Application of Derivatives — Class 12 Maths Solution

Since, volume of the box $= 1024\;{\rm{c}}{{\rm{m}}^3}$

Let length of the side of square base be $x\;{\rm{cm}}$ and height of the box be $y\,{\rm{cm}}$.

Therefore the volume of the box $(V) = {x^2} \cdot y = 1024$
Since, ${x^2}y = 1024 \Rightarrow y = \frac{{1024}}{{{x^2}}}$

Let $C$ denotes the cost of the box.

Therefore,$C = 2{x^2} \times 5 + 4xy \times 2.50$
$= 10{x^2} + 10xy = 10x(x + y)$

$= 10x\left( {x + \frac{{1024}}{{{x^2}}}} \right)$

$= \frac{{10x}}{{{x^2}}}\left( {{x^3} + 1024} \right)$

$\Rightarrow$ $C = 10{x^2} + \frac{{10240}}{x}$ …….(i)

On differentiating both sides w.r.t. $x$, we get
$\frac{{dC}}{{dx}} = 20x + 10240{( - x)^{ - 2}}$
$= 20x - \frac{{10240}}{{{x^2}}}$ …..(ii)

Now, $\frac{{dC}}{{dx}} = 0$

$\Rightarrow$ $20x = \frac{{10240}}{{{x^2}}}$
$\Rightarrow$ $20{x^3} = 10240$

$\Rightarrow$ ${x^3} = 512 = {8^3} \Rightarrow x = 8$

Again, differentiating Eq. (ii) w.r.t. $x$, we get

$\frac{{{d^2}C}}{{d{x^2}}} = 20 - 10240( - 2) \cdot \frac{1}{{{x^3}}}$
$= 20 + \frac{{20480}}{{{x^3}}} > 0$

$\therefore {\left( {\frac{{{d^2}C}}{{d{x^2}}}} \right)_{x = 8}} = 20 + \frac{{20480}}{{512}} = 60 > 0$

For $x = 8$, cost is minimum and the corresponding least cost of the box,

$C(8) = 10 \cdot {8^2} + \frac{{10240}}{8}$

$= 640 + 1280 = 1920$

Therefore the least cost =Rs.1920