Question

3. Copper is supplied to a company in rectangular sheets measuring 108 by 45 cm. The company wants to make open boxes in the shape of cuboids, with the maximum volume possible and the minimum of waste. If the height of the box is x use differentiation to find the dimensions and hence the maximum volume

Answer 1

Dimmension of copper sheets- 108 X 45 cm

Area of Copper sheet $= 108\times 45=4860$

Let the length, breadth and height of box be l, x , x ( as it has square base)

Area of open cuboid box-

$x^2+4xl=4860 \\[9pt] \Rightarrow l=\dfrac{4860-x^2}{4x}~~~~~~~~~~~~~~-(1)$

Volume of box formed $V= l\times x\times x$

$=x^2\dfrac{(4860-x^2)}{4x}\\[9pt]= \dfrac{4860x-x^3}{4}$

Differentiate V w.r.t. x-

$\dfrac{dV}{dx}=\dfrac{4860-3x^2}{4}$

Putting$\dfrac{dV}{dx}=0$

$\Rightarrow \dfrac{4860-3x^2}{4}=0\\[9pt]\Rightarrow x^2=1620\\[9pt]\Rightarrow x=40.24cm$

Again differentiate V'(x) w.r.t x-

$V''(x)=-6x \\[9pt] V''(x)_{x=40.24}=-241.44$

So $V''(x)<0, V(x)$ has local maxima at $x=40.61 cm$

Dimmension of box-

$l=\dfrac{4860-x^2}{4x}\\[9pt]=\dfrac{4860-(40.61)^2}{4\times 40.61}\\[9pt]=\dfrac{3210.82}{162.4}=19.77 cm$

Dimmension of box is - $19.77\times 40.61\times 40.61$

And Maximum volume is-

$V=\dfrac{4860x-x^3}{4x}\\[9pt]=\dfrac{4860\times 40.61-(40.61)^3}{4\times 40.61}\\[9pt]=\dfrac{130391.72}{162.44}\\[9pt]=802.706\text{ cm}^3$