Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ f(x, y)=x y^{2}-6 x^{2}-3 y^{2} $$

Step-by-Step Solution

Answer

The critical points are \((0, 0)\), \((3, 6)\), and \((3, -6)\).

Step 1: Understand the Function

First, look at the given function: \( f(x, y) = x y^2 - 6x^2 - 3y^2 \). This is a multivariable function involving both \(x\) and \(y\). It has terms involving \(x^2\), \(y^2\), and their product \(xy^2\). It is useful to identify each term to understand their impact on the function's behavior.

Step 2: Identify Critical Points

Find the critical points by setting the partial derivatives equal to zero. Calculate the partial derivative with respect to \(x\): \( \frac{\partial f}{\partial x} = y^2 - 12x = 0 \). Calculate the partial derivative with respect to \(y\): \( \frac{\partial f}{\partial y} = 2xy - 6y = 0 \). Solve these equations simultaneously to find the values of \(x\) and \(y\) for critical points.

Step 3: Solve Partial Derivative Equations

From the partial derivative with respect to \(x\), we have:\( y^2 = 12x \).From the partial derivative with respect to \(y\), factor the equation:\( 2y(x - 3) = 0 \).This gives two cases: \( y = 0 \) or \( x = 3 \). Substitute these into \( y^2 = 12x \) to get the critical points.

Step 4: Case 1: \( y = 0 \)

If \( y = 0 \), substitute into \( y^2 = 12x \), resulting in \( 0 = 12x \). Therefore, \( x = 0 \). So one critical point is \((0, 0)\).

Step 5: Case 2: \( x = 3 \)

For \( x = 3 \), substitute into \( y^2 = 12x \) to get \( y^2 = 36 \), thus \( y = 6 \) or \( y = -6 \). Critical points are \((3, 6)\) and \((3, -6)\).

Step 4: Short Summary of Critical Points

The critical points found are \((0, 0)\), \((3, 6)\), and \((3, -6)\). These points need to be further analyzed to determine their nature (i.e., whether they are minima, maxima, or saddle points). Further analysis can be done using the second partial derivative test.

Key Concepts

Multivariable Calculus

In multivariable calculus, we deal with functions that have more than one variable. These are functions like \( f(x, y) = x y^2 - 6x^2 - 3y^2 \), where both \( x \) and \( y \) are variables. Understanding these functions involves examining how they change with respect to each variable, individually and collectively.
Multivariable calculus is essential to model and solve problems arising in various fields such as physics, engineering, economics, and biology. It allows us to analyze systems in which multiple quantities are interdependent.

• Concepts such as gradients, divergence, and curl extend notions of derivative and integrals to multiple dimensions.
• Partial derivatives and multiple integrals play a crucial role in finding local maxima, minima, or saddle points.

These tools help simplify complex relationships between variables, making multivariable calculus a valuable asset in both theoretical and applied mathematics.

Partial Derivatives

Partial derivatives are a fundamental aspect of multivariable calculus, focusing on examining how a multivariable function changes with respect to one variable at a time while keeping the others constant. This partial differentiation allows us to study the behavior of each component of a function separately. For a function \( f(x, y) \), the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) denote the rate of change of the function with respect to \( x \) and \( y \) respectively.
Partial derivatives are critical in determining critical points of a function, which occur where these partial derivatives are zero. By setting \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) to zero, we can find potential points where a function may have maximum, minimum, or saddle point characteristics.

• They are computed similarly to single-variable derivatives, treating all other variables as constants.
• Partial derivatives are used extensively in optimization problems, especially in finding the optimal solution given constraints.

Understanding and calculating partial derivatives are vital steps in exploring and solving complex problems in multivariable settings.

Second Partial Derivative Test

The second partial derivative test is crucial in assessing the nature of critical points found using partial derivatives of a multivariable function. It helps determine whether a critical point is a local maximum, a local minimum, or a saddle point.
To use this test, we first compute the second partial derivatives of the function, which involves finding the derivatives of the initial partial derivatives. For a function \( f(x, y) \), these are \( \frac{\partial^2 f}{\partial x^2} \), \( \frac{\partial^2 f}{\partial y^2} \), and \( \frac{\partial^2 f}{\partial x \partial y} \).
Next, we calculate the determinant of the Hessian matrix, defined as:
\[ D(x, y) = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2 \]

• If \( D(x, y) > 0 \) and \( \frac{\partial^2 f}{\partial x^2} > 0 \), the function has a local minimum at the critical point.
• If \( D(x, y) > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), the function has a local maximum.
• If \( D(x, y) < 0 \), the point is a saddle point.

This test provides a systematic approach to determining the specific characteristics of critical points.