Making plot of f(x,y) shows that both Luca's answers are correct. Needles to say that:
1) at no point grad(f) equals to null (sorry, Johannes),
2) the set of extremal points is infinite (sorry, Keivan),
3) as the constraints are given as sharp inequalities then we are talking about supremum rather than maximum.
In this case, I don't think that we should recourse to the method of the Lagrange multipliers; the constraints are too simple, and actually the first one is not needed in order to answer to your problem.
The second constraint alone gives the answer, although we can only talk about of supremum, rather than maximum.
For me that looks like a simple extremal value problem. However you can just determine local extrema of k.
To do this you have to find your partial differentials d²k/dx² and d²k/dy²
You will then find a local maximum if: d²k/dx² > 0 and it has to hold simultaniously that: d²k/dx² * d²k/dy² - (d²k/(dxdy))² > 0
Here you might run in a little trouble as the second equation vanishes, but you can get a good start if you use: grad(k) =0. this gives you the points of extrema of your function (but it doesn't tell you if they are max or min!)
For the use of a Lagrange multiplier you would need some sonstraint, which you have in P/(Qx)-y < 1 - so you should be able to use it.
oh, Luca was faster than me - he is quite right, the equations are quite simple, so you can use Lagrange multipliers but you don't have to.
in range of x and y, min values of x and y give you maximum K in your formula.
Obviously all the points belonging to the hyperbola y = P/(Qx) - 1 are the suprema of f.
Making plot of f(x,y) shows that both Luca's answers are correct. Needles to say that:
1) at no point grad(f) equals to null (sorry, Johannes),
2) the set of extremal points is infinite (sorry, Keivan),
3) as the constraints are given as sharp inequalities then we are talking about supremum rather than maximum.
True - sorry, my mistake - the grad(f) is not zero. But the rest should work :)
thanks Marek
Johannes, don't worry too much. I myself usually go along your lines, except that I start with solving the equation grad(f)=0. Having those solutions I simply evaluate at all those points and - without even trying second derivativatives - I know which one corresponds to global minimum and which one to global maximum. Thit is because in the majority of my problems the function f is defined on the unit sphere; there are no active constraints. With constraints present, the Lagrange multiplier technique is the good choice, but also the red light must ne on as well. Especially, when there are several bounds and not all are active simultaneouskly, that is they intersect themelves and/or they are non-differentiable.
Sorry, but the rest (which "should work") contains a mistake again: your test is good for a minimum rather than maximum (inequalities are reversed). I understand it: in physics we are routinely interested mostly in minima (ground states, etc.). Besides, we have quite a hot weather today.
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