Question

Determine the maximum and minimum
OF Function F(x, y) = x² + y² + 6x +12

Answer 1

\begin{lgathered}{x}^{2} + {y}^{2} + 6x + 12 \\ = (x + 3) {}^{2} - 9 + {y}^{2} + 12 \\ = (x + 3) {}^{2} + {y}^{2} + 3 \\ now \: \: minimum \: value \: \\ of \: perfec t\: square \:is \: zero \: \: \\ = 0 + 0 + 3 \\ = 3\end{lgathered}

x

2

+y

2

+6x+12

=(x+3)

2

−9+y

2

+12

=(x+3)

2

+y

2

+3

nowminimumvalue

ofperfectsquareiszero

=0+0+3

=3

min value =3

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