An open topped box is to be constructed by removing equal squares from each corner of a $3$ metre by $8$ metre rectangular sheet of aluminium and folding up the sides. Find the volume of the largest such box.

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Solution

Let $x$ metre be the length of a side of the removed square
Length of the box $= 8 - x - x = 8 - 2 x$
Width of the box $= 3 - x - x = 3 - 2 x$
Height of the box $= x$
Now, the volume

$V = (8 - 2 x) (3 - 2 x) (x)$
$\Rightarrow V = 2 x (x - 4) (2 x - 3)$
$\Rightarrow V = 2 x (2 x^{2} - 11 x + 12)$
$\Rightarrow V = 2 (2 x^{3} - 11 x^{2} + 12 x)$

Differentating w.r.t $x$ ,
$\frac{d V}{d x} = 2 (6 x^{2} - 22 x + 12)$
$\Rightarrow \frac{d V}{d x} = 4 (3 x^{2} - 11 x + 6)$
$\Rightarrow \frac{d V}{d x} = 4 (3 x^{2} - 9 x - 2 x + 6)$
$\Rightarrow \frac{d V}{d x} = 4 (x - 3) (3 x - 2)$
Substituing $\frac{d V}{d x} = 0$ ,
$4 (x - 3) (3 x - 2) = 0$
$\Rightarrow x = \frac{2}{3}, 3$
When $x = 3,$ then
Width $= 3 - 2 x = - 3$

Which is not possible, so $x = \frac{2}{3}$
Again, Differentating w.r.t $x$
$\frac{d^{2} V}{d x^{2}} = 2 (12 x - 22)$
Substituting $x = \frac{2}{3}$ ,

$\frac{d^{2} V}{d x^{2}} = 2 (12 (\frac{2}{3}) - 22)$
$\Rightarrow \frac{d^{2} V}{d x^{2}} = 2 (8 - 22) = - 28 < 0$
So, $x = \frac{2}{3}$ is a point of maxima.
Therefore, the maximum volume is
$\Rightarrow V = 2 (2 x^{3} - 11 x^{2} + 12 x)$
$V = 2 x (x - 4) (2 x - 3)$

$\Rightarrow V = \frac{4}{3} (\frac{2}{3} - 4) (\frac{4}{3} - 3)$

$\Rightarrow V = \frac{4}{3} (- \frac{10}{3}) (- \frac{5}{3}) = \frac{200}{27} m^{3}$
Hence, volume of largest box is $\frac{200}{27} m^{3}$

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