Absolute Min/Max, Bounded region

[Iconate]

Homework Statement

Find the abs min/max values of the function

f(x,y) = e^1-2x2-y2

on the closed and bounded region x² + y² <= 1

The Attempt at a Solution

First I have to find the critical points

Df_x = (-4x)e^1-2x2-y2
Df_y = (-2y)e^1-2x2-y2

Clearly e^1-2x2-y2 cannot equal 0, therefore

x=y=0, Critical point is (0,0)

Now, my professor has done most of these problems by setting
\nablaf = \lambda\nablag, the gradients of each function

Even with this method yields x=0 and y=0 to be the only solution.

I am not too sure how to incorporate the bounded region into this question.

Do I just look at x=0 and y=0 of x² + y² = 1?
This would result in x² + 0 = 1
x=y= +/- 1 giving me two points, (-1, -1), and (1,1), however f(-1,-1) > 1 and

f(1,1) = e^-2 thus a min/max value for the function?

Is this correct? Any insight would be great, thanks.

 

[Mark44]

Neither of the points you found is in your region, which is a circular disk of radius 1, centered at (0, 0).

 

[Iconate]

How am I supposed to proceed then using my critical point and incorporating the bound?

I guess since it is a disk, I can use sin(t) and cos(t) as the coordinates to represent the boundary c(t)= (sint, cost) 0 <= t <= 2\pi

f(c(t)) = e^{1-2sin2t-cos2t} ?

At both of the boundaries 0 and 2\pi , f = 1.

Kinda confused otherwise on computing this

 

[Mark44]

You need to check f at all points on the boundary, not just at t = 0 and t = 2pi. Your exponent on e is 1 - 2sin² t - cos² t. I would use an identity to get this exponent in terms of sin or cos alone.

 

[Iconate]

Ah ok so now I have

e^-sin2x

Now the max and min values of the sin function are \pi/2 and 3\pi/2

These are the values I look at?

 

[Mark44]

The maximum and minimum values of the sine function are 1 and -1, and they are attained when x is pi/2 and 3pi/2, respectively. The maximum value of -sin² x is 0, which is attained at x = 0, x = pi, and x = 2pi. The minimum value of this function is -1, attained at x = pi/2 and x = 3pi/2.