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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?

User SpamapS
by
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1 Answer

5 votes

Answer:

At critical point in D

a


(x,y) = (0,0)

b


f(x,y) = f(x) =11 -x^2

where
-1 \le x \le 1

c

maximum value 11

minimum value 10

Explanation:

Given
f(x,y) =10x^2 + 11x^2

At critical point


f'(x,y) = 0

=>
[f'(x,y)]_x = 20x =0

=>
x =0

Also


[f'(x,y)]_y = 22y =0

=>
y =0

Now considering along the boundary


D = 1

=>
x^2 +y^2 = 1

=>
y =\pm √(1- x^2)

Restricting
f(x,y) to this boundary


f(x,y) = f(x) = 10x^2 +11(1-x^2)^{(2)/(1) *(1)/(2) }


= 11-x^2

At boundary point D = 1

Which implies that
x \le 1 or
x \ge -1

So the range of x is


-1 \le x \le 1

Now along this this boundary the critical point is at


f'(x) = 0

=>
f'(x) = -2x =0

=>
x=0

Now at maximum point
(i.e \ x =0)


f(0) =11 -(0)


= 11

For the minimum point x = -1 or x =1


f(1) = 11 - 1^2


=10


f(-1) = 11 -(-1)^2


=10

Now focus on the boundary of D, and solve for y2. Restricting f(x,y) to this boundary-example-1
User ToddR
by
6.5k points