Question

Find the extreme values, if any, of the following four functions. Check whether they are maxima or minima by the determinantal test. $$z=e^{2 x}+e^{-y}+e^{w^{2}}-\left(2 x+2 e^{w}-y\right)$$

Short answer

The function \( z \) has a local minimum at \((0, 0, 0)\).

Step-by-step solution

Find the first partial derivatives

First, we need to calculate the first partial derivatives of the function \( z = e^{2x} + e^{-y} + e^{w^2} - (2x + 2e^{w} - y) \) with respect to each variable (\(x, y, w\)).\[ \frac{\partial z}{\partial x} = 2e^{2x} - 2 \]\[ \frac{\partial z}{\partial y} = -e^{-y} + 1 \]\[ \frac{\partial z}{\partial w} = 2we^{w^2} - 2e^{w} \]

Set the first partial derivatives to zero

To find critical points, set each first partial derivative equal to zero and solve for the variables.1. \( 2e^{2x} - 2 = 0 \rightarrow e^{2x} = 1 \rightarrow x = 0 \)2. \( -e^{-y} + 1 = 0 \rightarrow e^{-y} = 1 \rightarrow y = 0 \)3. \( 2we^{w^2} - 2e^{w} = 0 \rightarrow we^{w^2} = e^{w} \rightarrow w = 0 \) (assuming this holds under specific conditions where \( w = 0 \) makes the equation satisfied).

Find the second partial derivatives

Calculate the second partial derivatives to use in the Hessian determinant (determinantal test).\[ \frac{\partial^2 z}{\partial x^2} = 4e^{2x} \]\[ \frac{\partial^2 z}{\partial y^2} = e^{-y} \]\[ \frac{\partial^2 z}{\partial w^2} = 2e^{w^2} + 4w^2e^{w^2} - 2e^{w} \]

Calculate mixed partial derivatives

Find the mixed partial derivatives necessary for the Hessian.\[ \frac{\partial^2 z}{\partial x \partial y} = 0 \]\[ \frac{\partial^2 z}{\partial x \partial w} = 0 \]\[ \frac{\partial^2 z}{\partial y \partial w} = 0 \]

Construct and evaluate the Hessian matrix

Form the Hessian matrix \( H \) using the second and mixed partial derivatives.\[ H = \begin{bmatrix} \frac{\partial^2 z}{\partial x^2} & 0 & 0 \ 0 & \frac{\partial^2 z}{\partial y^2} & 0 \ 0 & 0 & \frac{\partial^2 z}{\partial w^2} \end{bmatrix} = \begin{bmatrix} 4 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 2 \end{bmatrix} \]The determinant of \( H \) is the product of its diagonal elements:\[ \det(H) = 4 \times 1 \times 2 = 8 \]

Determine the nature of the critical point

Since \( \det(H) > 0 \), we look at the sign of \( \frac{\partial^2 z}{\partial x^2} \), \( \frac{\partial^2 z}{\partial y^2} \), and \( \frac{\partial^2 z}{\partial w^2} \), which are all positive at \( x=0, y=0, w=0 \). Therefore, the Hessian is positive definite, indicating a local minimum at \((0, 0, 0)\).

Key concepts

Partial Derivatives

A partial derivative represents how a multivariable function changes when one of its variables changes while the other variables are held constant. This is important in the calculus of several variables. It's akin to slicing a loaf of bread and examining just one slice, revealing only part of the story yet crucial in piecing together the entire picture of the function's behavior.

For example, in the function given, the partial derivative with respect to

• x is computed as \( \frac{\partial z}{\partial x} = 2e^{2x} - 2 \), showing how \( z \) changes as \( x \) changes.
• Similarly, for \( y \), we have \( \frac{\partial z}{\partial y} = -e^{-y} + 1 \).
• For \( w \), \( \frac{\partial z}{\partial w} = 2we^{w^2} - 2e^{w} \).

Each of these derivatives highlights how the function behaves individually with respect to each variable.

Critical Points

Critical points are the values of variables at which the first partial derivatives of a function are zero. These points are potential locations of maxima, minima, or saddle points on the graph of a function.
To find the critical points of our function:

• Set \( 2e^{2x} - 2 = 0 \) for \( x \), leading to \( x = 0 \).
• For \( y \), solve \( -e^{-y} + 1 = 0 \), resulting in \( y = 0 \).
• For \( w \), \( 2we^{w^2} - 2e^{w} = 0 \) simplifies to \( w = 0 \) under specific conditions.

Therefore, the critical point is \( (0, 0, 0) \), a central point to further analyze.

Hessian Matrix

The Hessian Matrix is a square matrix of second-order partial derivatives of a scalar-valued function. It provides critical insights into the local curvature of the function, thereby aiding in the classification of critical points.
For the given problem, the Hessian matrix \( H \) is formed as:\[H = \begin{bmatrix} \frac{\partial^2 z}{\partial x^2} & \frac{\partial^2 z}{\partial x \partial y} & \frac{\partial^2 z}{\partial x \partial w} \ \frac{\partial^2 z}{\partial y \partial x} & \frac{\partial^2 z}{\partial y^2} & \frac{\partial^2 z}{\partial y \partial w} \ \frac{\partial^2 z}{\partial w \partial x} & \frac{\partial^2 z}{\partial w \partial y} & \frac{\partial^2 z}{\partial w^2} \end{bmatrix}\]Simplified to:\[\begin{bmatrix} 4 & 0 & 0 \0 & 1 & 0 \0 & 0 & 2 \end{bmatrix}\]Each entry along the diagonal depicts the second partial derivatives, indicating how the slope of each variable changes as that particular variable changes.

Determinantal Test

The Determinantal Test uses the Hessian matrix to determine the nature of a critical point. Specifically, it looks at the determinant of the Hessian matrix to infer the type of extremum a critical point represents.

If \( \det(H) \) is positive, it suggests the Hessian is either positive definite (indicating a local minimum) or negative definite (indicating a local maximum). If negative, it typically indicates a saddle point.
For this function:

• \( \det(H) = 4 \times 1 \times 2 = 8 \) which is positive. Therefore, the Hessian matrix is positive definite.
• Since all second partial derivatives at the critical point (\( x = 0, y = 0, w = 0 \)) are positive (as computed), it confirms a local minimum at the critical point.

This provides a formal mathematical basis to identify how the function behaves around the critical points.