Inequality based on AM/GM Inequality

Question

Find the greatest value of $x^3y^4$ If $2x+3y=7 $ and $x≥0, y≥0$. (Probably based on weighted arithmetic and geometric mean)

Answer 1

Use this Hint: $$7=\frac23x+\frac23x+\frac23x + \frac34y+\frac34y+\frac34y+\frac34y\geqslant 7\sqrt[7]{\cdots}$$

Answer 2

it's pretty simple if you know about weighted means, proceed like this:

$(2x/3)*3+(3y/4)*4 ≥ 7[(2x/3)^³(3y/4)^4)^{1/7}$

i.e. $7≥7((x^3y^4)(3/32))^{1/7}$

or, $(x^3y^4)(3/32)≤1$

hence, $x^3y^4≤32/3$

so $32/3$ is the greatest value of $x^3y^4$

PS: sorry for the ill written answer, I'm on phone too, my friends out there please edit the answer ;)

Answer 3

My usual attempt to generalize.

To find the max of $x^m y^n$ subject to $ax+by = c$ where $m$ and $n$ are integers.

Following Macavity's solution, use the AGM inequality in the form $\sum_{i=1}^p a_i \ge p (\prod_{i=1}^p a_i)^{1/p} $.

Then

$\begin{array}\\ c &=m\frac{ax}{m} +n\frac{by}{n}\\ &\ge (m+n)((\frac{ax}{m})^m(\frac{by}{n})^n)^{1/(m+n)} \qquad (m \text{ copies of }\frac{ax}{m} \text{ and }n\text{ copies of }\frac{by}{n})\\ \end{array} $

with equality, by the AGM inequality, if and only if $\frac{ax}{m} =\frac{by}{n} $.

If this is true, then $y =\frac{anx}{bm} $ so

$\begin{array}\\ c &=m\frac{ax}{m} +n\frac{by}{n}\\ &=m\frac{ax}{m} +n\frac{ax}{m}\\ &=ax(1+\frac{n}{m})\\ &=ax\frac{m+n}{m}\\ \end{array} $

or $x =\frac{mc}{a(m+n)} $ and $y =\frac{mc}{a(m+n)}\frac{an}{bm} =\frac{nc}{b(m+n)} $.

Finally, $x^m y^n =(\frac{mc}{a(m+n)})^m(\frac{nc}{b(m+n)})^n =(\frac{m}{a})^m(\frac{n}{b})^n(\frac{c}{m+n})^{m+n} $.

If $m+n = c$, as in OP's case, $x =\frac{m}{a} $ and $y =\frac{n}{b} $ and the max is $x^m y^n =(\frac{m}{a})^m(\frac{n}{b})^n $.

In this case, $m=3, n=4, a=2, b=3, c=7$.

The result is $x = \frac{3}{2} $, $y = \frac{4}{3} $, and the max is $(\frac{3}{2})^3(\frac{4}{3})^4 =\frac{2^5}{3} $.

Note that this generalizes easily to maximize $\prod_{i=1}^m x_i^{n_i} $ subject to $c =\sum_{i=1}^m a_ix_i $ with $c, n_i,$ and $a_i$ being specified.

Answer 4

$y$ is a function of $x$. That is, $y=(7-2x)/3.$ And $\frac {dy}{dx}=y'=-2/3.$

Let $f(x)=x^3y^4.$ Then $f'(x)=3x^2 y^4+4x^3y^3y'=3x^2y^4-(8/3)x^3y^3.$

For $xy\ne 0$ we have $f(x)>0$ and $f'(x)=f(x)(\frac {3}{x}-\frac {8}{3y}).$

So for $xy\ne 0$ we have $f'(x)>0\iff \frac {3}{x}-\frac {8}{3y}>0\iff 9y>8x\iff21-6x>8x \iff 3/2>x.$ Similarly for $xy\ne 0$ we have $f'(x)<0\iff 3/2<x.$

And we have $xy=0\implies f'(x)=0.$

So $f(x)$ is increasing for $x<3/2$ and is decreasing for $3/2<x<7/2.$ ($7/2$ is the upper limit of the domain of $f,$ as $y=(7-2x)/3\geq 0.$) Therefore, since $f(x)$ is continuous, $\max f=f(3/2).$