Mathematics, 22.04.2020 22:18 AdoNice

We want to find the maximum and minimum values of f(x, y)=12x2+13y2 on the disk D: x2+y2≤1. What is the critical point in D? (x, y)=( , ) Now focus on the boundary of D, and solve for y2. Restricting f(x, y) to this boundary, we can express f(x, y) as a function of a single variable x. What is this function and its closed interval domain? f(x, y)=f(x)= where ≤x≤ What are the absolute maximum and minimum values of the function along the BOUNDARY of D? maximum value: minimum value: What are the absolute maximum and minimum values of f(x, y) over all of D? maximum value: minimum value:

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Answer from: youngsiron

Over the boundary: maximum:13, minimum:12

Over D: maximum:13, minimum:0

Step-by-step explanation:

We are given that f(x,y) = 12x^2+13y^2 and D is the disc of radius one. Namely, $x^2+y^2\leq 1$ .

First, we want to find a critical point of the function f. To do so, we want to find the values(x,y) such that

$\nabla f (x,y) =0$ .

Recall that $\nabla f (x,y) = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$ .

So, let us calculate $\nabla f (x,y)$ (the detailed calculation of the derivatives is beyond the scope of the answer.

$\frac{\partial f}{\partial x} = 24x$

$\frac{\partial f}{\partial y} = 26y$

When equalling it to 0, we get that the critical point is (0,0), which is in our region D. Note that the function f is the sum of the square of two real numbers multiplied by some constants. Hence, the function f fulfills that $f(x,y)\geq 0$ . Note that f(0,0)=0, so without further analysis we know that the point (0,0) is a minimum of f over D.

If we restrict to the boundary, we have the following equation x^2+y^2=1 . Main idea is to replace the value of one of the variables n the function f, so it becomes a function of a single variable. Then, we can find the critical values by using differential calculus:

Case 1:

Let us replace y. So, we have that y^2=1-x^2 . So, f(x,y) = 12x^2+13(1-x^2) = -x^2+13 .

So, we will find the derivative with respect to x and find the critical values. That is

$\frac{df}{dx} = -2x=0$

Which implies that x =0. Then, $y =\pm 1$ . So we have the following critical points (0,1), (0,-1). Notice that for both points, the value of f is f(0,1) = f(0,-1) = 13. If we calculate the second derivative, we have that at x=0

$\frac{d^2f}{dx^2} = -2$ . By the second derivative criteria, we know that this points are local maximums of the function f.

Case 2:

Let us replace x. So, we have that x^2=1-y^2 . So, f(x,y) = 12(1-y^2)+13y^2 = y^2+12 .

So, we will find the derivative with respect to y and find the critical values. That is

$\frac{df}{dx} = 2y=0$

Which implies that y =0. Then, $x =\pm 1$ . So we have the following critical points (-1,0), (1,0). Notice that for both points, the value of f is f(1,0) = f(-1,0) = 12. If we calculate the second derivative, we have that at y=0

$\frac{d^2f}{dx^2} = 20$ . By the second derivative criteria, we know that this points are local minimums of the function f.

So, over the boundary D, the maximum value of f is 13 and the minimum value is 12. Over all D, the maximum value of f is 13 and the minimum value is 0.

Answer from: Quest

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