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 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#