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

algebra-precalculus

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?

Best Answer

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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