Find 3A

Calculus Level 3

Let y = 3 x 2 + 6 x + 16 3 x 2 + 6 x + 6 y = \dfrac{3x^2+6x+16}{3x^2+6x+6} . If A A denotes the maximum possible value of y y for real x x , submit your answer as 3 A 3A .


The answer is 13.

Aly Ahmed by Aly Ahmed , 34, Egypt

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1 solution

Zee Ell
Jan 4, 2019

y = 3 x 2 + 6 x + 16 3 x 2 + 6 x + 6 = ( 3 x 2 + 6 x + 6 ) + 10 3 x 2 + 6 x + 6 = 3 x 2 + 6 x + 6 3 x 2 + 6 x + 6 + 10 3 x 2 + 6 x + 6 = 1 + 10 3 ( x 2 + 2 x + 1 ) + 3 = 1 + 10 3 ( x + 1 ) 2 + 3 y= \frac {3x^2+6x+16}{3x^2+6x+6} = \frac {(3x^2+6x+6) +10}{3x^2+6x+6} = \frac {3x^2+6x+6}{3x^2+6x+6} + \frac {10}{3x^2+6x+6} = 1 + \frac {10}{3(x^2+2x+1)+3} = 1 + \frac {10}{3(x+1)^2+3}

Now, our expression is at its maximum, when the denominator of the fraction is at its minimum. We can see from the completed square form, that this happens, when x = - 1

Hence:

A = y m a x = 1 + 10 3 = 13 3 A = y_{max} = 1 + \frac {10}{3} = \frac {13}{3}

3 A = 3 × 13 3 = 13 3A = 3 × \frac {13}{3} = \boxed {13}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Regarding the range, since the denominator of the fraction part can be infinitely large, therefore its minimum value is infinitely small, but positive (just bigger than 0, making the lower limit of y 1 + 0 = 1).

1 < y ≤ 13 3 1 < y \leq \frac {13}{3}

Zee Ell - 2 years, 5 months ago

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