Find the absolute maximum and minimum values of the following function

Question

Find the absolute maxima and minima of the function on the given domain.

f(x, y) = 2x² +9y² on the disk bounded by the circle x² + y² =9

Answer 1

Find the absolute maxima and minima of the function f(x, y) = 2x² +9y²  

on the given domain bounded by the circle x²+y² =9.

The domain of the circle is [-3,3].

The range of the circle is also [-3,3].

But the domain of the function f(x,y) are both the values of x and y that satisfies x²+y² ≤ 9 because the circle is just the boundary.

The extreme points of the circle are the intercepts and the origin:

(0,0), (3,0),(-3,0),(0,-3),(0,3)

f(0,0) = 2(0)² +9(0)² = 0

f(3,0) = 2(3)² +9(0)² = 18

f(-3,0) = 2(-3)² +9(0)² = 18

f(0,-3) = 2(0)² +9(-3)² = 81

f(0,3) = 2(0)² +9(3)² = 81

Therefore

The absolute max of f(x,y) = 81

The absolute min of f(x,y) = 0

Answer 2

First off draw the circle very simple!

It is in form (x^2) + (y^2) = ((radius)^2) so the radius = 3

The circle is centered at (0,0) and it ranges from [-3,3] in the x direction and [-3,3] in the y direction

They say the disk is bounded by this circle and which this disk it is 3D so you should definitely graph it first if possible!

It looks like it is a 3D version of a positive parabola on a basic x,y graph (2D) that has 2 intersection points with the circle given on the positive and negative side the the y axis.

To find these points just set the 2 functions equal to each other then plug in each respective solution x to find it corresponding y value in the respective intersection points of the 2 functions!!

By inspection, the max value has to be at point (3,3) since it the the bounded part of both functions you graph (it is much easier to see with a graph) and the min value is at point (0,0)

Thus the

absolute max = 3

and

absolute min = 0

are the answers!!