# If x and y are real numbers, the minimum possible value of the expression (x + 3)² + 2(y − 2)² + 4(x − 7)² + (y + 4)² is (A) 172 (B) 65 (C) 136 (D) 152 (E) 104

1
by Deleted account

2016-04-02T04:59:07+05:30

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Solutions

We expand the given expression to obtain

(x² + 6x + 9) + 2(y² − 4y + 4) + 4(x² − 14x + 49) + (y² + 8y + 16)

We expand further to obtain

x² + 6x + 9 + 2y² − 4y + 4+ 4x² − 14x + 49 + (y² + 8y + 16

We simplify to obtain 5x²− 50x + 3y² + 229

We remove a common factor of 5 from the first two terms

5(x²− 10x) + 3y² + 229

and then complete the square to obtain

5(x²− 10x + 52 − 5² ) + 3y² + 229

This gives

5(x − 5)² − 125 + 3y² + 229 or 5(x − 5)² + 3y² + 104

Since y ² ≥ 0 for all real numbers y, then the minimum value of 3y² + 24 is 24. Since the minimum value of (x−3)²+4(x−7)² is 80 and the minimum value of 2(y−2)²+(y+4)² is 24, then the minimum value of

(x − 3)² + 2(y − 2)² + 4(x − 7)²+ (y + 4)² is 80 + 24 = 104.