Problem 10 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The critical point of the function is $(3, -4)$.

Step 1: Write down the function

The given function is: $$f(x, y)=x^{2}-6 x+y^{2}+8 y$$

Step 2: Find the gradient of the function

The gradient of the function is a vector with the partial derivatives of the function as components. To find the gradient, we need to find the partial derivatives with respect to x and y: $$\frac{\partial f}{\partial x} = 2x -6$$ $$\frac{\partial f}{\partial y} = 2y +8$$ The gradient of the function is: $$\nabla f = \left( 2x-6, 2y+8\right)$$

Step 3: Set the gradient equal to zero

To find the critical points, we need to find the points where the gradient of the function is equal to zero. This means we need to solve the following system of equations: \begin{align*} 2x-6 &= 0\\ 2y+8 &= 0 \end{align*}

Step 4: Solve the system of equations

To solve the system of equations, we can isolate x and y from the equations: \begin{align*} x &= \frac{6}{2} = 3\\ y &= \frac{-8}{2} = -4 \end{align*}

Step 5: Write down the critical points

The critical points of the function are the values of x and y that satisfy both equations. In this case, the critical point is: $$(x, y) = (3, -4)$$

Key Concepts

Gradient

When working with multivariable functions, the gradient plays a key role in understanding how the function changes. The gradient is a vector that points in the direction of the greatest rate of increase of the function. For a function \(f(x, y)\), its gradient \(abla f\) is composed of the partial derivatives of the function:

• \(\frac{\partial f}{\partial x}\) - the rate of change of \(f\) with respect to \(x\)
• \(\frac{\partial f}{\partial y}\) - the rate of change of \(f\) with respect to \(y\)

In the context of finding critical points, the gradient is essential because it reveals where the slope of the function is flat. This occurs where \(abla f = 0\), indicating potential maxima, minima, or saddle points.

Partial Derivatives

Partial derivatives are a cornerstone of calculus involving functions of multiple variables. They give us information about how a function changes as each variable changes, independently of the others. Consider the function \(f(x, y) = x^2 - 6x + y^2 + 8y\). Here, you compute the partial derivative:

• With respect to \(x\), \(\frac{\partial f}{\partial x} = 2x - 6\), measures how the function changes as \(x\) alone changes, with \(y\) held constant.
• With respect to \(y\), \(\frac{\partial f}{\partial y} = 2y + 8\), measures how the function changes as \(y\) alone changes, with \(x\) held constant.

Finding these derivatives is the first step in determining how the function's surface is shaped, leading to identifying critical points.

System of Equations

To find the critical points of a function, we need to solve a system of equations composed of its gradient set to zero. From the partial derivatives of our example function, the system of equations is:

• \(2x - 6 = 0\)
• \(2y + 8 = 0\)

Each equation in the system represents a surface in the function's domain. The solutions to these equations are the points where these surfaces intersect, thus revealing the critical points. This step is pivotal as it identifies where potential maxima, minima, or other interesting features of the function occur.

Solve Critical Points

The solution of the system of equations gives the critical points of the function, which are crucial for determining its behavior. Continuing from the equations \(2x - 6 = 0\) and \(2y + 8 = 0\):

• Solve for \(x\), \(x = \frac{6}{2} = 3\)
• Solve for \(y\), \(y = \frac{-8}{2} = -4\)

Together, these values \((3, -4)\) are the coordinates of the critical point. Understanding where these points occur helps to explore the kind of extrema the function may possess, whether it's a peak (maximum), a dip (minimum), or a flat plain (saddle point). This forms the basis for further analysis, such as second derivative tests, to classify the nature of these points.