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Let $x$ and $y$ be positive real numbers such that $4x + 9y = 60.$ Find the maximum value of $xy.$


I think this problem has to do with the use of AM-GM. Can someone help me find the solution?

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Yes, you are right. In this case AM-GM is the most elementary method.

$$AM \ge GM\\ \frac{4x+9y}{2} \ge \sqrt{4x\times 9y}\\ 25 \ge xy$$

Hence maximum value of $xy$ is 25.

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    $\begingroup$ I think your answer is the best one ; $+1$ $\endgroup$ – J. W. Tanner May 15 '20 at 3:12
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If $4x+9y=60,$ then $y=\dfrac{60-4x}9$, so $xy=\dfrac{(60-4x)x}9=\dfrac{-4x^2+60x}9$

$=\dfrac{-(4x^2-60x+225)+225}{9}=\dfrac{225-(2x-15)^2}9\le\dfrac{225}9=25.$

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Different ways to do this problem. From a geometric stand point (I deliberately avoid a typical Calculus approach because you tagged Pre-Calculus), you can consider $y=c/x$ and $4x+9y=60$ in the first quadrant and you want these graphs to be tangential at a point. Subbing $y=c/x$ into the linear equation and multiplying through by $x$ results in $4x^2-60x+9c=0$ and using the Discriminant $b^2-4ac=0$ (You want ONE solution!) gives you $c=25$. From here you can find the point of tangency and all the rest.

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Solving for $x$ gives us

$$x=15-\frac94 y$$

Then $xy$ is

$$f(y)=xy=y\left(15-\frac 94 y\right)=-\frac94 y^2+15y$$

This has a critical point at

$$0=f'(y)=-\frac92 y+15$$

$$\Rightarrow y=\frac{30}{9}$$

Since this is the only critical point and $xy$ is a quadratic with a negative leading coefficient, we may conclude that maximum occurs at this $y$. Thus, the maximum of $xy$ is

$$xy=\frac{30}{9}\left(15-\frac 94 \cdot \frac{30}{9}\right)=25$$

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$$y=(60-4x)/9$$ $$xy=\frac{x(60-4x)}{9}$$ $$=\frac{4x(15-x)}{9}$$ $$=\frac{-4(x^2-15x)}{9}$$ $$=\frac{-4((x-7.5)^2-225/4)}{9}$$ $$=\frac{-4(x-7.5)^2+225)}{9}$$ max value = 225/9=25, which occurs for $x=7.5,y=10/3.$

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