Let x and y be positive real numbers such that 1/(x+2) + 1/(y+2) = 1/3. Find the minimum value of x + 2y.
We get the minimum at x = y. So 2/(x + 2) = 1/3 so x = 4. The minimum is 4 + 2(4) = 12.
Let x +2 = a
Let y + 2 = b
So we have that
1/a + 1/b = 1/3
(a + b) / ab = 1/3
3(a + b) = ab
3a + 3b = ab
ab - 3b = 3a
b (a - 3) = 3a
b = 3a
(a - 3)
These are the only integer solutions for a, b
a = 4, x = 2 and when b = 12, y = 10 so x + 2y = 22
a = 6, x = 4 and when b = 6, y = 4 so x + 2y = 12
a = 12, x =10 and when b = 4, y = 2 so x + 2y = 18
So....the minimum is when x = y = 4 and the minimum of x + 2y is 12 [ as the guest found !!! ]
The question says that x and y are real numbers, it does not say that they are integers (necessarily).
The minimum is 3 + 6sqrt(2), approximately 11.585, at x = 1 + 3sqrt(2), y = 1 + 3sqrt(2)/2, approximately (5.243, 3.121).
Sorry, but it's wrong. (Hint: It contains a square root)
Forget your condescending "Hint"
You are, by choice, a nobody.
You are addressing acknowledge Mathematicians.
If you believe you know the correct answer then spell out your reasons and give the answer, as you see it, in full.
Or if you are incapable of doing that then state why you believe a different answer is correct.
Like you could day. "The answer that the teacher gave me is .... but i do not know how she got it."