What is the range of f. Support your answer.?

Question

f(x,y)=sqrt(1-x^2) + sqrt(4-y^2).

Answer 1

f(x,y)=sqrt(1-x^2) + sqrt(4-y^2)

The definition domain is 1-x^2≥0 and 4-y^2≥0 <=>
-1≤x≤1 and -2≤y≤2

it's obvious that:

0 ≤ 1-x^2 ≤ 1
0 ≤ 4-y^2 ≤ 4

or that:

0 ≤ sqrt(1-x^2) ≤ 1
0 ≤ sqrt(4-y^2) ≤ 2

adding both relations will result that

0 ≤ sqrt(1-x^2) +sqrt(4-y^2) ≤ 3

or

0 ≤ f(x,y) ≤ 3

The range will be [0,3]

Answer 2

Solving for y gives y = +/-.5sqrt(1-x^2)
Taking the derivative of first the positive function we find the max to be .5 at x =0
Takingthe derivative of the negative function, we finf a min of -.5 at x= 0.
Therefor range is .5=< y=<.5