Maximum along the Boundary Intuition. Setting up the Statement

University
Massachusetts Institute of Technology
Course
Multivariable Calculus
Pages

2

Academic year

2023
Author

Sporkz
Views

74

Maximum along the boundary intuition. Setting up the statement The goal of this course is to determine where the maximum value of a functionexists. We will first write down the formula for it, then compute the actual values atwhich this maximum occurs. Before we begin our main topic, we have a quick warm-up. We'll practice findingvectors that are normal to the boundaries of regions. At each point of the boundaryof this region, find a vector that is normal to that boundary like so or like so.Finding a vector normal to a curve. One way is to think about gradients, because weknow that gradients—the gradient is perpendicular to the level curves. So if we havea curve, and if it's the level curve of something, we take the gradient of thatsomething, and that's perpendicular to the level curve. What's our boundary? Ourboundary is this curve, which is a circle with equation x^2+y^2=1. The gradient ofthis function g(x, y) is perpendicular to the boundary. The gradient of g is 2x − 4. Let us conduct a sanity check, in order to ensure that we have not made anyarithmetic errors or overlooked important details. An ideal point to test first would beone of the simplest parts of the problem. Thus, the suggestion is to take (x, commay) to be (0, 0). However, since (0, 0) does not lie on the curve, this is not areasonable choice.The vector (2, 0) is inside the region but not on the boundary. The boundary ishighlighted in purple.

The point on the boundary that is easiest to reach is (1, 0). By calculating thegradient of g and plugging in x = 1 and y = 0, we see that the first component of thegradient is negative 2 and the second component of the gradient is 0. The pointwhere these two components meet is (1, 0).The function f of (x, y)—that is, the value of f when x = 0 and y = 0—is 5 minus xsquared minus y minus 1 squared. The gradient is negative 2y plus 2. We are going to find the maximum of this function in this region. And we have animportant clue that the maximum point is roughly over here, and at that point, thegradient is perpendicular to the boundary. To find the maximum point of a sentence,we must convert it into an equation about x and y. We can then solve this equationby manipulating x and y until we arrive at a solution.