Use Lagrange multipliers to find the max & min

[smize]

Homework Statement

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

f(x,y) = e^xy; g(x,y) = x³ + y³ = 16

Homework Equations

∇f(x,y) = λ∇g(x,y)

f_x = λg_x
f_y = λg_y

The Attempt at a Solution

∇f(x,y) = < ye^xy, xe^xy >
∇g(x,y) = < 3x², 3y² >

f_x = λg_x \Rightarrow ye^xy = λ3x²

f_y = λg_y \Rightarrow xe^xy = λ3y²

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

Since x = y, x^{3} + y^{3} = 2x^{3} \Rightarrow 2x^{3} = 16, x = 2, y = 2

f(2,2) = e^{(2)(2)} = e^{4}

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.

 

[Ray Vickson]

Homework Statement

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

f(x,y) = e^xy; g(x,y) = x³ + y³ = 16

Homework Equations

∇f(x,y) = λ∇g(x,y)

f_x = λg_x
f_y = λg_y

The Attempt at a Solution

∇f(x,y) = < ye^xy, xe^xy >
∇g(x,y) = < 3x², 3y² >

f_x = λg_x \Rightarrow ye^xy = λ3x²

f_y = λg_y \Rightarrow xe^xy = λ3y²

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

Since x = y, x^{3} + y^{3} = 2x^{3} \Rightarrow 2x^{3} = 16, x = 2, y = 2

f(2,2) = e^{(2)(2)} = e^{4}

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
p^T H p &gt; 0 for all nonzero vectors p = (p_x,py) that satisfy p \cdot \nabla g(x,y) \equiv p_x \partial g/\partial x + p_y \partial g / \partial y = 0.

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

 

[smize]

I am sorry, but can you elaborate more on what a Hessian is? Currently we have only covered Lagrange multipliers.

 

[Ray Vickson]

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:
\text{Hessian} F(x,y) =<br /> \begin{pmatrix} \frac{\partial^2 F}{\partial x ^2}&amp; \frac{\partial^2 F}{\partial x \partial y}\\<br /> \frac{\partial^2 F}{\partial y \partial x}&amp;\frac{\partial^2 F}{\partial y ^2}<br /> \end{pmatrix},
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
H_{i,j} = \frac{\partial^2 F}{\partial x_i \partial x_j}.

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:
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.

RGV

 

[ehild]

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

 

[smize]

The Hessian of a function F(x,y) of two variables is the matrix of second partial derivatives:
\text{Hessian} F(x,y) =<br /> \begin{pmatrix} \frac{\partial^2 F}{\partial x ^2}&amp; \frac{\partial^2 F}{\partial x \partial y}\\<br /> \frac{\partial^2 F}{\partial y \partial x}&amp;\frac{\partial^2 F}{\partial y ^2}<br /> \end{pmatrix},
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
H_{i,j} = \frac{\partial^2 F}{\partial x_i \partial x_j}.

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:
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.

RGV

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