Question

The extreme values of f(x,y) = y² – x²  are:
1. 3, -2
2. 0
3. -1, 1
4. f has no extreme value

Answer 1

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 f_x,

-2x = 0 ⇒ x = 0

Differentiating with respect to f_y,

2y = 0 ⇒ y = 0

Again differentiating for second time, we get

r = f_xx = -2

s = f_yy = -2

Now, f_xy = 0 = s

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

∴ Neither maximum nor minimum