Problem 31 - Calculus Early Transcendentals

Question

Find the critical points of the following functions. Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, local minimum, or saddle point. Confirm your results using a graphing utility. $$f(x, y)=x^{4}+4 x^{2}(y-2)+8(y-1)^{2}$$

Step-by-Step Solution

Answer

#Question# Determine the critical points of the function \(f(x, y) = 2x^{4} + 4x^{2}(y - 2)^{2} - 16y + 32\) and classify them as a local minimum, local maximum, or saddle point by following the given step-by-step solution. #Answer# By following the given step-by-step solution, the critical point of the function is (0, 2), and it corresponds to a local minimum.

Step 1: Find the gradient of f(x, y)

First, let's calculate the gradient of the function given, which is the vector containing partial derivatives. $$\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$$ Now, let's find the partial derivatives of the function: $$\frac{\partial f}{\partial x} = 4x^{3} + 8x(y - 2)$$ $$\frac{\partial f}{\partial y} = 8x^{2} + 16(y - 1)$$ Thus, the gradient is: $$\nabla f = \left(4x^{3} + 8x(y - 2), 8x^{2} + 16(y - 1)\right)$$

Step 2: Find the critical points

To find the critical points, we set each component of the gradient to zero and solve the system of equations. $$4x^{3} + 8x(y - 2) = 0$$ $$8x^{2} + 16(y - 1) = 0$$ First, let's put \(4x^3+8x(y-2)=0\) into the form \(y=2-\frac{x^3}{2}\). Now, substitute this equation into the second equation of the gradient. $$8x^{2} + 16\left(1-\frac{3x^3}{2}\right) = 0$$ Solving this equation yields: \(x=0\). Now, we can find the corresponding y-value by plugging x back into the equation for y: \(y=2\). Thus, the critical point is \((0, 2)\).

Step 3: Perform the Second Derivative Test

For the Second Derivative Test, we need the Hessian matrix, which consists of the second partial derivatives of f. $$H(f) = \begin{bmatrix}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix}$$ Now, let's find the second partial derivatives: $$\frac{\partial^2 f}{\partial x^2} = 12x^{2} + 8(y-2)$$ $$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 8x$$ $$\frac{\partial^2 f}{\partial y^2} = 16$$ Now, we evaluate the Hessian matrix at the critical point (0, 2): $$H(f(0, 2)) = \begin{bmatrix}16 & 0 \\ 0 & 16\end{bmatrix}$$ Now, we compute the determinant of the Hessian matrix: $$\text{det}(H(f(0, 2)))=16^2 - 0^2 = 256$$ Since the determinant is positive and the second partial derivative with respect to x is positive, \((0,2)\) corresponds to a local minimum.

Step 4: Confirm the results using a graphing utility

Finally, we will use a graphing utility to obtain a 3D plot of the function and confirm that we have a local minimum at \((0, 2)\). Comparing the plot to the solution we found, we can see that our answer is correct and matches the graph.

Key Concepts

Second Derivative Test

In calculus, the Second Derivative Test is a handy tool to classify critical points of a function. When you have determined the critical points by setting the gradient to zero, this test helps identify whether these points are local maxima, local minima, or saddle points.
To perform the Second Derivative Test, you calculate the second derivatives and form the Hessian matrix. The Hessian matrix, in two variables, looks like this:

• \[ H(f) = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix} \]

Once you've evaluated the Hessian at the critical points, calculate the determinant. This value, alongside individual second partial derivatives, will guide you:

• If the determinant is positive and \( \frac{\partial^2 f}{\partial x^2} \) is positive, it indicates a local minimum.
• If the determinant is positive and \( \frac{\partial^2 f}{\partial x^2} \) is negative, it suggests a local maximum.
• If the determinant is negative, the point is a saddle point.

Using this systematic approach ensures you accurately categorize the critical points of multivariable functions.

Local Maximum

A local maximum of a function is a point where the function's value is higher than all nearby points. It is like standing on the peak of a hill in terms of elevation.
To find a local maximum using the Second Derivative Test, the Hessian matrix at the critical point should have a positive determinant. Additionally, the second partial derivative with respect to \(x\) should be negative:

• The logic is that the curvature of the function is bending downwards, similar to a bowl upside-down.
• The function f(x, y) decreases as you move away from the maximum point.

In graph plots, local maxima look like the highest points in a region, but not necessarily the highest overall (that would be a global maximum).
Understanding local maxima helps in optimizing tasks, finding peaks in data, and more advanced applied mathematics.

Local Minimum

A local minimum is the opposite of a local maximum; it is a point where the function is lower than nearby points. Picture it as standing at the bottom of a valley.
For a function to have a local minimum at a critical point,

• The second derivative test shows a positive determinant in the Hessian matrix.
• The second derivative with respect to \(x\) should also be positive.

This occurs because the function's slope curves upwards, resembling a right-side-up bowl.
When you see graphs, local minima are troughs or dips. These points can be quite important in various fields like physics and economics, where finding the lowest constraints of a system is crucial.
In the given exercise, the critical point (0, 2) was identified as a local minimum using this test.

Saddle Point

Unlike local maxima or minima, a saddle point does not exhibit a relative peak or valley.
At a saddle point, the function has a surface that curves upwards in one direction and downwards in another, resembling a horse's saddle.
To identify a saddle point using the Second Derivative Test:

• The Hessian matrix designates a saddle point if its determinant is negative.

This means the function behaves differently along different axes. The curvature changes direction, lacking the consistent behavior seen in maxima or minima.
Saddle points are crucial in understanding the behavior of complex systems. They often hint at a change in trends or directions, making them invaluable in mathematical modeling and dynamic systems analysis.