Question

Find any extrema of the function `f(x,y)=e^{-xy/4}` subject to the constraint `x^2 + y^2 <=1` ? I desperately need help with these problems as I do not understand them. Thanks!

Find any extrema of the function `f(x,y)=e^{-xy/4}` subject to the constraint `x^2 + y^2 <=1`

I desperately need help with these problems as I do not understand them. Thanks!

Answer 1

Find the extrema of `f(x,y) = e^(-(xy)/4)` with the constraint `C(x,y) = x^2+y^2<=1`

But we must write the constraint function so that the constant on the right is 0:

`C(x,y) = x^2+y^2-1<=0`

We can use a Lagrange multiplier with an inequality constraint

NOTE: I recommend that you read the above references that I have provided.

The Lagrange equation for a single constraint is:

`\mathcal L(x,y,lambda) = f(x,y)+lambdaC(x,y)`

Substitute the values of the functions:

`\mathcal L(x,y,lambda) = e^(-(xy)/4)+lambda(x^2+y^2-1)`

Compute the partial derivatives:

`(del(\mathcal L(x,y,lambda)))/(delx) = -y/4e^(-(xy)/4)+2lambdax`

`(del(\mathcal L(x,y,lambda)))/(dely) = -x/4e^(-(xy)/4)+2lambday`

`(del(\mathcal L(x,y,lambda)))/(del lambda) = x^2+y^2-1`

Set the partial derivatives equal to 0:

`0 = -y/4e^(-(xy)/4)+2lambdax" [1]"`

`0 = -x/4e^(-(xy)/4)+2lambday" [2]"`

`0 = x^2+y^2-1" [3]"`

Rewrite the equations in the following form:

`y/4e^(-(xy)/4)=2lambdax" [1.1]"`

`x/4e^(-(xy)/4)=2lambday" [2.1]"`

`x^2+y^2= 1" [3.1]"`

Divide equation [1.1] by equation [2.1]:

`y/x=x/y`

`y^2 = x^2`

Substitute into [3.1]:

`x^2+x^2= 1`

`x = +-sqrt2/2`

This happens at 4 points on the unit circle.

The minima occur at `(sqrt2/2,sqrt2/2) and (-sqrt2/2,-sqrt2/2)`:

`f(sqrt2/2,sqrt2/2) = f(-sqrt2/2,-sqrt2/2)= 1/(4sqrte)`

The maxima occur at `(sqrt2/2,-sqrt2/2) and (-sqrt2/2,sqrt2/2)`:

`f(sqrt2/2,-sqrt2/2) = f(-sqrt2/2,sqrt2/2) = sqrte/4`