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:

$$f(x,y)=3x-2y$$

Subject to the constraint:

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

Using Lagrange multipliers, we obtain:

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

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

And this implies:

$$x=-6y$$

Substituting into the constraint, we find:

$$y=\pm\frac{1}{\sqrt{10}}\implies x=\mp\frac{6}{\sqrt{10}}$$

We then find:

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

 

[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

$$3x-2y=m\implies y=\frac{3x-m}{2}$$

$$x^2+4y^2=4\Rightarrow10x^2-6mx+m^2-4=0$$

$$\Delta=160-4m^2$$

If $$m>2\sqrt{10}$$ $$x$$ is not real, hence

$$m=2\sqrt{10}$$

 

[MarkFL]

Spoiler

$$3x-2y=m\implies y=\frac{3x-m}{2}$$

$$x^2+4y^2=4\Rightarrow10x^2-6mx+m^2-4=0$$

$$\Delta=160-4m^2$$

If $$m>2\sqrt{10}$$ $$x$$ is not real, hence

$$m=2\sqrt{10}$$

Very nicely done! (Yes)