Problem about find the extreme of a function (Multipliers of Lagrange)

Question

Good morning. I have a problem with this:

Find the maximum and minimum distances from the origin to the curve* $$g\left(x,y\right)=5x^{2}+6xy+5y^{2}$$

I have done this:

Function to optimize:$f\left(x,y\right)=x^{2}+y^{2}$

Restriction: $g\left(x,y\right)=5x^{2}+6xy+5y^{2}=8 $

Applying Lagrange multipliers: $\nabla f\left(x,y\right)=\lambda\nabla g\left(x,y\right)$

Then, $\nabla f\left(x,y\right)=2x\hat{i}+2y\hat{j}$ and $\lambda\nabla g(x,y)=\lambda(2x+\frac{6}{5}y)\hat{i}+\lambda\left(2y+\frac{6}{5}x\right)\hat{j} $

Making the ecuation system:

$\begin{cases} 2x=(2x+\frac{6}{5}y)\lambda\\ 2y=(2y+\frac{6}{5}x)\lambda\\ x^{2}+\frac{6}{5}xy+y^{2}=8 \end{cases}$

But I have serious problem solving the system. Any suggestions?

Answer 1

Have you tried a software like Mathematica? There are four solutions to your system: $x=y=Sqrt[5/2]$ and the negative of that root (both with $λ=5/8$), and $x=-y=Sqrt[10]$ and the negative of that root (both with $λ=5/2$).

Answer 2

It turns out that this system of Lagrange equations doesn't really warrant the use of Mathematica. We are faced with

$$ \ 2x \ = \ (2x \ + \ \frac{6}{5}y)\lambda \ \ , \ \ 2y \ = \ (2y \ + \ \frac{6}{5}x)\lambda \ \ . $$

Since bringing all the terms to one side in each equation won't help us to factor the expressions, we might instead solve each equation for $ \ \lambda \ $ to obtain

$$ \lambda \ \ = \ \ \frac{x}{x \ + \ \frac{3}{5} y} \ \ = \ \ \frac{y}{y \ + \ \frac{3}{5} x} \ \ . $$

Assuming for the moment that neither denominator is equal to zero (it will turn out that they are not), we can "cross-multiply" the ratios to produce

$$ xy \ + \ \frac{3}{5} x^2 \ \ = \ \ xy \ + \ \frac{3}{5} y^2 \ \ \Rightarrow \ \ x^2 \ = \ y^2 \ \ \Rightarrow \ \ y \ = \ \pm x \ \ . $$

We have two cases now for insertion into the curve equation:

$$ \mathbf{y = x :} \quad 5x^2 \ + \ 6xy \ + \ 5y^2 \ = \ 8 \ \ \rightarrow \ \ 5x^2 \ + \ 6·x·x \ + \ 5x^2 \ = \ 8 \ \ \Rightarrow \ \ 16x^2 \ = \ 8 $$ $$ \Rightarrow \ \ x^2 \ = \ \frac{1}{2} \ \ ; $$ $$ \mathbf{y = -x :} \quad \quad \quad \rightarrow \ \ 5x^2 \ + \ 6·x·(-x) \ + \ 5x^2 \ = \ 8 \ \ \Rightarrow \ \ 4x^2 \ = \ 8 \ \ \Rightarrow \ \ x^2 \ = \ 2 \ \ . $$

If we only need the "distance-squared" from the origin for the extremal points, then the cases are $ \ \mathbf{y = x :} \ \ x^2 \ + \ y^2 \ = \ \frac{1}{2} \ + \ \frac{1}{2} \ = \ 1 \ \ $ and $ \ \ \mathbf{y = -x :} \ \ x^2 \ + \ y^2 \ = \ 2 \ + \ 2 \ = \ 4 \ , $ making the extremal distances from the origin $ \ 1 \ $ and $ \ 2 \ . $ [The points at these distances are $ \ \left( \pm \frac{1}{\sqrt{2}} \ , \ \pm \frac{1}{\sqrt{2}} \right) \ $ and $ \ \left( \pm \sqrt{2} \ , \ \mp \sqrt{2} \right) \ , $ respectively.

The constraint curve $ \ 5x^2 \ + \ 6xy \ + \ 5y^2 \ = \ 8 \ $ has symmetry about the origin, so it is reasonably to expect that there should be a critical point in each quadrant (with extremal points of the same type in opposite quadrants). The curve is a "rotated ellipse" for which we have found the lengths of its semi-major and semi-minor axes and the locations of their endpoints.