Recent content by Mike s

Thanks a lot!

I have already solved it, but I need confirmation: Are there other ways of proving this? Thanks in advance!

You are right. I will notify my teacher. However, when you take X = 1/x, \; Y = 1/y., can you really say it's a circle centered in (0,0)? X and Y cannot be equal to zero, therefore the constraint is not compact, so you cannot guarantee the existence of global extrema.

I have solved this problem by using Lagrange multipliers. However, the answers that my teacher has published, say that the solution to the Lagrange system of equations is neither minimum nor maximum because the function has no extrema. Proving that this function does not have extrema is the...

Hello, I need to find (if there are) minimum and maximum values of the following function: z=\frac{1}{x}+\frac{1}{y} subject to constraint: \frac{1}{{x}^{2}}+\frac{1}{{y}^{2}}=\frac{1}{{a}^{2}} a\neq 0 I think there are no extrema, but I do not know how to show it.

I understand, thanks for both of you.

Hello, Is x=-1 local maximum of F(x)=x\sqrt{1+x}? On the one hand, F(-1+\delta)<F(-1) for 0<\delta<1. However, F(-1-\delta) is not defined. As far a I know, the point x=a is considered local maximum, if there exists small neighborhood \delta such that F(a)>F(x) for every...

For all the other directions except (-1,y) and (x,-1), I can find the other critical points simply by solving the two equations: Zx=0 Zy=0 right? I specifically chose to check (-1,y) and (x,-1) because the gradient is not defined in them.

Thanks! Since we haven't learned yet how to find extreme points under constraints, I've tried to solve this in a different way: Z\left(x,-1\right)=-{\left(1+x\right)}^{\frac{1}{2}} Z_x\left(x,-1\right)=-\frac{1}{2}{\left(1+x\right)}^{-\frac{1}{2}} Since Z_x is negative for all x\ge -1 ...

Hello, As far as I know, critical points are the point in which the gradient equals zero or not defined. In the following functionZ=X(1+Y)^{1/2}+Y(1+X)^{1/2}, the partial derivatives are not defined for all the points: (-1,Y) or (X,-1) in which X,Y are bigger or equal to -1. Why is the point...

Alright, thanks a lot!

Well, as far as I know, the limit does not have to be defined at the point itself (0,0), but there must exist at least one disc that contains it.

Can you think of another way? If you do, please post

I need to prove this limit does NOT exist. I know that the denominator cancels out, however my lecturer says the limit does not exist because we cannot find a circle with radius \delta that contains the point (0,0), as the function is not defined for all y=x. Here is a little drawing of the...