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Question Number 144777 by Nhozie last updated on 29/Jun/21

Answered by som(math1967) last updated on 29/Jun/21

$${(x+3)}^2+y^2+3$$$$ifx,y\in Rthen{(x+3)}^2⩾0,y^2⩾0$$$$\therefore {(x+3)}^2+y^2+3⩾3$$$$\therefore x^2+y^2+6x+12⩾3$$$$minimumvalue3$$

Answered by imjagoll last updated on 29/Jun/21

$$\left(1\right)\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=2\mathrm{x}+6=0\Rightarrow \mathrm{x}=-3$$$$\left(2\right)\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=2\mathrm{y}=0\Rightarrow \mathrm{y}=0$$$$\Rightarrow \mathrm{f}{(\mathrm{x},\mathrm{y})}_{\min }=\mathrm{f}(-3,0)=9-18+12=3$$

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