Finding extrema of a function subject to constraint

[Mike s]

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.

 

[Ray Vickson]

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.

There are two standard methods: (1) use the constraint to solve for one of the variables in terms of the other, then substitute that expression into z; or (2) use the method of Lagrange multipliers. If, for example, you solve for y in terms of x, you will have two solutions because you have a quadratic constraint. You need to look at both of the resulting problems.

There _are_ extrema in this problem.

RGV

 

[HallsofIvy]

I guess my question would be why do you need to find minimum and maximum values? You seem to be saying you have no idea how to approach this problem so apparently it is not for a Calculus class where one would expect to see such problems. Using the Lagrange multiplier method leads to y= x which should also be obvious from the symmetry of the two formulas. As Ray Vickson says, there is one minimum and one maximum.

 

[Mike s]

I guess my question would be why do you need to find minimum and maximum values? You seem to be saying you have no idea how to approach this problem so apparently it is not for a Calculus class where one would expect to see such problems. Using the Lagrange multiplier method leads to y= x which should also be obvious from the symmetry of the two formulas. As Ray Vickson says, there is one minimum and one maximum.

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 part where I am stuck.

Here is the solution posted by my teacher (sorry for my English):

 

[Ray Vickson]

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 part where I am stuck.

Here is the solution posted by my teacher (sorry for my English):

Here is another twist on the problem. Assume throughout that a > 0. Suppose we look at $X = 1/x, \; Y = 1/y.$ Now the problem is to maximize or minimize $z = X + Y,$ subject to the constraint $X^2 + Y^2 = 1/a^2.$ The constraint region is a circle of radius 1/a centered at (0,0) in (X,Y)-space, and in this problem the GLOBAL max occurs at $$(X,Y) = \left(\frac{1}{a\sqrt{2}},\frac{1}{a \sqrt{2}} \right),$$ giving $z_{\max} = \sqrt{2}/a,$ while the global min is at
$$(X,Y) = \left(-\frac{1}{a\sqrt{2}},-\frac{a}{a \sqrt{2}} \right),$$ giving
$z_{\min} = - \sqrt{2}/a.$ Since these optimal are global, it is absolutely impossible to have (X,Y) giving z > z_max or z < z_min.

Your posted solution given above contains errors. The correct equation for y should be
$$\left(n^2 - \frac{1}{a^2} \right) y^2 - 2ny + 2 = 0,$$ which does NOT have real roots for $|n| > \sqrt{2}/a.$

RGV

 

[Mike s]

Here is another twist on the problem. Assume throughout that a > 0. Suppose we look at $X = 1/x, \; Y = 1/y.$ Now the problem is to maximize or minimize $z = X + Y,$ subject to the constraint $X^2 + Y^2 = 1/a^2.$ The constraint region is a circle of radius 1/a centered at (0,0) in (X,Y)-space, and in this problem the GLOBAL max occurs at $$(X,Y) = \left(\frac{1}{a\sqrt{2}},\frac{1}{a \sqrt{2}} \right),$$ giving $z_{\max} = \sqrt{2}/a,$ while the global min is at
$$(X,Y) = \left(-\frac{1}{a\sqrt{2}},-\frac{a}{a \sqrt{2}} \right),$$ giving
$z_{\min} = - \sqrt{2}/a.$ Since these optimal are global, it is absolutely impossible to have (X,Y) giving z > z_max or z < z_min.

Your posted solution given above contains errors. The correct equation for y should be
$$\left(n^2 - \frac{1}{a^2} \right) y^2 - 2ny + 2 = 0,$$ which does NOT have real roots for $|n| > \sqrt{2}/a.$

RGV

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.

 

[Ray Vickson]

Wrong. In (X,Y) space the function to be optimized is X+Y and the constraint region is a circle centered at (0,0). Of course I am allowed to have X=0, but such an X is not the reciprocal of a finite x. Do not confuse two separate issues: SOME X and Y are related to finite values of x and y, but I am allowed to look at other values of X and Y as well.

Also: do not ever assume that a continuous function on a non-compact set lacks maxima or minima. Compactness is a _sufficient_ condition for existence of a max or min, but it is not a _neccessary_ condition. Many smooth functions on non-compact regions have both maxima and minima.

RGV