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# Help quickk

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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.

Nov 27, 2019

#1
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We get the minimum at x = y.  So 2/(x + 2) = 1/3 so x = 4.  The minimum is 4 + 2(4) = 12.

Nov 27, 2019
#2
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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)

a       b

4      12

6       6

12     4

These are the only integer solutions for  a, b

When

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 !!! ]   Nov 27, 2019
#3
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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).

Nov 28, 2019
#4
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Sorry, but it's wrong. (Hint: It contains a square root)

Nov 28, 2019
#5
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You are, by choice, a nobody.