Find The Local Maximum And Minimum Values And Saddle Points Of The Following Functions:a) $f(x, Y)=x^4+y^4-4xy+1$b) $f(x, Y)=x^3y+12x^2-8y$c) $f(x, Y)=e^{4y-x^2-y^2}$

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In calculus, finding the local maximum, minimum, and saddle points of a function is crucial in understanding its behavior and properties. This article will guide you through the process of finding these critical points for three given multivariable functions.

Function a) f(x,y)=x4+y4−4xy+1f(x, y)=x^4+y^4-4xy+1

To find the local maximum, minimum, and saddle points of the function f(x,y)=x4+y4−4xy+1f(x, y)=x^4+y^4-4xy+1, we need to find the critical points by taking the partial derivatives of the function with respect to xx and yy and setting them equal to zero.

Partial Derivatives

The partial derivatives of the function f(x,y)f(x, y) with respect to xx and yy are:

  • ∂f∂x=4x3−4y\frac{\partial f}{\partial x} = 4x^3 - 4y
  • ∂f∂y=4y3−4x\frac{\partial f}{\partial y} = 4y^3 - 4x

Critical Points

To find the critical points, we set the partial derivatives equal to zero and solve for xx and yy:

  • 4x3−4y=04x^3 - 4y = 0
  • 4y3−4x=04y^3 - 4x = 0

Solving these equations simultaneously, we get:

  • x=0x = 0
  • y=0y = 0

Therefore, the only critical point of the function f(x,y)=x4+y4−4xy+1f(x, y)=x^4+y^4-4xy+1 is (0,0)(0, 0).

Second Derivative Test

To determine the nature of the critical point (0,0)(0, 0), we need to compute the second partial derivatives of the function and evaluate the discriminant:

  • ∂2f∂x2=12x2\frac{\partial^2 f}{\partial x^2} = 12x^2
  • ∂2f∂y2=12y2\frac{\partial^2 f}{\partial y^2} = 12y^2
  • ∂2f∂x∂y=−4\frac{\partial^2 f}{\partial x \partial y} = -4

Evaluating the discriminant at the critical point (0,0)(0, 0), we get:

  • D(0,0)=∂2f∂x2(0,0)⋅∂2f∂y2(0,0)−(∂2f∂x∂y(0,0))2=0D(0, 0) = \frac{\partial^2 f}{\partial x^2}(0, 0) \cdot \frac{\partial^2 f}{\partial y^2}(0, 0) - \left(\frac{\partial^2 f}{\partial x \partial y}(0, 0)\right)^2 = 0

Since the discriminant is zero, the second derivative test is inconclusive, and we need to use other methods to determine the nature of the critical point.

Conclusion

In conclusion, the function f(x,y)=x4+y4−4xy+1f(x, y)=x^4+y^4-4xy+1 has only one critical point at (0,0)(0, 0). However, the second derivative test is inconclusive, and further analysis is required to determine the nature of this critical point.

Function b) f(x,y)=x3y+12x2−8yf(x, y)=x^3y+12x^2-8y

To find the local maximum, minimum, and saddle points of the function f(x,y)=x3y+12x2−8yf(x, y)=x^3y+12x^2-8y, we need to find the critical points by taking the partial derivatives of the function with respect to xx and yy and setting them equal to zero.

Partial Derivatives

The partial derivatives of the function f(x,y)f(x, y) with respect to xx and yy are:

  • ∂f∂x=3x2y+24x\frac{\partial f}{\partial x} = 3x^2y + 24x
  • ∂f∂y=x3−8\frac{\partial f}{\partial y} = x^3 - 8

Critical Points

To find the critical points, we set the partial derivatives equal to zero and solve for xx and yy:

  • 3x2y+24x=03x^2y + 24x = 0
  • x3−8=0x^3 - 8 = 0

Solving these equations simultaneously, we get:

  • x=2x = 2
  • y=0y = 0

Therefore, the only critical point of the function f(x,y)=x3y+12x2−8yf(x, y)=x^3y+12x^2-8y is (2,0)(2, 0).

Second Derivative Test

To determine the nature of the critical point (2,0)(2, 0), we need to compute the second partial derivatives of the function and evaluate the discriminant:

  • ∂2f∂x2=6xy+24\frac{\partial^2 f}{\partial x^2} = 6xy + 24
  • ∂2f∂y2=0\frac{\partial^2 f}{\partial y^2} = 0
  • ∂2f∂x∂y=3x2\frac{\partial^2 f}{\partial x \partial y} = 3x^2

Evaluating the discriminant at the critical point (2,0)(2, 0), we get:

  • D(2,0)=∂2f∂x2(2,0)⋅∂2f∂y2(2,0)−(∂2f∂x∂y(2,0))2=48D(2, 0) = \frac{\partial^2 f}{\partial x^2}(2, 0) \cdot \frac{\partial^2 f}{\partial y^2}(2, 0) - \left(\frac{\partial^2 f}{\partial x \partial y}(2, 0)\right)^2 = 48

Since the discriminant is positive, the second derivative test indicates that the critical point (2,0)(2, 0) is a local minimum.

Conclusion

In conclusion, the function f(x,y)=x3y+12x2−8yf(x, y)=x^3y+12x^2-8y has only one critical point at (2,0)(2, 0), which is a local minimum.

Function c) f(x,y)=e4y−x2−y2f(x, y)=e^{4y-x^2-y^2}

To find the local maximum, minimum, and saddle points of the function f(x,y)=e4y−x2−y2f(x, y)=e^{4y-x^2-y^2}, we need to find the critical points by taking the partial derivatives of the function with respect to xx and yy and setting them equal to zero.

Partial Derivatives

The partial derivatives of the function f(x,y)f(x, y) with respect to xx and yy are:

  • ∂f∂x=−2xe4y−x2−y2\frac{\partial f}{\partial x} = -2xe^{4y-x^2-y^2}
  • ∂f∂y=4e4y−x2−y2\frac{\partial f}{\partial y} = 4e^{4y-x^2-y^2}

Critical Points

To find the critical points, we set the partial derivatives equal to zero and solve for xx and yy:

  • −2xe4y−x2−y2=0-2xe^{4y-x^2-y^2} = 0
  • 4e4y−x2−y2=04e^{4y-x^2-y^2} = 0

Solving these equations simultaneously, we get:

  • x=0x = 0
  • y=0y = 0

Therefore, the only critical point of the function f(x,y)=e4y−x2−y2f(x, y)=e^{4y-x^2-y^2} is (0,0)(0, 0).

Second Derivative Test

To determine the nature of the critical point (0,0)(0, 0), we need to compute the second partial derivatives of the function and evaluate the discriminant:

  • ∂2f∂x2=−2e4y−x2−y2+4x2e4y−x2−y2\frac{\partial^2 f}{\partial x^2} = -2e^{4y-x^2-y^2} + 4x^2e^{4y-x^2-y^2}
  • ∂2f∂y2=4e4y−x2−y2+8ye4y−x2−y2\frac{\partial^2 f}{\partial y^2} = 4e^{4y-x^2-y^2} + 8ye^{4y-x^2-y^2}
  • ∂2f∂x∂y=−8xe4y−x2−y2\frac{\partial^2 f}{\partial x \partial y} = -8xe^{4y-x^2-y^2}

Evaluating the discriminant at the critical point (0,0)(0, 0), we get:

  • D(0,0)=∂2f∂x2(0,0)⋅∂2f∂y2(0,0)−(∂2f∂x∂y(0,0))2=16D(0, 0) = \frac{\partial^2 f}{\partial x^2}(0, 0) \cdot \frac{\partial^2 f}{\partial y^2}(0, 0) - \left(\frac{\partial^2 f}{\partial x \partial y}(0, 0)\right)^2 = 16

Since the discriminant is positive, the second derivative test indicates that the critical point (0,0)(0, 0) is a local minimum.

Conclusion

In conclusion, the function f(x,y)=e4y−x2−y2f(x, y)=e^{4y-x^2-y^2} has only one critical point at (0,0)(0, 0), which is a local minimum.

In the previous article, we discussed the process of finding local maximum, minimum, and saddle points of multivariable functions. In this article, we will answer some frequently asked questions related to this topic.

Q: What is the difference between a local maximum and a global maximum?

A: A local maximum is a point where the function has a maximum value in a small neighborhood, while a global maximum is a point where the function has a maximum value over its entire domain.

Q: How do I find the local maximum, minimum, and saddle points of a multivariable function?

A: To find the local maximum, minimum, and saddle points of a multivariable function, you need to find the critical points by taking the partial derivatives of the function with respect to each variable and setting them equal to zero. Then, you need to compute the second partial derivatives and evaluate the discriminant to determine the nature of the critical points.

Q: What is the second derivative test?

A: The second derivative test is a method used to determine the nature of a critical point by computing the second partial derivatives of the function and evaluating the discriminant.

Q: How do I use the second derivative test to determine the nature of a critical point?

A: To use the second derivative test, you need to compute the second partial derivatives of the function and evaluate the discriminant at the critical point. If the discriminant is positive, the critical point is a local minimum or maximum. If the discriminant is negative, the critical point is a saddle point. If the discriminant is zero, the second derivative test is inconclusive.

Q: What is a saddle point?

A: A saddle point is a point where the function has a maximum value in one direction and a minimum value in another direction.

Q: How do I determine the nature of a saddle point?

A: To determine the nature of a saddle point, you need to compute the second partial derivatives of the function and evaluate the discriminant. If the discriminant is negative, the saddle point is a local maximum in one direction and a local minimum in another direction.

Q: Can a function have multiple local maximum, minimum, and saddle points?

A: Yes, a function can have multiple local maximum, minimum, and saddle points.

Q: How do I find the local maximum, minimum, and saddle points of a function with multiple variables?

A: To find the local maximum, minimum, and saddle points of a function with multiple variables, you need to find the critical points by taking the partial derivatives of the function with respect to each variable and setting them equal to zero. Then, you need to compute the second partial derivatives and evaluate the discriminant to determine the nature of the critical points.

Q: What is the importance of finding local maximum, minimum, and saddle points of a function?

A: Finding local maximum, minimum, and saddle points of a function is important in many applications, such as optimization problems, game theory, and economics.

Q: Can you provide an example of a function with multiple local maximum, minimum, and saddle points?

A: Yes, consider the function f(x,y)=x2+y2−2x−2y+1f(x, y) = x^2 + y^2 - 2x - 2y + 1. This function has multiple local maximum, minimum, and saddle points.

Q: How do I visualize the local maximum, minimum, and saddle points of a function?

A: You can visualize the local maximum, minimum, and saddle points of a function using contour plots or 3D plots.

Q: What are some common mistakes to avoid when finding local maximum, minimum, and saddle points of a function?

A: Some common mistakes to avoid when finding local maximum, minimum, and saddle points of a function include:

  • Not checking for critical points in all directions
  • Not computing the second partial derivatives correctly
  • Not evaluating the discriminant correctly
  • Not considering the domain of the function

By following these tips and avoiding common mistakes, you can accurately find the local maximum, minimum, and saddle points of a function.