Relative Extrema of h(x,y) = 62x + 62y - x^2 - y^2 Calculus III

TL;DR

Finding critical points using partial derivatives, then applying second derivative test for extrema in multivariable functions.

Transcript

hey everyone so in this video we're going to find the relative extrema of this multivariable function so step one is to find the critical points to do that you start by hitting the partial derivatives with respect to x and y and you set them both equal to zero so solution a partial derivative of H with respect to X so the derivative of 62 X is 62 a... Read More

Key Insights

• 😥 Critical points are found by setting partial derivatives to zero.
• 😥 Second derivative test helps identify extrema at critical points.
• 😥 Positive D and positive Hxx indicate a minimum at the critical point.

Questions & Answers

Q: What is the first step in finding relative extrema for multivariable functions?

The first step is to find critical points by setting partial derivatives with respect to x and y equal to zero.

Q: How do you apply the second derivative test in multivariable functions?

The second derivative test involves calculating a quantity D, then analyzing the sign of Hxx to determine if the critical point is a max, min, or saddle point.

Q: What does it mean when D is positive and Hxx is positive in the second derivative test?

If D is positive and Hxx is positive, it indicates a minimum at the critical point based on the second derivative test.

Q: How do you find the maximum value in multivariable functions at a critical point?

To find the maximum value, plug the critical point back into the original function to get the maximum value and its corresponding point in space.

Summary & Key Takeaways

• Finding critical points by setting partial derivatives with respect to x and y equal to zero.

• Using second derivative test to determine extrema at critical points.

• Maximum value found at critical point (31, 31) in the original function.