A calculus problem by A Former Brilliant Member

$12x+12x+18y=5400$ $\implies$ $24x+18y=5400$ $\implies$ $18y=5400-24x$ $\implies$ $y=300-\dfrac{4x}{3}$

The area of the fence is $A=xy=x\left(300-\dfrac{4x}{3} \right)=300x-\dfrac{4x^2}{3}$

Differentiate both sides with respect to $x$ ,

$A=300x-\dfrac{4x^2}{3}$ $\implies$ $\dfrac{dA}{dx}=300-\dfrac{4}{3}(2x)=300-\dfrac{8x}{3}$

set $\dfrac{dA}{dx}=0$

$300-\dfrac{8x}{3}=0$

$\dfrac{8x}{3}=300$

$x=\dfrac{3}{8}(300)=112.5$

Solving for $y$ , we get

$y=300-\dfrac{4}{3}(112.5)=150$

The largest possible area is

$A=xy=112.5(150)=$ $\large \color{#D61F06}\boxed{16875~ft^2}$