How do you minimize and maximize #f(x,y)=(x-y)/((x-2)^2(y-4))# constrained to #xy=3#?

Question

Samantha Adkison · Accepted Answer

Local maximum located at #(x = -4.89184, y =-0.613266)#

We will be looking for stationary points with posterior qualification.
This technique consists in finding points such that the normal vector to the objective function

#f(x,y)=(x - y)/((x - 2)^2 (y - 4))#

and the restriction function

#g(x,y)=x y - 3 = 0#

are aligned. Formally speaking, there exists #lambda# such that

#grad f(x,y)+lambda grad g(x,y) = vec 0#

proceeding this way we obtain the equation set

#{ (-((2 + x - 2 y)/((x-2)^3 (y-4))) +lambda y=0),( ( 4 - x)/((x-2)^2 (y-4)^2) + lambda x=0), (x y-3=0) :}#

Without much effort can be obtained the unique real solution

#(x = -4.89184, y =-0.613266, lambda= 0.00179817)#

Qualifying this point must be done considering the restriction so the easiest way to do that is substituting the restriction #g(x,y)# into the
objective function #f(x,y)# obtaining

#(f_g) (x) =(3 - x^2)/((x-2)^2 (4 x-3)) #

then we can verify that #d/(dx)(f_g) (-4.89184) = 0#
and

#d^2/(dx^2)(f_g) (-4.89184)= -0.000965087#

qualifying this point as a local maximum.

Attached are two figures. One representing the surface intersection of #f(x,y)# and #g(x,y)# and the other showing the function #(f_g) (x)#

Axel Alvarez · Answer

undefined To minimize and maximize $ f(x,y) = \frac{x - y}{(x - 2)^2(y - 4)} $ constrained to $ xy = 3 $, you can use the method of Lagrange multipliers.

First, write the Lagrangian:

$$ L(x, y, \lambda) = \frac{x - y}{(x - 2)^2(y - 4)} + \lambda(xy - 3) $$

Next, find the partial derivatives of $ L $ with respect to $ x $, $ y $, and $ \lambda $, and set them equal to zero to find critical points.

$$ \frac{\partial L}{\partial x} = \frac{y(x - 4) + \lambda y}{(x - 2)^2(y - 4)^2} = 0 $$

$$ \frac{\partial L}{\partial y} = \frac{x(x - 4) - \lambda x}{(x - 2)^2(y - 4)^2} = 0 $$

$$ \frac{\partial L}{\partial \lambda} = xy - 3 = 0 $$

Solve these equations simultaneously to find the critical points. Then, evaluate $ f(x, y) $ at each critical point and compare the values to find the minimum and maximum.