Problem 10 - Calculus Early Transcendentals

Question

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=2 x^{3}+x y^{2}+5 x^{2}+y^{2}$$

Step-by-Step Solution

Answer

The function has a local minimum at (0,0), a local maximum at (-5/3,0), and saddle points at (-1,2) and (-1,-2).

Step 1: Find the First Partial Derivatives

To find the critical points, we first need to calculate the partial derivatives of the function with respect to both variables. The partial derivative with respect to \(x\) is:\[\frac{\partial f}{\partial x} = 6x^2 + y^2 + 10x\]The partial derivative with respect to \(y\) is:\[\frac{\partial f}{\partial y} = 2xy + 2y\]

Step 2: Set the First Derivatives to Zero

To find critical points, set the first partial derivatives equal to zero and solve the resulting system of equations:\[6x^2 + y^2 + 10x = 0\]\[2xy + 2y = 0\]Simplifying the second equation gives:\[2y(x+1) = 0\]which suggests that either \(y = 0\) or \(x = -1\).

Step 3: Analyze the System of Equations

Consider \(y = 0\):Plug it into the first equation:\[6x^2 + 10x = 0\]Factor out \(x\):\[x(6x + 10) = 0\]So, \(x = 0\) or \(x = -\frac{5}{3}\).Consider \(x = -1\):Plug it into the first equation:\[6(-1)^2 + y^2 + 10(-1) = 0\]This simplifies to:\[6 + y^2 - 10 = 0 \Rightarrow y^2 = 4\]Hence, \(y = 2\) or \(y = -2\).

Step 4: List the Critical Points

The critical points from the analysis above are \((0, 0)\), \((-\frac{5}{3}, 0)\), \((-1, 2)\), and \((-1, -2)\).

Step 5: Find the Second Partial Derivatives

Next, compute the second partial derivatives:\[\frac{\partial^2 f}{\partial x^2} = 12x + 10\]\[\frac{\partial^2 f}{\partial y^2} = 2x + 2\]\[\frac{\partial^2 f}{\partial x \partial y} = 2y\]

Step 6: Use the Second Derivative Test

For a function \(f(x, y)\), the Hessian matrix is used to apply the second derivative test, which is:\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix}\]The determinant of \(H\) is:\[D = \left(\frac{\partial^2 f}{\partial x^2}\right)\left(\frac{\partial^2 f}{\partial y^2}\right) - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2\]

Step 7: Evaluate the Hessian at Each Critical Point

Evaluate \(D\) at each of the critical points:1. At \((0, 0)\): \[\frac{\partial^2 f}{\partial x^2} = 10, \frac{\partial^2 f}{\partial y^2} = 2, \frac{\partial^2 f}{\partial x \partial y} = 0\] \[D = 10 \cdot 2 - 0^2 = 20\] Since \(D > 0\) and \( \frac{\partial^2 f}{\partial x^2} > 0\), \((0, 0)\) is a local minimum. 2. At \((-\frac{5}{3}, 0)\): \[\frac{\partial^2 f}{\partial x^2} = 12(-\frac{5}{3}) + 10 = -10\] \[\frac{\partial^2 f}{\partial y^2} = 2(-\frac{5}{3}) + 2 = -\frac{4}{3}\] \[\frac{\partial^2 f}{\partial x \partial y} = 0\] \[D = (-10)(-\frac{4}{3}) - 0^2 > 0\] and \( \frac{\partial^2 f}{\partial x^2} < 0\), \((-\frac{5}{3}, 0)\) is a local maximum. 3. At \((-1, 2)\) and \((-1, -2)\): Since the determinant involves the squares of non-zero mixed derivative terms, they both result with negative discriminants indicating saddle points.

Key Concepts

Critical Points in Multivariable Calculus

When examining functions of multiple variables, identifying critical points is key. A critical point occurs where the gradient, or vector of partial derivatives, equals zero or is undefined. This helps in finding potential maxima, minima, or saddle points of the function.
For the given function \( f(x, y) = 2x^3 + xy^2 + 5x^2 + y^2 \), we determined the critical points by first finding the partial derivatives of \( f \) with respect to \( x \) and \( y \), then setting them equal to zero to solve the equations.

• Solving \( \frac{\partial f}{\partial x} = 6x^2 + y^2 + 10x = 0 \)
• Solving \( \frac{\partial f}{\partial y} = 2xy + 2y = 0 \)

This process unveiled the potential critical points at \( (0, 0), (-\frac{5}{3}, 0), (-1, 2), \) and \( (-1, -2) \).

Understanding Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They measure how a function changes as one variable changes, while the other variables are held constant.
For the function \( f(x, y) = 2x^3 + xy^2 + 5x^2 + y^2 \), the partial derivatives are:

• \( \frac{\partial f}{\partial x} = 6x^2 + y^2 + 10x \), which considers changes in \( x \) while keeping \( y \) constant.
• \( \frac{\partial f}{\partial y} = 2xy + 2y \), which considers changes in \( y \) while keeping \( x \) constant.

By setting these derivatives to zero, we determine where the rate of change of the function is zero in every direction. These points are critical in identifying where potential extrema or saddle points occur within a multivariable function.

Applying the Second Derivative Test

The second derivative test for functions of two variables is an extension of the single-variable test for concavity. It uses the Hessian matrix, a 2x2 matrix of the second partial derivatives.
The determinant of this matrix \( D \) is used to identify the nature of each critical point.

• If \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} > 0 \), the point is a local minimum.
• If \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), the point is a local maximum.
• If \( D < 0 \), the point is a saddle point, indicating a mixed curvature.

This test helps us confirm whether the critical points identified are maxima, minima, or saddle points. It provides a concrete basis for describing the local geometry of the function.

Decoding the Hessian Matrix

The Hessian matrix is a powerful tool in determining the nature of critical points in multivariable functions. It is composed of the second partial derivatives of the function.
For the given function, the Hessian is:
\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix}\]
Where:

• \( \frac{\partial^2 f}{\partial x^2} = 12x + 10 \)
• \( \frac{\partial^2 f}{\partial y^2} = 2x + 2 \)
• \( \frac{\partial^2 f}{\partial x \partial y} = 2y \)

By evaluating this Hessian at each critical point and calculating its determinant \( D \), we are empowered to classify each point. Computing \( D \) and assessing the terms allows one to understand the type of curvature present at each point, leading to a qualified categorization of each point as a local minimum, maximum, or saddle point.