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If $3x+4y=12$ $\forall x,y \ge 0$, the maximum value of $(x^2) \times (y^3)$ is

  1. $6 \times (6/5)^5$
  2. $3 \times (6/5)^5$
  3. $ (6/5)^5 $
  4. $7 \times (6/5)^5$

How to approach this problem? I thought of using the approach for finding maxima-minima for two independent variable but I am not sure about that as the $x$ and $y$ vanishes after the first partial derivative itself.

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    Rewrite $x^2y^3$ in one variable using a substitution derived from the constraint. Also, it's a bit funky to use $*$ for multiplication - mostly people just put letters next to each other or they use $\cdot$ (`\cdot`).2011-08-02

3 Answers 3

10

AM-GM gives

$$12 = \frac{3}{2} x + \frac{3}{2} x + \frac{4}{3} y + \frac{4}{3} y + \frac{4}{3} y \ge 5 \left( \frac{16}{3} x^2 y^3 \right)^{1/5}$$

hence

$$x^2 y^3 \le \frac{3}{16} \left( \frac{12}{5} \right)^5 = 6 \cdot \left( \frac{6}{5} \right)^5$$

with equality when $\frac{3}{2} x = \frac{4}{3} y$, which is attainable.

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Lagrange multipliers are pretty fast, and don't require seeing any tricks:

$$2xy^3 = 3\lambda$$ $$3x^2y^2 = 4\lambda$$ Divide the two equations, and you get ${\displaystyle {2 \over 3} {y \over x} = {3 \over 4}}$ or ${\displaystyle y = {9 \over 8} x}$. Putting back into the equation you get $$3x + {9 \over 2} x = 12$$ This solves as $x = 24/15 = 8/5$ and $y = 9/8\times 8/5 = 9/5$. So the minimum value of $x^2y^3$ is $(8/5)^2(9/5)^3$.

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    Yes, that would be a general method, but I can never resist pointing out that many optimization problems don't actually require calculus to solve because of the simple nature of the functions involved. By the way, I wouldn't consider what I did a trick; the underlying principle here is convexity.2011-08-03
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    I do think that recognizing that one should use the convexity of the logarithm (or exponential) and then recognizing how to divide the expression into five terms might be tricky if the student had a limited amount of time for the problem.2011-08-03
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    well, think about it this way. Two basic tools in the hands of the student of inequalities are AM-GM and Cauchy-Schwarz. The latter doesn't seem likely to be useful, so once you have it in your mind to try applying the former, there's really only one way to do it (in such a way that the equality case is attainable).2011-08-03
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    If he's restricted to high school level math I agree with you.. but when I read the question it really looked like a typical multivariable calculus problem, and I inferred from his comments after stating it that he was doing multivariable calc.2011-08-03
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Here's a bit of a tricky way to do it: instead of maximizing $x^2 y^3$, we'll maximize its logarithm $2 \log x + 3 \log y$, or alternatively $2 \log x + 3 \log (3 - 3x/4)$.

But that denominator of $4$ is annoying; since $$ \log (3 - 3x/4) = \log (12-3x) - \log 4 $$ we can minimize $$ f(x) = 2 \log x + 3 \log (12-3x) $$ instead.

Differentiate: $f^\prime(x) = 2/x - 9/(12-3x)$. This is zero when $2(12-3x) = 9x$, or $24-6x = 9x$, or $x = 8/5$. Since $3x+4y=12$ we can solve to get $y = 9/5$. The maximum value of $x^2 y^3$ is therefore

$$ (8/5)^2 (9/5)^3 = (4/3 \times 6/5)^2 (3/2 \times 6/5)^3 = (4/3)^2 (3/2)^3 (6/5)^5 = 6 (6/5)^5. $$

This is probably not the most efficient solution to this particular problem. (This is made worse by the fact that once you get the answer, it still takes a while to figure out which of the choices it is; I suspect it's not what the writers of whatever test you're preparing for had in mind.) But I give it because it illustrates a general principle: products are annoying, so if you want to maximize one sometimes it's easier to maximize its log, which is a sum.