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# (1) x > 0. Find the max value of y = x / (x^2 + x + 2)? (2) a > 0, b > 0 and x >= 0. Show that y = x^3 - 3abx + a^3 + b^3 >= 0. How?

### 2 Answers

- la consoleLv 71 year ago
y = x/(x² + x + 2) ← this is a function, you can obtain the maximum when the derivative is zero

The function looks like (u/v), so the derivative looks like: [(u'.v) - (v'.u)]/v² → where:

u = x → u' = 1

v = x² + x + 2 → v' = 2x + 1

y' = [(u'.v) - (v'.u)]/v²

y' = [(x² + x + 2) - (2x + 1).x]/(x² + x + 2)²

y' = [x² + x + 2 - 2x² - x]/(x² + x + 2)²

y' = (- x² + 2)/(x² + x + 2)² → then you solve for x the equation: y' = 0

(- x² + 2)/(x² + x + 2)² = 0 → the denominator cannot be zero

y = [4√2 - 2] / [16 - 4√2 + 4√2 - 2]

y = (4√2 - 2) / 14

y = (2√2 - 1)/7 ← this is the maximum

- husoskiLv 71 year ago
To show you how far I'll go to avoid the quotient rule... :^)

First I notice that x^2 + x + 2 > 0 always (negative discriminant and positive at x=0). That means that y>0 for all x>0 and maximizing y is the same as minimizing 1/y.

1/y = (x^2 + x + 2)/x = x + 1 + 2/x

(1/y)' = 1 - 2/x^2

Solving for (1/y)' = 0 gets x^2 = 2 and only the positive square root of 2 fits the x>0 restriction.

That's problem 1, the easy way.

For problem 2, find the minimum value for fixed a,b and variable x.

y'(x) = 3x^2 - 3ab = 0

x^2 = ab

x = sqrt(ab)

The minimum y value on [0, oo) is:

y(sqrt(ab)) = (ab)^(3/2) - 3ab*sqrt(ab) + a^3 + b^3

= (ab)^(3/2) - 3*(ab)^(3/2) + a^3 + b^3

= a^3 - 2*a^(3/2)*b^(3/2) + b^3

That's the square of [a^(3/2) - b^(3/2)], and a square of a real number is never is never negative. The minimum value of y(x) on [0,oo) can't be negative, y(x) >= 0 for all x>=0.

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I got it. Thanks!