Problem 11 - 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. $$f(x, y)=x^{3}-12 x y+8 y^{3}$$

Step-by-Step Solution

Answer

Local minimum at (2, 1); saddle points at (0, 0) and (-2, 1).

Step 1: Find the First Partial Derivatives

To find the critical points of the function, we first need to compute the partial derivatives with respect to \(x\) and \(y\). Calculate \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\):\[\frac{\partial f}{\partial x} = 3x^2 - 12y\] \[\frac{\partial f}{\partial y} = -12x + 24y^2\].

Step 2: Solve for Critical Points

Set each of the partial derivatives equal to zero to find the critical points: 1. \(3x^2 - 12y = 0\)2. \(-12x + 24y^2 = 0\).From equation (1), \(y = \frac{x^2}{4}\). Substitute into equation (2) to get: \(-12x + 24\left(\frac{x^2}{4}\right)^2 = 0\). Solving this gives the critical points.

Step 3: Determine Critical Points

Substitute \(y = \frac{x^2}{4}\) into the second equation and solve:\(-12x + 24\frac{x^4}{16} = 0\). Simplify to isolate and solve for \(x\): \(x(-12 + \frac{3}{2}x^3) = 0\). Solve for \(x\) obtaining \(x = 0, \pm2\). Substitute back to find \(y\): When \(x = 0\), \(y = 0\).When \(x = 2\), \(y = 1\).When \(x = -2\), \(y = 1\). Thus, the critical points are \((0,0), (2,1), (-2,1)\).

Step 4: Analyze Each Critical Point Using the Hessian Matrix

Use the second derivative test to classify each critical point. Compute the second partial derivatives:\[ \frac{\partial^2 f}{\partial x^2} = 6x, \quad \frac{\partial^2 f}{\partial y^2} = 48y, \quad \text{and}\quad \frac{\partial^2 f}{\partial x \partial y} = -12. \]The Hessian matrix, \(H\), is:\[H = \begin{bmatrix}6x & -12\-12 & 48y\end{bmatrix}.\] Calculate the determinant of \(H\) at each critical point.

Step 5: Classify the Critical Points

Evaluate the determinate of the Hessian matrix at each critical point:- At \((0, 0)\): \(\det(H) = 6(0) \cdot 48(0) - (-12)^2 = -144\), indicating a saddle point.- At \((2, 1)\): \(\det(H) = 6(2) \cdot 48(1) - (-12)^2 = 576 - 144 = 432\), with \(6 > 0\), indicating a local minimum.- At \((-2, 1)\): \(\det(H) = 6(-2) \cdot 48(1) - (-12)^2 = -576 - 144 = -720\), indicating another saddle point.

Step 6: Summarize the Findings

The function \(f(x, y) = x^{3} - 12xy + 8y^{3}\) has a saddle point at \((0, 0)\) and another saddle point at \((-2,1)\). There is a local minimum at \((2, 1)\).

Key Concepts

Partial Derivatives

Partial derivatives are fundamental in multivariable calculus, helping us understand how functions change with respect to one variable while keeping others constant. This is particularly useful for functions of more than one variable, like in our exercise where we have the function \(f(x, y) = x^3 - 12xy + 8y^3\). To find partial derivatives, we focus our calculations on one variable at a time.

• The partial derivative with respect to \(x\), denoted \(\frac{\partial f}{\partial x}\), indicates how the function \(f\) changes as \(x\) changes, keeping \(y\) constant.
• Similarly, \(\frac{\partial f}{\partial y}\) tells us about changes in \(f\) as \(y\) changes, keeping \(x\) constant.

In our solution, we computed these partial derivatives as follows:\[\frac{\partial f}{\partial x} = 3x^2 - 12y\]\[\frac{\partial f}{\partial y} = -12x + 24y^2\]These derivatives are crucial for finding critical points, which provide deeper insight into the behavior of multivariable functions.

Critical Points

Critical points occur where all partial derivatives of a function are zero or undefined, showing potential places of maxima, minima, or saddle points on its graph. In the given function \(f(x, y) = x^3 - 12xy + 8y^3\), we set its partial derivatives to zero to find these critical points.Setting \(\frac{\partial f}{\partial x} = 0\) and \(\frac{\partial f}{\partial y} = 0\), we obtained the following equations:1. \(3x^2 - 12y = 0\)2. \(-12x + 24y^2 = 0\)From the first equation, \(y = \frac{x^2}{4}\). Substituting into the second equation and solving gives us the critical points:

• \((0, 0)\)
• \((2, 1)\)
• \((-2, 1)\)

These points are essential to classify, using the Hessian matrix, to ascertain whether they correspond to maxima, minima, or saddle points.

Hessian Matrix

The Hessian matrix provides a way to classify critical points in multivariable calculus by using second-order partial derivatives. For the function \(f(x, y) = x^3 - 12xy + 8y^3\), the Hessian matrix \(H\) is constructed using the following:\[H = \begin{bmatrix}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix}\] Calculating these, we have:

• \(\frac{\partial^2 f}{\partial x^2} = 6x\)
• \(\frac{\partial^2 f}{\partial y^2} = 48y\)
• \(\frac{\partial^2 f}{\partial x \partial y} = -12\)

Thus, the Hessian matrix becomes:\[H = \begin{bmatrix} 6x & -12 \ -12 & 48y \end{bmatrix}\]To determine the nature of each critical point, we calculate the determinant of this matrix at each point. - At \((0, 0)\), \(\det(H) = -144\) indicating a saddle point.- At \((2, 1)\), \(\det(H) = 432\) suggesting a local minimum.- At \((-2, 1)\), \(\det(H) = -720\) also indicating a saddle point.These calculations clarify the behavior of the function around each critical point, showing us how the surface curves in these areas.