What is the minimum value of f(x,y) = x^2 + y^2 + 6x + 12?
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min value =3
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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).
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