f_{\max}=f\left(\frac{6}{\sqrt{10}},-\frac{1}{\sqrt{10}}\right)=2\sqrt{10}\)

 

[Albert1]

Spoiler

Using Cauchy-Schwarz inequality:
$40=(3^2+(-1)^2)\times 4=(3^2+(-1)^2)\times (x^2+(2y)^2)$
$\geq (3x-2y)^2$
$\therefore (3x-2y)\leq \sqrt {40}=2\sqrt {10}$

 

[Greg]

Spoiler

\(\displaystyle 3x-2y=m\implies y=\frac{3x-m}{2}\)

\(\displaystyle x^2+4y^2=4\Rightarrow10x^2-6mx+m^2-4=0\)

\(\displaystyle \Delta=160-4m^2\)

If \(\displaystyle m>2\sqrt{10}\) \(\displaystyle x\) is not real, hence

\(\displaystyle m=2\sqrt{10}\)

 

[MarkFL]

Spoiler

\(\displaystyle 3x-2y=m\implies y=\frac{3x-m}{2}\)

\(\displaystyle x^2+4y^2=4\Rightarrow10x^2-6mx+m^2-4=0\)

\(\displaystyle \Delta=160-4m^2\)

If \(\displaystyle m>2\sqrt{10}\) \(\displaystyle x\) is not real, hence

\(\displaystyle m=2\sqrt{10}\)

Very nicely done! (Yes)