An open box is made from a $3$ m by $8$ m sheet by cutting squares of side $x$ from each corner. The volume is largest when $x$ equals

1. $V(x)=x(3-2x)(8-2x)=4x^3-22x^2+24x$. 2. $V'(x)=12x^2-44x+24=4(x-3)(3x-2)$. 3. $V'(x)=0 \Rightarrow x=3$ or $x=\tfrac{2}{3}$. 4. $x=3$ is rejected (width $3-2x$ would be negative). 5. $V''(\tfrac{2}{3})=24(\tfrac{2}{3})-44=-28<0$, so $x=\tfrac{2}{3}$ gives the maximum. _Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.35_