Maximize x^2+y^2 subject to the condition ax^2+2hxy+by^2=1.

Question

freckles · Answer

$$P=x^2+y^2 \ ax^2+2hxy+by^2=1 \a  \cdot x^2+2hy \cdot x+by^2-1=0 \ \ \text{ Solve the second equation for x} \ x=\frac{-a \pm \sqrt{(2hy)^2-4(a)(by^2-1)}}{2(a)}$$
That's really ugly...
So you can replace x with that and differentiate P w.r.t. y

freckles · Answer

well -b would be -2hy

freckles · Answer

$$x=\frac{-\color{red}{2hy} \pm \sqrt{(2hy)^2-4(a)(by^2-1)}}{2(a)}  $$

paxpolaris · Answer

** correction:$$x=\frac{-\color{red}{2hy} \pm \sqrt{(2hy)^2-4(a)(by^2-1)}}{2(a)}$$

$$\implies x={-hy \pm \sqrt{\left( h^2-ab \right) y^2+a}\over a}$$

freckles · Answer

By the way pax and I are just using the quadratic formula. I know it looks scarier than that.

anonymous · Answer

There is Lagrange's method to solve , which may be easier I think, but its huge.

freckles · Answer

$$\text{ Let } P=x^2+y^2 \ \text{ and } G=ax^2+2hxy+by^2 \ P_x=\lambda G_x \ P_y=\lambda G_y \ax^2+2hxy+by^2=1$$
We need to solve those 3 equations...

freckles · Answer

solve this system of 3 equations*