Question

Define \(f\) in \(R^{2}\) by $$ f(x, y)=2 x^{3}-3 x^{2}+2 y^{3}+3 y^{2} $$ (a) Find the four points in \(R^{2}\) at which the gradient of \(f\) is zero. Show that \(f\) has exactly one local maximum and one local minimum in \(R^{2}\). (b) Let \(S\) be the set of all \((x, y) \in R^{2}\) at which \(f(x, y)=0\). Find those points of \(S\) that have no neighborhoods in which the equation \(f(x, y)=0\) can be solved for \(y\) in terms of \(x\) (or for \(x\) in terms of \(y\) ). Describe \(S\) as precisely as you can.

Short answer

Question: Determine the exact locations of the local maximum and local minimum of the function \(f(x, y) = 2x^3 - 3x^2 + 2y^3 + 3y^2\). Additionally, find the equation of the set \(S\) where the given function is equal to zero and discuss the expressibility of the equation in terms of \(x\) and \(y\). Answer: The function \(f(x, y)\) has exactly one local maximum at the point \((0, -1)\) and one local minimum at the point \((1, 0)\). The set \(S\) is given by the equation \(2x^3 - 3x^2 + 2y^3 + 3y^2 = 0\), and it cannot be expressed in terms of \(x\) or \(y\) for any point in the set.

Step-by-step solution

Calculate the gradient

We first compute the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\). For \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\), we have: $$ \frac{\partial f}{\partial x} = 6x^2 - 6x, $$ and $$ \frac{\partial f}{\partial y} = 6y^2 + 6y. $$ The gradient of \(f\) is given by \(\nabla f(x, y)=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})\).

Find the critical points where the gradient is zero

We need to find the points \((x, y)\) where the gradient of \(f\) is zero. This requires solving the following system of equations: $$ 6x^2 - 6x = 0, $$ and $$ 6y^2 + 6y = 0. $$ For the first equation, we can factor \(6x\) to get \(6x(x - 1) = 0\), so \(x = 0\) or \(x = 1\). For the second equation, we can factor \(6y\) to get \(6y(y + 1) = 0\), so \(y = 0\) or \(y = -1\). Thus, we have four critical points: \((0, 0)\), \((1, 0)\), \((0, -1)\), and \((1, -1)\).

Determine local extrema using the second derivative test

We will find the second derivatives of \(f\) and use the second derivative test to determine if the critical points are local maxima, minima, or saddle points. The second partial derivatives of \(f\) are: $$ \frac{\partial^2 f}{\partial x^2} = 12x - 6, $$ $$ \frac{\partial^2 f}{\partial y^2} = 12y + 6, $$ and $$ \frac{\partial^2 f}{\partial x\partial y} = 0. $$ The determinant of the Hessian matrix is given by \(D(x, y) = (\frac{\partial^2 f}{\partial x^2})(\frac{\partial^2 f}{\partial y^2}) - (\frac{\partial^2 f}{\partial x\partial y})^2\). Since \(\frac{\partial^2 f}{\partial x\partial y} = 0\), we have: $$ D(x, y) = (12x - 6)(12y + 6). $$ Now, we can evaluate the determinant \(D(x, y)\) at the critical points from step 2. 1. At \((0, 0)\), we have \(D(0,0) = -6 \cdot 6 <0\), which means it's a saddle point. 2. At \((1, 0)\), we have \(D(1,0) = 6 \cdot 6 >0\) and \(\frac{\partial^2 f}{\partial x^2}(1,0) = 6 > 0\), so it's a local minimum. 3. At \((0, -1)\), we have \(D(0,-1) = -6 \cdot -6 >0\) and \(\frac{\partial^2 f}{\partial x^2}(0,-1) = -6 < 0\), so it's a local maximum. 4. At \((1, -1)\), we have \(D(1,-1) = 6 \cdot -6 <0\), which means it's a saddle point. Thus, we found that there is exactly one local maximum \((0, -1)\) and one local minimum \((1, 0)\) in \(R^2\).

Finding the equation of the set \(S\)

We are given that the set \(S\) consists of all \((x, y) \in R^2\) such that \(f(x, y) = 0\). Therefore, we need to find the equation of the set \(S\) by setting the function equal to zero: $$ 2x^3 - 3x^2 + 2y^3 + 3y^2 = 0. $$

Analyzing the equation of set \(S\) for expressibility in terms of \(x\) and \(y\)

With the equation of set \(S\), we will look for points where it is impossible to express \(y\) in terms of \(x\) or \(x\) in terms of \(y\). Looking at the equation, it seems very difficult to isolate either variable without having a highly complicated or impossible expression. Given this, we can conclude that there are no neighborhoods in the set \(S\) where the equation can be solved for \(y\) in terms of \(x\) or \(x\) in terms of \(y\). Therefore, every point on the set will have no neighborhood with an explicit expression for \(x\) or \(y\). As a conclusion, the set of all \((x, y) \in R^2\) at which \(f(x, y) = 0\) is given by the equation \(2x^3 - 3x^2 + 2y^3 + 3y^2 = 0\), and none of the points in this set have neighborhoods in which this equation can be expressed in terms of either \(x\) or \(y\).

Key concepts

Gradient of a Function

Understanding the gradient of a function is crucial for exploring the landscape of a mathematical surface. In essence, the gradient vector captures the steepest ascent direction at a particular point on a surface. It's a directional measure pointing towards the highest increase of the function's value. Given a function like \(f(x, y) = 2x^3 - 3x^2 + 2y^3 + 3y^2\), finding the gradient involves taking the partial derivatives with respect to each variable.

Calculating \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) gives us components of the gradient vector, \( abla f(x, y) \). Each component represents the rate of change of the function along their respective axes. Zero gradient, where both partial derivatives are zero, indicates we've found a point where there is no instantaneous change in the function's value—potentially a peak, trough, or saddle point.

It's worth noting that the gradient's utility isn't just in finding critical points. It's also a workhorse in optimization problems and machine learning algorithms, where it helps to minimize or maximize functions.

Critical Points of a Function

Critical points of a function are where first derivatives (or gradients, in the case of multivariable functions) are zero, indicating a possible maximum, minimum, or saddle point. When evaluating \(f(x, y) = 2x^3 - 3x^2 + 2y^3 + 3y^2\), we find critical points by setting the gradient components to zero and solving the resulting system of equations.

By factoring the derived equations for \(x\) and \(y\), we unravel the critical points. This systematic approach helps us to explore the function's landscape, analogous to finding the tops and bottoms of hills or points of inflection on a rolling terrain. Remember, not all critical points will correspond to absolute maxima or minima; some might be local extremes or even flatlands (saddles).

Given the intertwined nature of concepts in mathematics, one often finds that a deep understanding of critical points paves the way for grasping other areas such as optimization and dynamical systems, reinforcing the importance of mastering this concept.

Second Derivative Test

The second derivative test is a method used in mathematical analysis to help classify critical points found via the gradient of a function. For functions of two variables like \(f(x, y)\), the test involves the Hessian matrix—a matrix of second-order partial derivatives—and its determinant.

In our case study, the determinant of the Hessian matrix at a critical point, \(D(x, y)\), helps us decide the nature of the point: a local maximum, minimum, or a saddle point. The rules are simple but potent:

• If \(D(x, y) > 0\) and \(\frac{\partial^2 f}{\partial x^2} > 0\), the function has a local minimum at \((x, y)\).
• If \(D(x, y) > 0\) and \(\frac{\partial^2 f}{\partial x^2} < 0\), it's a local maximum.
• If \(D(x, y) < 0\), the point is a saddle point, signifying a dip in one direction and a rise in another.

The elegance of this test is that it takes the guesswork out of identifying critical points after they're found. Such methods are not mere academic exercises—they're tools widely used in economics, engineering, and the physical sciences to model reality in a way that is simple and approachable.