Using partial derivatives to find a maximum

[Signifier]

Okay, I have a question that I am not good enough at multivariable calculus yet to answer myself.

Basically, say I have the following inequality:

a < sin(xy)/y < b

with a < b. How can I find out what the maximum value of y is on the interval 0 < y < 1 such that the above inequality holds for some x > 0?

That is, what is the biggest value of y between 0 and 1, such that for some x > 0, sin(xy)/y is between two limits a and b with a < b?

Is this even possible...?

Any information, hints, outright answers, scathing insults etc. would be appreciated! Thank you.

 

[Office_Shredder]

Given complete control of x, xy can be whatever you want. So imagine instead xy = z, we can assume z is completely independent of y (since we can pick x to satisfy what we need). So you really have:

a<sin(z)/y<b

Keeping in mind that sin(z) is between 0 and 1, can you work from there?

 

[jbusc]

Hopefully this will be enough to solve your problem:

Theorem: If a function has local extrema at a point, and the first partial derivatives of the function exist at that point, then the first derivatives of the function vanish at that point.

Theorem: For every function f(x, y), define D(a, b) = $$\frac{\partial ^2 f}{\partial x^2} \frac{\partial^2 f}{\partial y^2} - \frac{\partial^2 f}{\partial x \partial y} \frac{\partial^2 f}{\partial y \partial x}$$ everywhere the first partial derivatives vanish and the second partial derivatives exist.

Then, if D(a, b) < 0, then (a, b) is a saddle point of f
If D(a, b) > 0, and the second derivatives are positive, then (a, b) is a local minimum.
If D(a, b) > 0, and the second derivatives are negative, then (a, b) is a local maximum.
If D(a, b) = 0, then any of saddle point, local maximum or minimum, or no extrema could occur at that point - no new information is known.