Question

11
find minimum value of x^2+y^2 +6x +12.​

Answer 1

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.

Answer 2

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,

$\frac{df}{dx}$ = 2x + 6 and $\frac{df}{dy}$ = 2

Putting, $\frac{df}{dx}$ and $\frac{df}{dy}$ = 0

We get,

2x+6 =0

x = -3

The point will be (x,y) = (-3,0)

Now,

r= $\frac{d^2f}{dx^2}$ = 2>0 and t= $\frac{d^2f}{dy^2}$ = 2 and s= $\frac{d^2f}{dxdy}$ = 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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