Problem 10 - Applied Calculus

Question

Find all the critical points and determine whether each is a local maximum, local minimum, a saddle point, or none of these. $$f(x, y)=x^{3}+y^{2}-3 x^{2}+10 y+6$$

Step-by-Step Solution

Answer

The critical point (0, -5) is a saddle point, and the critical point (2, -5) is a local minimum.

Step 1: Find the First Partial Derivatives

The first step in finding critical points is to calculate the first partial derivatives of the function. For \( f(x, y) = x^3 + y^2 - 3x^2 + 10y + 6 \), we calculate the partial derivatives with respect to \( x \) and \( y \):\[ f_x = \frac{\partial f}{\partial x} = 3x^2 - 6x \]\[ f_y = \frac{\partial f}{\partial y} = 2y + 10 \]

Step 2: Set the Partial Derivatives to Zero

Next, we find the critical points by setting each of the partial derivatives to zero and solving the resulting equations:\[ 3x^2 - 6x = 0 \]\[ 2y + 10 = 0 \]Solving these, we get the equations:1. \( x(3x - 6) = 0 \) which gives \( x = 0 \) or \( x = 2 \).2. \( 2y + 10 = 0 \) which gives \( y = -5 \).

Step 3: Find the Critical Points

Combining the solutions from Step 2, the critical points are:- \((x, y) = (0, -5)\)- \((x, y) = (2, -5)\)

Step 4: Calculate the Second Partial Derivatives

To classify each critical point, we need to calculate the second partial derivatives:\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x - 6 \]\[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2 \]\[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \]

Step 5: Use the Second Derivative Test

The second derivative test for functions of two variables uses the Hessian determinant:\[ D = f_{xx}f_{yy} - (f_{xy})^2 = (6x - 6)(2) - (0)^2 = 12x - 12 \]Evaluate \( D \) at each critical point:- For \((x, y) = (0, -5)\): \[ D = 12(0) - 12 = -12 \] Since \( D < 0 \), this is a saddle point.- For \((x, y) = (2, -5)\): \[ D = 12(2) - 12 = 24 - 12 = 12 \] Since \( D > 0 \) and \( f_{xx} = 6(2) - 6 = 6 > 0 \), this is a local minimum.

Key Concepts

Partial Derivatives

Partial derivatives are like taking regular derivatives, but they’re done on functions with more than one variable. For our function, \(f(x, y) = x^3 + y^2 - 3x^2 + 10y + 6\), we want to find how the function changes when we change just \(x\) or just \(y\), treating the other variable as a constant. This is important because it allows us to understand the behavior of the function on surfaces rather than just lines. - The partial derivative with respect to \(x\) is calculated by differentiating \(f\) while treating \(y\) as a constant: \[ f_x = \frac{\partial f}{\partial x} = 3x^2 - 6x. \] - Similarly, for \(y\), we treat \(x\) as a constant and differentiate: \[ f_y = \frac{\partial f}{\partial y} = 2y + 10. \] These results are used in the next step to find the critical points by setting them to zero. This is how we determine where the function’s rate of change is zero in any given direction.

Second Derivative Test

The second derivative test is used to classify the nature of critical points found from the first partial derivatives. Once we have these points, we need to check how the function behaves around them. The test involves calculating the second partial derivatives.For our function:

• \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x - 6 \)
• \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2 \)
• \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \)

These second derivatives are key because they form part of the Hessian, which we use to evaluate what kind of turning point we have at each critical point. The role of the second derivatives can be thought of as a way to check the concavity or convexity of the function in each direction.

Hessian Determinant

The Hessian determinant, \(D\), is a specific calculation used to classify the nature of critical points more definitively. It uses the second partial derivatives in its formula. For the function we're analyzing, \(D\) is determined using:\[ D = f_{xx}f_{yy} - (f_{xy})^2. \]This determinant tells us important information:

• If \( D > 0 \), and \( f_{xx} > 0 \), the critical point is a local minimum.
• If \( D > 0 \), and \( f_{xx} < 0 \), the critical point is a local maximum.
• If \( D < 0 \), the critical point is a saddle point.
• If \( D = 0 \), the test is inconclusive.

In our problem, at the critical point \((0, -5)\), \(D = -12\) which indicates a saddle point. For \((2, -5)\), \(D = 12\) and \(f_{xx} = 6\), which confirms a local minimum.

Local Maximum and Minimum

Understanding local maxima and minima is crucial in determining the behavior of a function at given points. - A **local maximum** is a point where the function value is higher than at nearby points. It resembles a peak.- A **local minimum** is when the function value is lower than at surrounding points, like the bottom of a valley.Using our function's Hessian determinant results, we identified:- At \((0, -5)\), the critical point is a **saddle point**. This means it’s neither a peak nor a valley, but rather the function changes direction in a complex way.- At \((2, -5)\), there is a **local minimum** since \(D > 0\) and \(f_{xx} > 0\). It represents a trough point in our function surface.Understanding these concepts helps graph the function and anticipate its behavior easily.