Problem 312 - Calculus Volume 3

Question

For the following exercises, find all critical points. $$f(x, y)=x^{4}+y^{4}-16 x y$$

Step-by-Step Solution

Answer

The critical points are (0, 0), (2, 2), and (-2, -2).

Step 1: Find Partial Derivatives

To find the critical points of \[f(x, y) = x^{4} + y^{4} - 16xy\]we first need to calculate the partial derivatives with respect to both variables. The partial derivative with respect to \(x\) is found by differentiating while treating \(y\) as a constant. Similarly, we find the partial derivative with respect to \(y\).

Step 2: Calculate f_x and f_y

The partial derivative \(f_x\) is:\[f_x = \frac{\partial}{\partial x}(x^4 + y^4 - 16xy) = 4x^3 - 16y\]The partial derivative \(f_y\) is:\[f_y = \frac{\partial}{\partial y}(x^4 + y^4 - 16xy) = 4y^3 - 16x\]

Step 3: Set Partial Derivatives to Zero

To find the critical points, set the partial derivatives equal to zero:\[4x^3 - 16y = 0\]\[4y^3 - 16x = 0\]

Step 4: Simplify the System of Equations

Simplify each equation by dividing by 4:\[x^3 - 4y = 0\]\[y^3 - 4x = 0\]

Step 5: Solve the System for Critical Points

From \(x^3 = 4y\) and \(y^3 = 4x\), substitute \(y = \frac{x^3}{4}\) into \(y^3 = 4x\):\[\left(\frac{x^3}{4}\right)^3 = 4x\]which simplifies to:\[\frac{x^9}{64} = 4x\]Solve for \(x\):\[x^9 = 256x\]\[x(x^8 - 256) = 0\]Thus, \(x = 0\) or \(x^8 = 256\).

Step 6: Solve for y Based on x Values

When \(x = 0\), substitute into \(x^3 = 4y\) to find:\[0 = 4y \Rightarrow y = 0\]When \(x^8 = 256\), solve for \(x\):\[x = 2\] or \(x = -2\)For these \(x\) values, \(y\) will be:\[y = \frac{x^3}{4}\]Thus, for \(x = 2\), \(y = 2\); for \(x = -2\), \(y = -2\).

Step 7: List Critical Points

The critical points of the function are \((0, 0), (2, 2), (-2, -2)\).

Key Concepts

Partial Derivatives

When working with multivariable functions, partial derivatives play a significant role. They help determine how a function changes when only one variable is altered, keeping the others constant. They are similar to single-variable derivatives but are taken separately for each variable.
To find the partial derivative of a multivariable function like \(f(x, y) = x^4 + y^4 - 16xy\):

• Differentiate with respect to \(x\) while treating \(y\) as a constant. This gives the partial derivative \(f_x\).
• Differentiate with respect to \(y\) while treating \(x\) as a constant. This gives the partial derivative \(f_y\).

Both \(f_x\) and \(f_y\) tell us about the slope of the function surface in the directions parallel to the \(x\)- and \(y\)-axes respectively. Critical points are usually where these partial derivatives both equal zero, indicating potential peaks, troughs, or saddle points.

System of Equations

A critical step in finding critical points is solving the system of equations derived from setting partial derivatives equal to zero. This process involves simplifying and solving the resulting simultaneous equations.
For the function \(f(x, y) = x^4 + y^4 - 16xy\), we set the partial derivatives equal to zero:

• \(4x^3 - 16y = 0\)
• \(4y^3 - 16x = 0\)

Simplifying these equations helps in reducing their complexity, leading to simplified relations such as \(x^3 - 4y = 0\) and \(y^3 - 4x = 0\).
The goal is to solve these simplified equations to find the variable values that make both equations true simultaneously. Substitution methods or algebraic manipulations can be used to isolate variables and solve this system.

Multivariable Function Analysis

Analyzing multivariable functions involves understanding not only how each variable affects the function but also how they interact with each other. Critical points are important since they often indicate locations where interesting behavior occurs in a function.
For the function \(f(x, y) = x^4 + y^4 - 16xy\), the critical points are determined by solving: - The resulting critical points, \((0, 0)\), \((2, 2)\), and \((-2, -2)\) point to strategic locations on the function's surface.- By interpreting these points, one can determine whether they correspond to local maxima, minima, or saddle points using further calculus techniques like the second derivative test or Hessian matrix analysis.
These analyses are crucial in applications such as optimizing profit, minimizing cost, or analyzing physical phenomena.

Symmetry in Equations

Symmetry in equations can significantly simplify the process of finding solutions in multivariable calculus. When functions exhibit symmetry, it often reduces the amount of work needed to find critical points.
In the example function \(f(x, y) = x^4 + y^4 - 16xy\), symmetry is evident from:

• The terms \(x^4\) and \(y^4\) treat \(x\) and \(y\) in the same manner, indicating symmetric roles.
• The cross term \(-16xy\) also suggests an interplay between the variables that maintains symmetry.

Therefore, solutions like \((x, y) = (2, 2)\) and \((-2, -2)\) show that the critical points are symmetric too. Recognizing symmetry can guide you to attempt solving with substitutions, knowing similar calculations or reasoning applies along parallel or analogous paths.