How do you minimize and maximize #f(x,y)=x^2+y^3# constrained to #0<x+3xy<4#?

Answer 1

A local minimum at #{x= 1.12872, y= 0.847942}# and
a local minimum also at #{x = 0, y = 0}#

Introducing the so called slack variables #s_1, s_2# the optimization problem is transformed into an equivalent one

Find local minima, maxima of

#f(x,y)= x^2+y^3#

subject to

#g_1(x,y,s_1) = x + 3 x y - s_1^2 = 0#
#g_2(x.y,s_2)=x+3 x y + s_2^2-4=0#

The lagrangian is

#L(x,y,s_1,s_2,lambda_1,lambda_2) = f(x,y)+lambda_1 g_1(x,y,s_1)+lambda_2g_2(x,y,s_2)#

#L# is analytical so the stationary points include the relative maxima and minima.

The determination of stationary points is done solving for #x,y,s_1,lambda_1,s_2,lambda_2# the system of equations given by

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

or

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

Solving we get

#(x=0., y = 0., lambda_1 = 0., s_1 = 0., lambda_2= 0., s_2= 2.)#

and

# (x= 1.12872, y= 0.847942, lambda_1=0., s_1 = 2., lambda_2 = -0.637008, s_2 = 0.)#

Both points are at the boundaries of #g_1(x,y,0)=0# and #g_2(x,y,0)=0# respectively.

Their qualification must be done with #f_(g_1)# and #f_{g_2}# respectively. So,

#f_{g_1}(x) = x^2-1/27#
#f_{g_2}(x) = -(x-4)^3/(27 x^3) + x^2#

#d^2/(dx^2)f_{g_1}(0)=2# qualifying this point as a local minimum
#d^2/(dx^2)f_{g_2}(1.12872) =11.5726# qualifying this point as a local minimum also

Attached a figure with a contour mapping with the found points

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To minimize and maximize ( f(x, y) = x^2 + y^3 ) constrained to ( 0 < x + 3xy < 4 ), we use the method of Lagrange multipliers. The Lagrangian function is:

[ L(x, y, \lambda) = x^2 + y^3 + \lambda(x + 3xy - 4) ]

We then solve the following system of equations:

[ \frac{\partial L}{\partial x} = 2x + 3y\lambda = 0 ] [ \frac{\partial L}{\partial y} = 3y^2 + 3x\lambda = 0 ] [ \frac{\partial L}{\partial \lambda} = x + 3xy - 4 = 0 ]

Solving these equations simultaneously will give us the critical points. We then evaluate ( f(x, y) ) at these points and check the boundaries of the constraint for possible extrema.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7