# The extreme values of f(x,y) = y2 – x2 are:

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The extreme values of f(x,y) = y2 – xare:
1. 3, -2
2. 0
3. -1, 1
4. f has no extreme value

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Correct Answer - Option 4 : f has no extreme value

Concept:

$r = {\left( {\frac{{{\partial ^2}f}}{{\partial {x^2}}}} \right)_{\left( {a,b} \right)}},\;s = {\left( {\frac{{{\partial ^2}f}}{{\partial x\partial y}}} \right)_{\left( {a,b} \right)}},\;t = \;{\left( {\frac{{{\partial ^2}f}}{{\partial {y^2}}}} \right)_{\left( {a,b} \right)}}$

• If f will be maximum or minimum, if rt – s2 is +ve → (> 0), s2 – rt is –ve → (<0)
• Based on r → -ve f is maximum at (a, b)
• Based on r → +ve f is minimum at (a, b)
• If rt – s2 is –ve, (a, b) saddle point i.e. neither maximum nor minimum

Calculation:

Given:

$f\left( {x,y} \right) = \;{y^2} - {x^2}$

Differentiating with respect to fx,

-2x = 0 x = 0

Differentiating with respect to fy,

2y = 0 y = 0

Again differentiating for second time, we get

r = fxx = -2

s = fyy = -2

Now, fxy = 0 = s

$rt - {s^2} = \; - 4\;\left( {negative} \right)$

∴ Neither maximum nor minimum