Yahoo Answers: Answers and Comments for (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? [Mathematics]
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From Yaba
enUS
Tue, 16 Jul 2019 15:28:51 +0000
3
Yahoo Answers: Answers and Comments for (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? [Mathematics]
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From la console: y = x/(x² + x + 2) ← this is a function, you c...
https://answers.yahoo.com/question/index?qid=20190716152851AAbraWN
https://answers.yahoo.com/question/index?qid=20190716152851AAbraWN
Tue, 16 Jul 2019 16:59:48 +0000
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

From husoski: To show you how far I'll go to avoid the q...
https://answers.yahoo.com/question/index?qid=20190716152851AAbraWN
https://answers.yahoo.com/question/index?qid=20190716152851AAbraWN
Tue, 16 Jul 2019 16:26:18 +0000
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.