3

I've been struggling with this equation and how to find the maximum value it can take:

Maximise $xy^2z^3$ given that $x^2+y^2+z^2 = 1$

The question is from the book Introduction to Inequalities - CJ Bradley.

Any help would be much appreciated!

Thanks

Aryabhata
  • 82,206
Hummus
  • 563
  • 3
    Lagrange Multipliers? – Varun Iyer Jul 15 '14 at 09:45
  • 1
    The question is from the AM-GM inequality chapter in the book I am reading, and Lagrange multipliers has not been mentioned before. The question could probably be solved using that method, however I would like to know of the way using the AM-GM inequality. Thanks for the help though! Any ideas with the AM-GM? @VarunIyer – Hummus Jul 15 '14 at 09:47
  • Maybe this link could give you some inspiration :): http://math.stackexchange.com/questions/839433/minimizing-fx-y-z-dfracxyzxyz-on-a-sphere/844037#844037 – Surb Jul 15 '14 at 10:06
  • Thanks @Surb. How should I modify the solution to find the maximum value? – Hummus Jul 15 '14 at 10:11
  • Hummus, could you please write the title of your book? – Mohammad W. Alomari Jul 15 '14 at 14:11
  • 1
    Edited! :) @mwomath – Hummus Jul 15 '14 at 14:40

3 Answers3

10

To use AM=GM. rewrite the condition as $$x^2+(1/2)y^2+(1/2)y^2+(1/3)z^2+(1/3)z^2+(1/3)z^2=1.$$ Applying AM=GM to the left side, we get that $$\frac{1}{6}=\frac{x^2+(1/2)y^2+(1/2)y^2+(1/3)z^2+(1/3)z^2+(1/3)z^2}{6}\ge \left((1/4)(1/27)x^2y^4z^6 \right)^{1/6}.$$ From this we can write down the answer. Note that we have equality when $|x|=\sqrt{1/2}\,|y|=\sqrt{1/3}\,|z|$.

André Nicolas
  • 507,029
4

Since $$ \frac16\left(6x^2\right)+\frac13\left(3y^2\right)+\frac12\left(2z^2\right)=1 $$ The AM-GM says that $$ \left(6x^2\right)^{1/6}\left(3y^2\right)^{1/3}\left(2z^2\right)^{1/2}\le1 $$ which, upon raising to the third power, becomes $$ xy^2z^3\le\frac1{12\sqrt3} $$ The maximum is attained when $6x^2=3y^2=2z^2=1$.

robjohn
  • 345,667
0

Another non-AM/GM approach, also avoiding Lagrange multipliers, is to observe that the problem is equivalent to maximizing

$$f(x,z)=x(1-x^2-z^2)z^3$$

in the square region $0\le x,z\le1$.

(Note: The "Another" referred to an answer that has since been deleted. André Nicolas's marvelous answer is, of course, the "right" one for the OP's purpose.)

Barry Cipra
  • 79,832