91Ó°ÊÓ

Consider the function $$f(x, y)=\left(18 x-x^{2}\right)\left(18 y-y^{2}\right)$$ Find and classify all critical points of the function. If there are more blanks than critical points, leave the remaining entries blank. \(f_{x}=\) _________. \(f_{y}=\) _________. \(f_{x x}=\) _________. \(f_{x y}=\) _________. \(f_{y y}=\) _________. There are several critical points to be listed. List them lexicograhically, that is in ascending order by x-coordinates, and for equal x-coordinates in ascending order by y-coordinates (e.g., (1,1),(2,-1),(2,3) is a correct order) The critical point with the smallest x-coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined) The critical point with the next smallest x-coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined) The critical point with the next smallest \(\mathrm{x}\) -coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined) The critical point with the next smallest x-coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined) The critical point with the next smallest \(x\) -coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined)

Step by step solution

01

Compute partial derivatives \(f_{x}\) and \(f_y\)

Compute the partial derivatives of the given function with respect to x and y: \(f_x = \frac{d}{dx}(18x-x^2)(18y-y^2)\) \(f_y = \frac{d}{dy}(18x-x^2)(18y-y^2)\) Using the product rule, we get: \(f_x = (18 - 2x)(18y - y^2)\) \(f_y = (18x - x^2)(2y - 18)\)

02

Set partial derivatives to 0

To find the critical points, we need to set \(f_x\) and \(f_y\) to 0 and solve for x and y: \(f_x= (18 - 2x)(18y-y^2) = 0\) \(f_y = (18x - x^2)(2y - 18) = 0\) Solve these equations for x and y to get the critical points.

03

Compute second order partial derivatives

Compute \(f_{xx}\), \(f_{yy}\), and \(f_{xy}\): \(f_{xx} = \frac{d^2}{dx^2}(18x - x^2)(18y - y^2)\) \(f_{yy} = \frac{d^2}{dy^2}(18x - x^2)(18y - y^2)\) \(f_{xy} = \frac{d^2}{dxdy}(18x - x^2)(18y - y^2)\) We have: \(f_{xx} = -2(18y - y^2)\) \(f_{yy} = -2(18x - x^2)\) \(f_{xy} = \frac{d}{dx}(18 - 2x)(2y - 18) = (18 - 2x)(-2)\)

04

Determine the type of critical points

To determine the type of critical points, use the second derivative test formula D = \(f_{xx}f_{yy} - (f_{xy})^2\). If D > 0 and \(f_{xx}\) > 0, then it's a local minimum; if D > 0 and \(f_{xx}\) < 0, then it's a local maximum; if D < 0, then it's a saddle point; if D = 0, the test is inconclusive. Calculate the D values for each critical point found in Step 2 and classify the critical points.

05

Listing critical points and their classifications

List the critical points lexicographically, in ascending order by x-coordinates and, for equal x-coordinates, in ascending order by y-coordinates. For each critical point, write its type (local minimum, local maximum, saddle point, or cannot be determined) based on the computed D values. Present the results in a step-by-step manner as described in the problem statement.

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Key Concepts

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

Partial Derivatives

Partial derivatives play a seminal role when you're analyzing multi-variable functions, like the given function,
\(f(x, y) = (18x - x^2)(18y - y^2)\). To find the rate of change of the function in a single variable's direction while keeping others constant, we compute its partial derivatives.

For instance, to find \(f_x\), the partial derivative of \(f\) with respect to \(x\), you treat \(y\) as a constant and differentiate with respect to \(x\). The significance of this process is substantial; it hooks us up with valuable insights about the function's behavior along the \(x\) direction. In our exercise, the partial derivative \(f_x\) is \((18 - 2x)(18y - y^2)\), which demonstrates how \(f\)'s rate of change pivots on both \(x\) and \(y\).

Similarly, \(f_y\) is discerned by seeing \(x\) as a constant while differentiating with regard to \(y\). So, \(f_y = (18x - x^2)(2y - 18)\) delivers the rate of change across the \(y\) axis.

By setting these partial derivatives to zero, we can pinpoint where the function plateaus or valleys—these points are known as 'critical points.' It's where the function's incline is momentarily null, hinting at potential minima, maxima, or saddle points, which are crucial for understanding the topography of the function.

Second-order Partial Derivatives

Venturing further, we encounter second-order partial derivatives, the offspring of partial derivatives. They are the derivatives of the partial derivatives, and they come in various combinations—\(f_{xx}\), \(f_{yy}\), and \(f_{xy}\) or \(f_{yx}\).

In the context of our exercise, these second derivatives depict how the change rate of the function’s slope evolves as you move along one variable, while bearing in mind the influence of the other. To elaborate, \(f_{xx} = -2(18y - y^2)\) illustrates how the slope along the \(x\) direction changes as we stroll along the \(x\) direction itself, and similarly for \(f_{yy}\).

As for \(f_{xy} = (18 - 2x)(-2)\), it's a mixed partial derivative illustrating how the slope along \(x\) changes as you take steps in the \(y\) direction. This cross-examination between \(x\) and \(y\) unveils interaction effects that are invisible to simple first-order derivatives—the curvature, the twists and twirls in the function's terrain. These second-order derivatives are key elements in the second derivative test, a formidable tool used to classify the critical points we previously discovered.

Second Derivative Test

Now buckle up for the litmus test for determining the nature of our critical points: the second derivative test. This tool adds a layer of scrutiny to our previously found critical points—beyond merely spotting them. It helps us categorize each as a local minimum, local maximum, or a saddle point.

Let's focus on how it works. We bring into play a discriminant \(D\), defined as \(D = f_{xx}f_{yy} - (f_{xy})^2\). Here, if \(D\) comes out positive and \(f_{xx} > 0\), then the function grins at a local minimum. Alternatively, if \(D > 0\) but \(f_{xx} < 0\), a local maximum is in play. And if \(D\) dips below zero, you're dealing with a saddle point—the terrain takes on a horse saddle shape, curving up in one direction and swooping down in the other.

For the inconclusive scenario where \(D\) is zero, it's the analytical equivalent of a shrug—a more in-depth investigation is needed to reveal the point's true nature. By applying this test to our function, we can confidently assign each critical point its proper tag, crystalizing our conceptual understanding of the function's landscape.