Problem 13 - APEX Calculus

Question

Find the critical points of the given function. Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, or saddle point. $$ f(x, y)=\sqrt{16-(x-3)^{2}-y^{2}} $$

Step-by-Step Solution

Answer

The critical point (3, 0) is a local maximum.

Step 1: Find the First Partial Derivatives

To find the critical points, we first need to find the first partial derivatives with respect to both variables, \(x\) and \(y\). Let \( z=f(x,y) = \sqrt{16-(x-3)^2-y^2} \). The partial derivative with respect to \(x\) is:\[\frac{\partial z}{\partial x} = \frac{-2(x-3)}{2\sqrt{16-(x-3)^2-y^2}}\]Simplify it to:\[\frac{\partial z}{\partial x} = \frac{-(x-3)}{\sqrt{16-(x-3)^2-y^2}}\]For \(y\):\[\frac{\partial z}{\partial y} = \frac{-2y}{2\sqrt{16-(x-3)^2-y^2}}\]Simplify it to:\[\frac{\partial z}{\partial y} = \frac{-y}{\sqrt{16-(x-3)^2-y^2}}\]

Step 2: Set the Partial Derivatives to Zero

To find the critical points, set the first partial derivatives equal to zero.1. \(\frac{-(x-3)}{\sqrt{16-(x-3)^2-y^2}} = 0\)This implies \(x = 3\).2. \(\frac{-y}{\sqrt{16-(x-3)^2-y^2}} = 0\)This implies \(y = 0\).Both conditions must hold simultaneously, so the only critical point is at \((x,y) = (3,0)\).

Step 3: Compute the Second Partial Derivatives

To classify the critical point, we need to compute the second partial derivatives. \(f_{xx} = \frac{4(16-(x-3)^2-y^2) + 4(x-3)^2}{[16-(x-3)^2-y^2]^{3/2}}\)At the point (3,0), this simplifies to \(f_{xx} = -\frac{1}{2}\).\(f_{yy} = \frac{4(16-(x-3)^2-y^2) + 4y^2}{[16-(x-3)^2-y^2]^{3/2}}\)At the point (3,0), this simplifies to \(f_{yy} = -\frac{1}{2}\).\(f_{xy} = 0\) always because the mixed derivative simplifies to zero.

Step 4: Apply the Second Derivative Test

The Second Derivative Test for functions of two variables is determined by the Hessian matrix determinant:\[D = f_{xx}f_{yy} - (f_{xy})^2\]Calculate it using the derivatives found at \( (3,0) \):\[D = \left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right) - (0)^2 = \frac{1}{4}\]Since \( D > 0 \) and \( f_{xx} < 0 \), the critical point \( (3,0) \) is a local maximum.

Key Concepts

Partial Derivatives

In calculus, finding partial derivatives is crucial when dealing with functions of two or more variables. They allow us to understand how a function changes as one of the variables shifts, keeping the others constant.

For the function \( f(x, y)=\sqrt{16-(x-3)^{2}-y^{2}} \), take the derivative of \( f \) with respect to \( x \) and \( y \). This is done by treating all other variables as constants. The partial derivative \( \frac{\partial z}{\partial x} \) was calculated as:
\[ \frac{-(x-3)}{\sqrt{16-(x-3)^2-y^2}} \]
and for \( y \), \( \frac{\partial z}{\partial y} \) as:
\[ \frac{-y}{\sqrt{16-(x-3)^2-y^2}}. \]
By setting these derivatives to zero, critical points are identified, revealing locations where the function does not change with small movements in \( x \) or \( y \). This foundational step is integral to further analyzing a function using other mathematical tests.

Second Derivative Test

The Second Derivative Test helps in determining the nature of critical points identified by the first derivatives. It uses second-order derivatives to evaluate the concavity of the function around those critical points.

We compute the second derivatives \( f_{xx} \), \( f_{yy} \), and \( f_{xy} \) to analyze these concavities. The steps to find the derivatives were already executed in the example, leading to:

• \( f_{xx} = -\frac{1}{2} \) at \( (3,0) \)
• \( f_{yy} = -\frac{1}{2} \) at \( (3,0) \)
• \( f_{xy} = 0 \)

These results play a significant role when constructing the Hessian matrix and applying the test, providing insights into whether the critical point is a local maximum, minimum, or saddle point.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a function, crucial for applying the Second Derivative Test. It helps in understanding how the curvature of the function behaves in the vicinity of the critical points.

For a function of two variables, the Hessian matrix is:
\[ H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix} \]
For the given function and at the point \( (3,0) \), the Hessian matrix becomes:
\[ H = \begin{bmatrix} -\frac{1}{2} & 0 \ 0 & -\frac{1}{2} \end{bmatrix} \]
The determinant of the Hessian \( D = f_{xx}f_{yy} - (f_{xy})^2 \) turns out to be positive \( \left(\frac{1}{4}\right) \), providing critical information for classifying the critical points.

Local Maximum

A local maximum is a point where a function attains its highest value within a neighboring set of points. At this point, small changes in any variable will lead to a decrease in function value.

To confirm a local maximum, we verify using the Second Derivative Test results: if \( D > 0 \) and \( f_{xx} < 0 \), the point is indeed a local maximum.

In our exercise, the critical point \( (3, 0) \) was found to be a local maximum, as the determinant \( D \) was positive and \( f_{xx} \) was negative. This result tells us that near \( (3, 0) \), the function behaves like a hill, reaching its peak at this specific point. Understanding local maxima is often essential in optimization problems, where the goal is to find the highest or lowest points within a defined region.