0

If $x$ and $y$ are positive numbers, then what is the maximum value of the expression $x^4y^3$ where $4x + 6y = 28$?

One way to solve this question is to find the expression of $x$ or $y$ in terms of each other from the equation given and then put the value in $x^4y^3$ and then differentiate and find the maxima but that will be a lengthy process. Is there any other simpler way to solve this if this question comes up in a competitive exam?

Thanks in advance !!!

Ganit
  • 1,689

2 Answers2

4

Using AM-GM
$\displaystyle \frac{x+x+x+x+2y+2y+2y}{7} \geqslant \sqrt[7]{8x^{4} y^{3}} \ \Longrightarrow \ 4\geqslant \sqrt[7]{8x^{4} y^{3}} \Longrightarrow \ x^{4} y^{3} \leqslant 4^{7} /8 = 2048$

mrtechtroid
  • 795
  • 4
    There are a lot of ways to use AM-GM, many of which will not give the actual maximum. This answer would be complete after demonstrating the bound is in fact achievable for some allowable $(x, y)$. – Macavity May 31 '23 at 04:14
  • When (x,y)=(4,2) we see that it becomes equal to 2048. @Macavity – mrtechtroid May 31 '23 at 07:35
  • @mrtechtroid Please add that into your solution to make it complete (Comments are considered transient). Otherwise, you have only found an upper bound, and not shown that it can be achieved. – Calvin Lin May 31 '23 at 14:35
0

This is not difficult using differentiation. Let $V=x^4y^3$ then $$ V=(2/3)^3x^4(7-x)^3 $$ and \begin{align} {dV\over dx}&=(2/3)^3(4x^3(7-x)^3-3x^4(7-x)^2)\\ &=(2/3)^2x^3(7-x)^2(28-7x) \end{align} which has obvious minima at $x=0,7$, since $V$ is zero there, and what then has to be a maximum at $x=4$.

Suzu Hirose
  • 11,660