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find minimum value of x^2+y^2 +6x +12.
Answers
Step-by-step explanation:
Discuss minimum value of f(x,y)=x2 + y2 + 6x + 12. hence. f(x,y) has minimum value at (-3,0), which is f(x,y) = 12 + 9 – 18 = 3.
3 is the minimum value of the equation x²+ y² + 6x + 12 when the formula for the minimum value is rt – s².
Given that,
We have to find the minimum value of the equation x²+ y² + 6x + 12
We know that,
f(x,y) = x²+ y² + 6x + 12
Now,
= 2x + 6 and = 2
Putting, and = 0
We get,
2x+6 =0
x = -3
The point will be (x,y) = (-3,0)
Now,
r= = 2>0 and t= = 2 and s= = 0
So, formula is
rt – s² = 4>0 and r>0
f(x,y) has minimum value at (-3,0), which is f(x,y) = 12 + 9 – 18 = 3.
Therefore, 3 is the minimum value of the equation x²+ y² + 6x + 12.
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