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Chapter 12: Problem 10

Find all critical points of the following functions. \(f(x, y)=x^{2}-6 x+y^{2}+8 y\)

Short Answer

Expert verified
Answer: The critical point is a local minimum.

Step by step solution

01

Find the partial derivatives

We need to find the partial derivatives of \(f(x, y)\) with respect to both \(x\) and \(y\). Let's start with \(x\): \(\frac{\partial f}{\partial x} = 2x - 6\). Now let's find the partial derivative with respect to \(y\): \(\frac{\partial f}{\partial y} = 2y + 8\).
02

Find the critical points

To find the critical points, we'll set both partial derivatives equal to zero and solve the equations simultaneously: \begin{align*} 2x - 6 &= 0 \\ 2y + 8 &= 0 \end{align*} Solve for \(x\) and \(y\): \begin{align*} x &= 3 \\ y &= -4 \end{align*} So the critical point is at \((3, -4)\).
03

Use the second partial derivative test

Compute the following second partial derivatives: \begin{align*} \frac{\partial^2 f}{\partial x^2} &= 2 \\ \frac{\partial^2 f}{\partial y^2} &= 2 \\ \frac{\partial^2 f}{\partial x \partial y} &= \frac{\partial^2 f}{\partial y \partial x} = 0 \end{align*} Now, calculate the determinant \(D\) of the Hessian matrix: \[D = \frac{\partial^2 f}{\partial x^2} \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2\] \[D = (2)(2) - (0)^2 = 4\] Since \(D > 0\) and \(\frac{\partial^2 f}{\partial x^2} > 0\), the function has a local minimum at the critical point \((3, -4)\). So the critical point of the given function \(f(x, y) = x^2 - 6x + y^2 + 8y\) is a minimum at \((3, -4)\).

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Key Concepts

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

Partial Derivatives
Partial derivatives are an essential concept in multivariable calculus. They help us analyze how a function changes as each variable changes, keeping the others constant.
For a function of two variables, like our example function \(f(x, y) = x^2 - 6x + y^2 + 8y\), calculating partial derivatives allows us to understand how the function behaves in both the \(x\) direction and the \(y\) direction.
  • Partial Derivative with Respect to \(x\): Here, we hold \(y\) constant and differentiate with respect to \(x\). For this function, \(\frac{\partial f}{\partial x} = 2x - 6\).
  • Partial Derivative with Respect to \(y\): Similarly, holding \(x\) constant, we differentiate with respect to \(y\), leading to \(\frac{\partial f}{\partial y} = 2y + 8\).
Both partial derivatives are essential for discovering critical points, as they show where the function's slope is zero. These are key places where the function could potentially have minimums, maximums, or saddle points.
Hessian Matrix
The Hessian matrix is a square matrix of second-order partial derivatives. It provides critical information about the local curvature of functions in multivariable calculus.
For functions of two variables, like \(f(x, y)\), the Hessian matrix is:
  • \(\frac{\partial^2 f}{\partial x^2}\): Second partial derivative with respect to \(x\).
  • \(\frac{\partial^2 f}{\partial y^2}\): Second partial derivative with respect to \(y\).
  • \(\frac{\partial^2 f}{\partial x \partial y}\): Mixed partial derivative, \(\frac{\partial^2 f}{\partial y \partial x}\) should match due to equality of mixed partial derivatives with continuous functions.
In our example, these are:
  • \(\frac{\partial^2 f}{\partial x^2} = 2\)
  • \(\frac{\partial^2 f}{\partial y^2} = 2\)
  • \(\frac{\partial^2 f}{\partial x \partial y} = 0\)
The determinant of the Hessian, or \(D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - (\frac{\partial^2 f}{\partial x \partial y})^2\), helps classify critical points by indicating whether they are local minima, maxima, or saddle points. In our problem, \(D = 4\), which aids in concluding the type of critical point identified.
Local Minimum
A local minimum of a function is a point where the function value is lower than at any nearby points.
These points can be found by analyzing the partial derivatives and the Hessian matrix results. In the given exercise,
  • The critical condition, \(D > 0\), implies the function's quadratic form is definite.
  • The value of the second derivative \(\frac{\partial^2 f}{\partial x^2} = 2 > 0\) suggests that the critical point at \((3, -4)\) is indeed a local minimum.
Local minima are important because they can represent optimal solutions in various applied settings. The calculus of critical points and local minima becomes a powerful tool in fields like optimization, economics, and physics. Understanding these concepts ensures a robust foundation for advanced studies and practical applications involving optimization and analysis of complex, multifaceted functions.

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