AM-GM, AM-GM eveywhere

Algebra Level 3

For $x> 0$ , maximize $f(x)=(1+x)(1+x)(1-x)$ .

Give your answer up to 3 decimal places.

The answer is 1.185.

2 solutions

Akshat Sharda
Sep 25, 2015

Method $1 \rightarrow AM-GM$ inequality :

$f(x)=(1+x)(1+x)(1-x)=\frac{(2-2x)(1+x)^{2}}{2}$

Let, $P(x)=(2-2x)(1+x)^{2}$

By $AM-GM$ inequality,

$\Rightarrow \frac{2-2x+1+x+1+x}{3}≥(P(x))^{\frac{1}{3}}$

$\Rightarrow \left(\frac{4}{3}\right)^{3}≥P(x)$

Maximum value of $P(x)=\frac{64}{27}$

Therefore, $f(x)$ 's maximum value,

$=\frac{P(x)}{2}=\frac{\frac{64}{27}}{2}=\frac{32}{27}=\boxed{1.185}$

Method $2 \rightarrow$ A little $Calculus$ :

$f(x)=(1+x)(1+x)(1-x)=-x^{3}-x^{2}+x+1$

$f'(x)=-3x^{2}-2x+1$

Now, $f'(x)=0$ so as to find the critical points.

$\Rightarrow x=\frac{1}{3},-1$

Now we can go for the second derivative test to find the relative maximum,

$f''(x)=-6x-2$

Now, $f''(x)=0 \Rightarrow x=\frac{1}{3}$

Maximum value of $f(x)$ will be at $x=\frac{1}{3}$ ,

So,

$f(\frac{1}{3})=-\left(\frac{1}{3}\right)^{3}-\left(\frac{1}{3}\right)^{2}+\frac{1}{3}+1$

$=-\frac{1}{27}-\frac{1}{9}+\frac{1}{3}+1$

$=\frac{32}{27}=\boxed{1.185}$

In your calculus method you should solve f '(x)=0 to find that x=1/3 or x=-1 are local minimum or maximum values. You may then apply the second derivative test for x=1/3 (since x>0) to see that f "(1/3)<0 indicating that x=1/3 is a local maximum. To find the maximum of f(x) we then need to check the end points as x approaches 0 and positive infinity. f(0) approaches 1 and f(x) approaches negative infinity as x approaches infinity. Since f(1/3)=32/27>1, 32/27 is the maximum for f(x), x>0.

Jonathan Hocker - 5 years, 8 months ago

Yes ... I did it but in solution I just directly showed the result we wanted.

Akshat Sharda - 5 years, 8 months ago

Ok I just wanted to point that out because in your solution it looked like you were solving

f''(x)=-6x-2=0 --> x=-1/3

which is an inflection point and only a sign away from the maximum. You might edit it and make it more clear that that's not what is going on.

Jonathan Hocker - 5 years, 8 months ago

@Jonathan Hocker – I made an edit. Please check!!

Akshat Sharda - 5 years, 8 months ago

I edited the post because it was hard to see the ' or " after the f in f '(x) or f "(x). Hope that clears up any confusion between first and second derivatives. As for the solution, it is important that you realize the difference between finding inflection points and using the second derivative test to determine if a point is a local maximum or minimum. In the line where you equate f "(x)=0 that finds an inflection point. To determine whether a point is a local max or min you need to plug in the solutions to f '(x)=0 (the critical points) into the second derivative and check if it is positive or negative. If the value of the second derivative is positive, that point is a local minimum and vice versa. Here is a link to the wiki article about the second derivative test: https://en.wikipedia.org/wiki/Derivative test#Second derivative_test

Jonathan Hocker - 5 years, 6 months ago

For the calculus method, since you said for $x>0$ , you can automatically eliminate the negative value of $x$ .

Hobart Pao - 5 years, 6 months ago

Still have to check to make sure that the local max is the global max by checking the end points (zero and positive infinity).

Jonathan Hocker - 5 years, 6 months ago

I meant eliminate -1/3. That's beyond the endpoints

Hobart Pao - 5 years, 6 months ago

I used calculus

Hans Gabriel Daduya - 3 years, 7 months ago

Akash Singh
Nov 4, 2015

take am gm in

(1+x)/2 , (1+x)/2 , (1-x)

so [(1+x+1+x)/2+1-x]/3 >={[(1+x)^(2) * (1-x)]/4}^(1/3)

2/3 >=[t/4]^(1/3)]

where t=(1+x)(1+x)(1-x)

8/9>=t/4

t<=32/9

t<=1.185

$\frac{32}{9}\neq 1.185$ I got $\frac{32}{27}$ by using calculus.

Naren Bhandari - 3 years, 7 months ago

AM-GM, AM-GM eveywhere

The answer is 1.185.

2 solutions

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