Question

The rectangular sheet of metal sides a, b has four equal square portions removed at the corners and the sides are then turned up so as to form an open rectangular box. When the volume contained in the box is maximum, the depth of the box is

• A. $\frac{1}{6}\left[\right(a+b)+(a^2-ab+b^2{)}^{1/2}]$
• B. $\frac{1}{6}\left[\right(a+b)-(a^2-ab+b^2{)}^{1/2}]$
• C. $\frac{1}{6}\left[\right(a-b)+(a^2-ab+b^2{)}^{1/2}]$
• D. $\frac{1}{6}\left[\right(a-b)-(a^2-ab+b^2{)}^{1/2}]$

Answer 1

The correct option is B $\frac{1}{6}\left[\right(a+b)-(a^2-ab+b^2{)}^{1/2}]$

Let sqaure of edge d is removed
So side becomes $b-2d,a-2d$
and volume $=(b-2d)(a-2d)d$
$\frac{dv}{dd}=(b-2d)(a-2d)-2d(b-2d+a-2d)$
$0=12d^2-4d(a+b)+ab$

$d=\frac{4(a+b)\pm \sqrt{16(a+b{)}^2+48(ab)}}{2\times 12}$
$d=\frac{(a+b)\pm \sqrt{a^2+b^2-ab}}{6}$
V is max when $d=\frac{(a+b)-(a^2+b^2-ab{)}^{\frac{1}{2}}}{6}$