Question

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

Answer 1

Answer:

3

Step-by-step explanation:

⇒ x^2 + 6x + 12

⇒ x^2 + 6x + 9 + 3

⇒ x^2 + 6x + ( 3 )^2 + 3

⇒ x^2 + 2( 3 * x ) + ( 3 )^2 + 3

⇒ ( x + 3 )^2 + 3

 Square of any number can't be lesser than 0, so the minimum value of ( x +  3 )^2 can be 0.

⇒ min.: 0 + 3

⇒ min.: 3

   Hence the minimum value of the given expression is 3.

Answer 2

Correct question:-

$\textsf{What is the minimum value of f(x,y) =}$ $\sf{x\² + y\² + 6x + 12\:\:?}$

Solution:-

$\displaystile{x^2+y^2+6x+12}\\\\\displaystile{=(x+3)^2+y^2-9+12}$

$\textsf{The minimum value of a perfect square is 0.}\\\\\displaystile{f(x,y)=0+0+3}\\\\\boxed{\pink{=}\green{3}}$