Let $x$ and $y$ be positive real numbers such that $4x + 9y = 60.$  Find the maximum value of $xy.$
I can use the given equation to eliminate one of the variables and express $xy$ in terms of the other. For example, if I solve for $x$, I get $x = \frac{60 - 9y}{4}.$ Then, $xy = \frac{60 - 9y}{4} \cdot y = \frac{60y - 9y^2}{4}.$ Now, I have a quadratic function of $y$ that I want to maximize. In this case, $a = -\frac{9}{4}$ and $b = \frac{60}{4}.$ So, the vertex occurs at $y = -\frac{b}{2a} = -\frac{\frac{60}{4}}{-\frac{9}{2}} = \frac{10}{3}.$ To find the corresponding value of $x$, I can plug $y = \frac{10}{3}$ back into the equation $x = \frac{60 - 9y}{4}.$ I get $x = \frac{60 - 9 \cdot \frac{10}{3}}{4} = \frac{60 - 30}{4} = \frac{15}{2}.$ Therefore, the maximum value of $xy$ occurs when $x = \frac{15}{2}$ and $y = \frac{10}{3}.$ To find this value, I multiply $x$ and $y$ together: $xy = \frac{15}{2} \cdot \frac{10}{3} = \frac{150}{6} = 25.$ 