Question

A box open from top is made from a rectangular sheet of dimension $a\times b$ by cutting squares each of side $x$ from each of the four corners and folding up the flaps. If the volume of the box is maximum, then $x$ is equal to

• A. $\frac{a+b-\sqrt{a^2+b^2+ab}}{6}$
• B. $\frac{a+b-\sqrt{a^2+b^2-ab}}{6}$
• C. $\frac{a+b-\sqrt{a^2+b^2-ab}}{12}$
• D. $\frac{a+b+\sqrt{a^2+b^2-ab}}{6}$

Answer 1

The correct option is B $\frac{a+b-\sqrt{a^2+b^2-ab}}{6}$


$V=(a-2x)(b-2x)x$
For maximum volume
$\frac{dV}{dx}=0$
$\Rightarrow -2(b-2x)x+(a-2x)(-2)x+(a-2x)(b-2x)=0$
$\Rightarrow 12x^2+x(-2a-2b-2b-2a)+ab=0$
$\Rightarrow 12x^2-4(a+b)x+ab=0$
$\Rightarrow x=\frac{4(a+b)\pm \sqrt{16(a+b{)}^2-48ab}}{24}$
$\Rightarrow x=\frac{(a+b)\pm \sqrt{a^2+b^2-ab}}{6}$

$\frac{d^{2}V}{d{x}^2}=24x-4(a+b)$
$=4(6x-(a+b\left)\right)<0,$ for $x=\frac{a+b-\sqrt{a^2+b^2-ab}}{6}$

Hence$V_{\text{max}}\text{at}x=\frac{a+b-\sqrt{a^2+b^2-ab}}{6}$