Use Lagrange multipliers to find the max & min

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  • #1
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Homework Statement



Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint.

f(x,y) = exy; g(x,y) = x3 + y3 = 16

Homework Equations



∇f(x,y) = λ∇g(x,y)

fx = λgx
fy = λgy

The Attempt at a Solution



∇f(x,y) = < yexy, xexy >
∇g(x,y) = < 3x2, 3y2 >

fx = λgx [itex]\Rightarrow[/itex] yexy = λ3x2

fy = λgy [itex]\Rightarrow[/itex] xexy = λ3y2

λ = [itex]\frac{xe^{xy}}{3y^{2}}[/itex] = [itex]\frac{ye^{xy}}{3x^{2}}[/itex] [itex]\Rightarrow[/itex] [itex]3x^{3}e^{xy} = 3y^{3}e^{xy}[/itex] [itex]\Rightarrow[/itex] x = y

Since x = y, [itex]x^{3} + y^{3} = 2x^{3} \Rightarrow 2x^{3} = 16, x = 2, y = 2[/itex]

f(2,2) = [itex]e^{(2)(2)} = e^{4}[/itex]

Here is where I'm having troubles. How do I determine whether it is a maximum or minimum? Could you give me more than one example? If I did anything wrong, please point it out.
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint.

f(x,y) = exy; g(x,y) = x3 + y3 = 16

Homework Equations



∇f(x,y) = λ∇g(x,y)

fx = λgx
fy = λgy

The Attempt at a Solution



∇f(x,y) = < yexy, xexy >
∇g(x,y) = < 3x2, 3y2 >

fx = λgx [itex]\Rightarrow[/itex] yexy = λ3x2

fy = λgy [itex]\Rightarrow[/itex] xexy = λ3y2

λ = [itex]\frac{xe^{xy}}{3y^{2}}[/itex] = [itex]\frac{ye^{xy}}{3x^{2}}[/itex] [itex]\Rightarrow[/itex] [itex]3x^{3}e^{xy} = 3y^{3}e^{xy}[/itex] [itex]\Rightarrow[/itex] x = y

Since x = y, [itex]x^{3} + y^{3} = 2x^{3} \Rightarrow 2x^{3} = 16, x = 2, y = 2[/itex]

f(2,2) = [itex]e^{(2)(2)} = e^{4}[/itex]

Here is where I'm having troubles. How do I determine whether it is a maximum or minimum? Could you give me more than one example? If I did anything wrong, please point it out.

You could try a rough plot of g = 16 and a rough contour plot of f, to see whether the point you have is a maximum or a minimum. It might be easier to use f = x*y instead, because in the first quadrant x,y ≥ 0, x*y is a max or min if and only if exp(x*y) is a max or a min.

There are second-order tests for max or min in constrained problems: (1) A necessary condition for a min of f at a stationarty point (x,y) of the Lagrangian is that the Hessian of the Lagrangian (not f or g separately!) is positive semi-definite in the tangent space of the constraint at (x,y); and (2) a sufficient condition for a (strict) local min is that the Hessian of the Lagrangian is positive definite in the tangent space of the constraint. By positive (semi-)definite in the tangent space, I mean that the Hessian H of L must satisfy
[tex] p^T H p > 0 [/tex] for all nonzero vectors p = (px,py) that satisfy [tex] p \cdot \nabla g(x,y) \equiv p_x \partial g/\partial x + p_y \partial g / \partial y = 0.[/tex]

Tests for a max are the opposite: just test for negative (semi-) definite instead.

Note that to carry out the test you need a value of λ as well as values for x and y, because you need the Hessian of the Lagrangian at (x,y,λ).

RGV
 
  • #3
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I am sorry, but can you elaborate more on what a Hessian is? Currently we have only covered Lagrange multipliers.
 
  • #4
Ray Vickson
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I am sorry, but can you elaborate more on what a Hessian is? Currently we have only covered Lagrange multipliers.

The Hessian of a function F(x,y) of two variables is the matrix of second partial derivatives:
[tex]\text{Hessian} F(x,y) =
\begin{pmatrix} \frac{\partial^2 F}{\partial x ^2}& \frac{\partial^2 F}{\partial x \partial y}\\
\frac{\partial^2 F}{\partial y \partial x}&\frac{\partial^2 F}{\partial y ^2}
\end{pmatrix},[/tex]
where all partial derivatives are evaluated at (x,y).

If you have more than two variables, say ##F(x_1, x_2, \ldots, x_n),## the Hessian is an n×n matrix with elements
[tex] H_{i,j} = \frac{\partial^2 F}{\partial x_i \partial x_j}.[/tex]

In the multivariate case the Hessian plays a role similar to that of the second derivative in the univariate case, in the sense that it gives the second-order terms in the Taylor expansion of F:
[tex] F(x+u,y+v) \approx F(x,y) + uF_x(x,y) + v F_y(x,y) + \frac{1}{2} (u,v)^T H(x,y) \pmatrix{u \\v} + \text{ higher order terms in } u, \; v. [/tex]

RGV
 
Last edited:
  • #5
ehild
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Here is where I'm having troubles. How do I determine whether it is a maximum or minimum? Could you give me more than one example? If I did anything wrong, please point it out.

Choose an x value close to x=2, (say x=2.1) and the appropriate y value which satisfy the condition. (y~1.89) It should be less than e4 if (2,2) is a maximum point.

ehild
 
  • #6
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The Hessian of a function F(x,y) of two variables is the matrix of second partial derivatives:
[tex]\text{Hessian} F(x,y) =
\begin{pmatrix} \frac{\partial^2 F}{\partial x ^2}& \frac{\partial^2 F}{\partial x \partial y}\\
\frac{\partial^2 F}{\partial y \partial x}&\frac{\partial^2 F}{\partial y ^2}
\end{pmatrix},[/tex]
where all partial derivatives are evaluated at (x,y).

If you have more than two variables, say ##F(x_1, x_2, \ldots, x_n),## the Hessian is an n×n matrix with elements
[tex] H_{i,j} = \frac{\partial^2 F}{\partial x_i \partial x_j}.[/tex]

In the multivariate case the Hessian plays a role similar to that of the second derivative in the univariate case, in the sense that it gives the second-order terms in the Taylor expansion of F:
[tex] F(x+u,y+v) \approx F(x,y) + uF_x(x,y) + v F_y(x,y) + \frac{1}{2} (u,v)^T H(x,y) \pmatrix{u \\v} + \text{ higher order terms in } u, \; v. [/tex]

RGV

Ah! Yes, I know this function. I just didn't know it's name. Thank-you for reminding me of it!
 

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