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Question Number 2524 by 123456 last updated on 21/Nov/15

f : R^{2} \to R

f (x, y) = x^{2} - y^{2} y ⩾ 0

f (x, y) = x^{2} + y^{2} y ⩽ 0

find

(x, y) for \min f (x, y)

(x, y) for f (x, y) = 1

Answered by prakash jain last updated on 22/Nov/15

f (x, y) = 1

x^{2} - y^{2} = 1 \Rightarrow x^{2} = 1 + y^{2} \Rightarrow x = \pm \sqrt{1 + y^{2}}, y > 0

x^{2} + y^{2} = 1 \Rightarrow x^{2} = 1 - y^{2} \Rightarrow x = \pm \sqrt{1 - y^{2}}, - 1 ⩽ y ⩽ 0

\min f (x, y)

There is no minimum value for f (x, y) since y can be

made arbitarilary large .