Problem 20 - Calculus Early Transcendentals

Question

Find the critical points of the following functions. Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, local minimum, or saddle point. Confirm your results using a graphing utility. $$f(x, y)=(4 x-1)^{2}+(2 y+4)^{2}+1$$

Step-by-Step Solution

Answer

Question: Determine the critical point(s) of the given function and identify whether it is a local minimum, local maximum, or saddle point - $$f(x, y) = (4x - 1)^2 + (2y + 4)^2 + 1$$. Answer: The function has a critical point at \((\frac{1}{4}, -2)\) which corresponds to a local minimum.

Step 1: Find the partial derivatives of the function with respect to x and y

First, find the partial derivatives of the function with respect to x and y. For the given function $$f(x, y) = (4x - 1)^2 + (2y + 4)^2 + 1$$, we have: $$ f_x(x, y) = \frac{\partial f}{\partial x} = 2(4x - 1)(4) = 32x - 8 \\ f_y(x, y) = \frac{\partial f}{\partial y} = 2(2y + 4)(2) = 8y + 16 $$

Step 2: Find the critical points

Find where both partial derivatives are equal to zero. $$ f_x(x, y) = 32x - 8 = 0 \\ f_y(x, y) = 8y + 16 = 0 $$ Solve these equations simultaneously: $$ x = \frac{1}{4} \\ y = -2 $$ We found a critical point at \((\frac{1}{4}, -2)\).

Step 3: Apply the Second Derivative Test

Calculate the second partial derivatives: $$ f_{xx}(x, y) = \frac{\partial^2 f}{\partial x^2} = 32 \\ f_{yy}(x, y) = \frac{\partial^2 f}{\partial y^2} = 8 \\ f_{xy}(x, y) = \frac{\partial^2 f}{\partial x \partial y} = 0 \\ f_{yx}(x, y) = \frac{\partial^2 f}{\partial y \partial x} = 0 $$ Compute the discriminant: $$ D(x, y) = f_{xx}(x, y)f_{yy}(x, y) - (f_{xy}(x, y))^2 = (32)(8)-(0)^2 = 256 $$ Since D > 0 and $$f_{xx}(\frac{1}{4}, -2) = 32 > 0$$, the critical point \((\frac{1}{4}, -2)\) is a local minimum.

Step 4: Confirm the results with a graphing utility

Plot the function in a graphing utility (not provided here) and observe that the function has a local minimum at the point \((\frac{1}{4}, -2)\), confirming our results. In conclusion, the given function $$f(x, y) = (4x - 1)^2 + (2y + 4)^2 + 1$$ has a critical point at \((\frac{1}{4}, -2)\), and this point corresponds to a local minimum.

Key Concepts

Partial Derivatives

Partial derivatives are a key tool in calculus when dealing with functions of two or more variables. They help us understand how the function changes as we change one variable while keeping others constant. For the function \( f(x, y) = (4x - 1)^2 + (2y + 4)^2 + 1 \), we find partial derivatives with respect to \(x\) and \(y\).

• The partial derivative with respect to \(x\), denoted \( f_x \), measures the rate of change of \( f \) as \( x \) changes, holding \( y \) constant. In our function, we have \( f_x = 32x - 8 \).
• Similarly, the partial derivative with respect to \( y \), denoted \( f_y \), measures the rate of change of \( f \) as \( y \) changes, keeping \( x \) fixed. Here, it is \( f_y = 8y + 16 \).

These derivatives are fundamental in locating critical points, which involve solving for when these derivatives equal zero.

Critical Points

Critical points are values of \( x \) and \( y \) where the partial derivatives of a function simultaneously equal zero. These points can correspond to local maxima, minima, or saddle points of the function.
By setting our partial derivatives \( 32x - 8 = 0 \) and \( 8y + 16 = 0 \) equal to zero and solving, we find:

• For \( x \), we solve \( 32x - 8 = 0 \) to find \( x = \frac{1}{4} \).
• For \( y \), \( 8y + 16 = 0 \) leads us to \( y = -2 \).

Hence, the critical point is \( (\frac{1}{4}, -2) \), a point where the surface represented by the function has a potential extremum.

Local Minimum

The distinction between a local minimum and other types of critical points often involves applying the Second Derivative Test.
This test uses the second partial derivatives to analyze the nature of the critical point. In our exercise:

• We computed \( f_{xx} = 32 \) and \( f_{yy} = 8 \), which are the second partial derivatives with respect to \( x \) and \( y \) respectively.
• The mixed partial derivatives \( f_{xy} \) and \( f_{yx} \) are both zero in this case.
• To determine if our critical point is a local minimum, we calculate the discriminant \( D(x, y) = f_{xx} \cdot f_{yy} - (f_{xy})^2 = 256 \).

Because \( D > 0 \) and \( f_{xx} = 32 > 0 \), this confirms that the point \( (\frac{1}{4}, -2) \) is indeed a local minimum, indicating a low point in the surface in the neighborhood of this point.