Question

What is the minimum value of f(x,y) = x^2 + y^2 + 6x + 12?

Answer 1

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

min value =3

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Answer 2

Answer:

Step-by-step explanation:

the minimum value of f(x,y) = x^2 + y^2 + 6x + 12

As a function with 2 variables there is no minimum at any case but a minimum in x and y axis.

So the minimum for x: is d/dx(f(x,y)) (partial) => 2x+6 =0 x=-3 ,

the minimum for y : is d/dy(f(x,y)) (partial) => 2y = 0 => y = 0.

So it is the point (x,y) = (-3,0).