Question

A closed rectangular box is made of aluminium
sheet of negligible thickness and the length of th
box is twice its breadth. If the capacity of the bo
is 243 cm3, compute the dimensions of the box o
least surface area.​

Answer 1

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Answer 2

Given:

Length of rectangular box = twice its breadth

Volume of box = 243cm³

To find:

The dimensions of the box.

Solution:

Let $l$ be the length, $b$ be the breadth and $h$ be the height of the rectangular aluminium box which is in the form of a cuboid.

Then volume, V of the cuboid is given by:

$Volume=length*breadth*height$

$V=l*b*h$

Since length is twice the breadth, $l=2b$ and volume (capacity) is given as  243cm³,

$243=2b*b*h$

$243=2b^{2}h$

∴ $h=\frac{243}{2b^{2} }...(1)$

The surface area of a cuboid is given by:

$S = 2(lb+bh+lh)$

$S=2(2b^{2} +bh+2bh)$

$S=2(2b^{2}+3bh)$

$S=4b^{2}+6bh...(2)$

Substituting the value of $h$ from equation (1) in the above equation (2),

$S=4b^{2}+6b*\frac{243}{2b^{2} }$

$S=4b^{2}+\frac{729}{b}...(3)$

Differentiating equation (3) with respect to (w.r.t) $b$,

$\frac{dS}{db}=8b-\frac{729}{b^{2}}$

At maxima or minima, $\frac{dS}{db}=0$

$0=8b-\frac{729}{b^{2} }$

$8b=\frac{729}{b^{2} }$

$b^{3}=\frac{729}{8}$

Taking cube root on both sides,

∴ $b=\frac{9}{2}cm$

Differentiating equation (3) again with respect to (w.r.t) $b$,

$\frac{d^{2}S}{db^{2}}=8+\frac{2*729}{b^{3}}$

To have the least surface area, $b=\frac{9}{2}cm$

i.e., at $b=\frac{9}{2}cm$, $\frac{d^{2}S}{db^{2}}>0$

Now, $l=2b$

∴ $l=2*\frac{9}{2}=9cm$

$h=\frac{243}{2b^{2} }$

∴ $h=\frac{243}{2*\frac{81}{4} }=\frac{243}{\frac{81}{2} }=\frac{243*2}{81}=6cm$

Hence, the dimensions of the rectangular box are 9cm×$\frac{9}{2}$cm×6cm so that it has the least surface area.

The dimensions of the rectangular box are 9cm×$\frac{9}{2}$cm×6cm so that it has the least surface area.