Volume of the cardboard box

Since we've removed the four corner squares (of side length $x$ ) of the cardboard and converted it into a cuboid, the cuboid will have dimensions $x, 18-2x, 20-2x$ .

Now, we want to maximize the volume of this cuboid , $V = x(18-2x)(20-2x) = 4x^3 - 76x^2 + 360x$ .

Differentiating $V$ with respect to $x$ gives $\dfrac{dV}{dx} = 12x^2 - 152 x + 360$ .

At the extremal point, $\dfrac{dV}{dx} = 0$ . Applying the quadratic formula gives $x = \dfrac{19}3 \pm \dfrac{\sqrt{91}}3$ . But if $x = \dfrac{19}3 + \dfrac{\sqrt{91}}3$ , then $V< 0$ which is absurd, thus $x = \dfrac{19}3 - \dfrac{\sqrt{91}}3$ only.

Lastly, we need to show that $V$ is maximized when $x = \dfrac{19}3 - \dfrac{\sqrt{91}}3$ . Applying the second derivative test shows that $\dfrac{d^2 V}{dx^2} = 24x - 152 < 0$ when $x = \dfrac{19}3 + \dfrac{\sqrt{91}}3$ . This concludes that $V$ is indeed maximized when $x = \dfrac{19}3 - \dfrac{\sqrt{91}}3$ .

Upon substitution, we get the answer of $V = 4 \left( \dfrac{19}3 - \dfrac{\sqrt{91}}3 \right)^3 - 76\left( \dfrac{19}3 - \dfrac{\sqrt{91}}3 \right)^2 + 360 \left( \dfrac{19}3 - \dfrac{\sqrt{91}}3 \right) =\dfrac{8}{27} (836 + 91\sqrt{91} ) \approx \boxed{505} .$