Question

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?

Answer 1

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`