Shortest Distance Homework: Find Origin to Curve x^2+2xy+y^2=150

[Air]

Homework Statement

Find the shortest distance from the origin to the curve $$x^2+2xy+y^2=150$$.

Homework Equations

$$\frac{\partial f}{\partial x}$$, $$\frac{\partial f}{\partial y}$$


3. The problem I'm occurring
I'm not sure how to start is thus can't attempt it. I would have used the Lagrange theory but that would give the max point and I also don't have the constraint so cannot use it. For a start, can someone suggest how I would start. Any methods? I have to use partial differentiation.

 

[yyat]

You can use Lagrange's multiplier method to find maxima as well as minima. Apply it to the distance squared function f(x,y)=x^2+y^2 with the constraint x^2+2xy+y^2=150.

 

[Air]

I have reached a point of confusion...

$$f(x,y) = x^2+y^2$$
$$g(x,y) = x^2+2xy+y^2=150$$

$$\mathrm{Equation 1: }2x - \lambda (2x + 2y) = 0 \implies \lambda = \frac{2x}{2x+2y}$$
$$\mathrm{Equation 2: }2y - \lambda (2x + 2y) = 0 \rightarrow 2y - \left( \frac{2x}{2x+2y}\right) (2x + 2y ) = 0 \rightarrow 2x+2y = 0 \implies y=x$$

Into constraint equation to solve for value of $$x$$ and $$y$$.
$$\therefore y^2 + 2y^2 y^2 = 150 \implies y = \pm \sqrt{\frac{150}{4}} = x$$

Which gives many solutions. How can I saw which is the shortest distance?

 

[yyat]

Well, you got two solutions: $$(\sqrt{150}/2,\sqrt{150}/2)$$ and $$(-\sqrt{150}/2,-\sqrt{150}/2)$$. Both have the same distance from the origin, compute it and you have the answer.