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Chapter 13: Problem 13

Find all critical points of the following functions. $$f(x, y)=x^{4}+y^{4}-16 x y$$

Short Answer

Expert verified
Question: Find the critical points of the function \(f(x,y) = x^{4}+y^{4}-16xy\), and determine their nature (local maximum, local minimum, or saddle point). Answer: The critical points of the given function are (0,0), (2,2), and (-2,-2). To determine their nature, use the second derivative test or compute the eigenvalues of the Hessian matrix.

Step by step solution

01

1. Find the partial derivative with respect to x

To find the partial derivative of the function with respect to x, we treat y as a constant and differentiate only the terms involving x: $$\frac{\partial f}{\partial x} = \frac{d}{dx}(x^{4}+y^{4}-16 x y) = 4x^{3}-16y$$
02

2. Find the partial derivative with respect to y

To find the partial derivative of the function with respect to y, we treat x as a constant and differentiate only the terms involving y: $$\frac{\partial f}{\partial y} = \frac{d}{dy}(x^{4}+y^{4}-16 x y) = 4y^{3}-16x$$
03

3. Set the partial derivatives equal to 0

To find the critical points, we need to set both partial derivatives equal to 0 and solve for x and y: $$4x^{3}-16y = 0$$ $$4y^{3}-16x = 0$$
04

4. Solve for x and y

From the first equation, we can express x in terms of y: $$x^{3} = 4y \Rightarrow x = (4y)^{1/3}$$ Now, substitute this expression for x in the second equation: $$4y^{3} - 16(4y)^{1/3} = 0$$ Factor out a 4y to simplify: $$y( y^{2} - 4(4y)^{-2/3}) = 0$$ So, either \(y = 0\) or \(y^{2} = 4(4y)^{-2/3}\). Solving for y in the second equation gives us \(y = \pm 2\). Now, we find the corresponding x values. If \(y = 0\), then \(x^{3} = 4y \Rightarrow x = 0\). If \(y = 2\), then \(x = (4(2))^{1/3} = 2\). If \(y = -2\), then \(x = (4(-2))^{1/3} = -2\).
05

5. List the critical points

We found three possible critical points: $$(0,0),(2,2),(-2,-2)$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
In multivariable calculus, critical points are where the function's derivative is zero or undefined. This concept helps identify points where the function might achieve local maxima, minima, or saddle points. For our function
  • \(f(x, y) = x^4 + y^4 - 16xy\), finding critical points involves setting the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) to zero and solving for \(x\) and \(y\).
  • This means we are looking for values where the slope鈥攐r rate of change鈥攊n either \(x\) or \(y\) direction is zero.
These points are relevant in optimization problems and can be solved through systems of equations.
Partial Derivatives
To find critical points, partial derivatives are essential. They show how a function changes as we alter one variable, keeping others constant. By calculating these, we understand the behavior of multivariable functions:
  • A partial derivative of \(f(x, y)\) with respect to \(x\), written as \(\frac{\partial f}{\partial x}\), is the derivative while treating \(y\) as a constant.
    For \(x^4 + y^4 - 16xy\), this results in \(4x^3 - 16y\).
  • Similarly, the partial derivative of \(f(x, y)\) with respect to \(y\), shown as \(\frac{\partial f}{\partial y}\), involves differentiating the function with \(x\) held constant:
    Here, that's \(4y^3 - 16x\).
Partial derivatives give the slope of the function along each axis and are critical for solving systems of equations for critical points.
Systems of Equations
Once partial derivatives are found, setting them to zero forms a system of equations. Here's how it works:
  • Our function's derivatives \(4x^3 - 16y = 0\) and \(4y^3 - 16x = 0\) create simultaneous equations.
  • These equations are solved together to find common solutions for both \(x\) and \(y\).
    For instance, by expressing one variable in terms of the other using one equation, substitution into the second can simplify the system.
  • Our example, \(x^3 = 4y\) and \(y^3 = 4x\), leads us to identify values of \((x, y)\) like \((0, 0)\), \((2, 2)\), and \((-2, -2)\).
The solutions to these systems help identify where the function's gradient is zero.
Functions of Multiple Variables
Functions of multiple variables, like \(f(x, y) = x^4 + y^4 - 16xy\), extend single-variable calculus to higher dimensions. Unlike single-variable functions, these have inputs and outputs across different axes, visually represented as surfaces in three-dimensional space.
  • Each variable adds another dimension to the graph, requiring more complex analysis regarding changes and slopes.
  • For instance, changes in \(x\) and \(y\) influence results simultaneously.
  • Understanding these requires utilizing tools like partial derivatives and system solving鈥攁 staple in multivariable calculus.
    The critical points found by setting derivatives to zero illustrate where these multidimensional surfaces behave differently, such as peaking or dipping.
This area of calculus is vital for fields leveraging multidimensional data, including physics, engineering, and economics.

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Find an equation of the plane passing through the point (3,2,1) that slices off the region in the first octant with the least volume.

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