How do you minimize and maximize #f(x,y)=(2x-y)+(x-2y)/x^2# constrained to #1

Question

How do you minimize and maximize #f(x,y)=(2x-y)+(x-2y)/x^2# constrained to #1<yx^2+xy^2<3#?

Luke Alvarez · Accepted Answer

Two local maxima at
#{x = -2.43526, y= 0.716915}#
#{x = -1.75902, y = 0.426693}#
Two local minima at
#{x= 1.60753, y= -2.38878}#
#{x = 1.39063, y = -1.79193}#

With the so called slack variables #s_1, s_2# we transform the maximization/minimization with inequality constraints problem, into a formulation amenable for the Lagrange Multipliers technique.

Now the lagrangian formulation reads:

minimize/maximize
#f(x,y) = f(x,y)=2x-y+(x-2y)/x^2#

subjected to
#g_1(x,y,s_1) = x y^2+x^2y -s_1^2 -1=0#
#g_2(x,y,s_2) = x y^2+x^2y +s_1^2 -3=0#

forming the lagrangian

# L(x,y,s_1,s_2,lambda_1,lambda_2) = f(x,y)+lambda_1g_1(x,y,s_1)+lambda_2 g_2(x,y,s_2)#

The local minima/maxima points are included into the lagrangian stationary points found by solving

#grad L(x,y,s_1,s_2,lambda_1,lambda_2) = vec 0#

or

#{ ( (4 y + x^3 (2 + (lambda_1 + lambda_2) y (2 x + y)))/x^3 - 1/x^2=0), ((lambda_1 + lambda_2) x (x + 2 y) - 1 - 2/x^2=0), (1 + s_1^2 - x y (x + y)=0), (lambda_1 s_1 = 0), (s_2^2 + x y (x + y) -3=0), (lambda_2 s_2 = 0) :}#

Solving for #x,y,s_1,s_2,lambda_1,lambda_2# we obtain

#( (x = -2.43526, y= 0.716915, l1 = 0, s1= -1.41421, l2= 0.548335, s2 = 0), (x = -2.43526, y = 0.716915, l1= 0, s1 = 1.41421, l2 = 0.548335, s2= 0), (x= 1.60753, y= -2.38878, l1 = 0, s1 = -1.41421, l2= -0.348111, s2 = 0), (x = 1.60753, y = -2.38878, l1 = 0, s1 = 1.41421, l2= -0.348111, s2 = 0), (x = -1.75902, y = 0.426693, l1 = 1.03348, s1 = 0, l2= 0, s2= -1.41421), (x = -1.75902, y = 0.426693, l1 = 1.03348, s1 = 0, l2 = 0, s2 = 1.41421), (x = 1.39063, y = -1.79193, l1 = -0.666958, s1 = 0, l2 = 0, s2 = -1.41421), (x = 1.39063, y = -1.79193, l1 = -0.666958, s1= 0, l2 = 0, s2 = 1.41421) )#

Those eight points are truly four. The first and second activate constraint #g_2(x,y,0)# and the two other activate constraint #g_1(x,y,0)# and their qualification will be done with

#f@g_1 (x) = 2/x + (5 x)/2 + sqrt[4 + x^3]/x^(5/2) + sqrt[4 + x^3]/(2 sqrt[x])# and
#f@g_2(x) =2/x + (5 x)/2 + sqrt[12 + x^3]/x^(5/2) + sqrt[12 + x^3]/(2 sqrt[x])#

giving

#d^2/(dx^2)f@g_1 (-1.75902) = -8.29064# local maximum
#d^2/(dx^2)f@g_1 (1.39063) = 5.90567# local minimum
#d^2/(dx^2)f@g_2 (-2.43526) = -8.85699# local maximum
#d^2/(dx^2)f@g_2 ( 1.60753) = 5.10192# local minimum

Attached a figure with the #f(x,y)# contour map inside the feasible region, with the local maxima/minima points.

Elijah Abram · Answer

undefined To minimize and maximize the function $ f(x,y) = \frac{{2x-y}}{{x}} + \frac{{x-2y}}{{x^2}} $ constrained to $ 1 < yx^2 + xy^2 < 3 $, you can use the method of Lagrange multipliers.

First, you set up the Lagrangian:

$ L(x, y, \lambda) = \frac{{2x-y}}{{x}} + \frac{{x-2y}}{{x^2}} + \lambda(yx^2 + xy^2 - 1) - \mu(yx^2 + xy^2 - 3) $

Then, take the partial derivatives with respect to $ x $, $ y $, and $ \lambda $, and set them equal to zero:

$ \frac{{\partial L}}{{\partial x}} = 0 $,
$ \frac{{\partial L}}{{\partial y}} = 0 $,
$ \frac{{\partial L}}{{\partial \lambda}} = 0 $.

Solve the resulting system of equations to find the critical points. Then, evaluate the function at these critical points and the boundary points of the constraint $ 1 < yx^2 + xy^2 < 3 $ to find the minimum and maximum values of $ f(x, y) $.

Isabella Ackerman · Answer

undefined To minimize and maximize the function $ f(x, y) = (2x - y) + \frac{x - 2y}{x^2} $ constrained to $ 1 < yx^2 + xy^2 < 3 $, follow these steps:

1. Find the critical points of the function by taking the partial derivatives with respect to $ x $ and $ y $ and setting them equal to zero.
2. Use the second derivative test to determine whether the critical points are maxima, minima, or saddle points.
3. Evaluate the function at the critical points and at the boundary points of the constraint region.
4. Compare the values obtained to find the maximum and minimum values of the function within the constraint region.

How do you minimize and maximize #f(x,y)=(2x-y)+(x-2y)/x^2# constrained to #1&lt;yx^2+xy^2&lt;3#?

How do you minimize and maximize #f(x,y)=(2x-y)+(x-2y)/x^2# constrained to #1<yx^2+xy^2<3#?