JAM 2022 [ 51-60 ]

Let \mathbb{f} : \mathbb{R}^2 \rightarrow \mathbb{R} be a function defined by f(x,y) = x^2-12y . If M and m be the maximum value and the minimum value, respectively, of the function f on the circle x^2+y^2 =49 , then |M| and |m| equals __________________.

\text{Using the Lagrange's Method we define,} \\
\phi(x,y) = x^2-12y + \lambda ( x^2+y^2 -49)

Using partial differentiation, we have,

\frac{\delta }{\delta x} \phi(x,y) = 2x + 2\lambda x = 2x (1+\lambda) \\
\frac{\delta }{\delta y} \phi(x,y) = -12 + 2\lambda y = 2 (\lambda y - 6) \\ 
\phi_x = 0  \implies x=0 \quad or \quad \lambda =1  \\
\phi_y = 0 \implies \lambda y =6
\text{Now if, } x=0 \text{ using the constraint we have } y=7,-7 \\
and \\
\text{If } \lambda=1 \text{ then y=6 and } x^2=13

For these three points we have,

f(0,7)= -84 , f(0,-7) = 84 , f(\sqrt{13},6) = 85

Thus the M=85 and m=-84 and |M|+|m|=169.

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