Maximising the Volume of a Box

Cut a square of side $x$ cm from each corner of a 24 cm by 15 cm sheet, then fold up the sides to make an open box.

Each base dimension loses $x$ at both ends. The folded height is $x$. A cuboid’s volume is base length × base width × height:

All three dimensions must be positive. The tighter restriction is $15-2x>0$, giving $0<x<7.5$. At either endpoint the box has zero volume.

Expand the brackets before differentiating.

The quadratic solver quickly gives 3 and 10. These convenient roots let you write exact working immediately:

Reject 10: it makes the 15 cm base dimension negative. At $x=3$, $V''=24x-156=-84<0$. The derivative is positive before 3 and negative after 3 within the physical domain, so this gives the largest volume.

For awkward roots, use the quadratic formula and keep the exact expression until the final calculation. A calculator result alone does not show the model, domain or maximum argument.