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

Question

Isabella Ackerman · Accepted Answer

The function:

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

constrained by #xy=2#

has two local maximums for #(x,y) = (-1,-2)# and #(x,y) = (1,2)# From the constraint equation: #xy = 2# we can derive: #y=2/x# and substitute it in the original equation: #g(x) = f(x,y(x)) = 2/((x-2)(2/x-4)) = (2x)/((x-2)(2-4x))# We can now find the local extrema for #g(x)#: #g(x) = (2x)/( 2x-4-4x^2+8x) = - (x)/(2x^2-5x+2)# #(dg)/dx = - ( 2x^2-5x+2 - x(4x-5))/(2x^2-5x+2)# #(dg)/dx = - ( 2x^2-5x+2 - 4x^2+5x)/(2x^2-5x+2)# #(dg)/dx = ( 2x^2-2)/(2x^2-5x+2) = (2(x^2-1))/((x-2)(2x-1))# The critical points are determined by the equation: #(dg)/dx = 0 => x=+-1# Looking at the sign of the derivative we can see that: #(dg)/dx > 0# for #x<-1# #(dg)/dx < 0# for #-1 < x< 1/2# So that #x=-1# is a local maximum and: #(dg)/dx > 0# for #1/2 < x < 1# #(dg)/dx < 0# for #1 < x < 2# so that also #x=1# is a local maximum.

Luke Alvarez · Answer

undefined To minimize and maximize the function $ f(x, y) = \frac{xy}{(x - 2)(y - 4)} $ constrained to $ xy = 2 $, you can follow these steps:

1. Use the constraint $ xy = 2 $ to express one variable in terms of the other. Let's solve for $ y $ in terms of $ x $:
   $$ y = \frac{2}{x} $$

2. Substitute $ y = \frac{2}{x} $ into the objective function $ f(x, y) $ to express it as a function of one variable $ x $ only.

3. Take the derivative of the resulting function with respect to $ x $ and find the critical points by setting the derivative equal to zero and solving for $ x $.

4. Test the critical points to find which ones correspond to local minima, local maxima, or saddle points.

5. Check the endpoints of the domain of $ x $ to see if they yield any extreme values.

6. Finally, compare the values of $ f(x, y) $ at the critical points and endpoints to determine the absolute minimum and maximum values.