An open box with square base is to be made out of a given quantity of cardboard of area \(\mathrm{p}^{2}\) sq. units. The maximum volume of the box is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{\mathrm{p}^{3}}{6\sqrt{3}}\) cubic units
Let the side of the square base be x and the height be h. An open box has a base and four sides but no top, so
\(x^2 + 4xh = p^2\)
\( \displaystyle h = \frac{p^2 - x^2}{4x} \)
The volume is
\( \displaystyle V = x^2 h = \frac{p^2 x - x^3}{4} \)
For the maximum volume:
\( \displaystyle \frac{dV}{dx} = \frac{p^2 - 3x^2}{4} = 0 \)
\( \displaystyle x = \frac{p}{\sqrt{3}} \)
\( \displaystyle \frac{d^2V}{dx^2} = -\frac{3x}{2} \lt 0 \)
so this x gives a maximum. Then
\( \displaystyle p^2 - x^2 = p^2 - \frac{p^2}{3} = \frac{2p^2}{3} \)
\( \displaystyle V = \frac{x\,(p^2 - x^2)}{4} = \frac{1}{4} \times \frac{p}{\sqrt{3}} \times \frac{2p^2}{3} \)
\( \displaystyle = \frac{p^3}{6\sqrt{3}} \)
So the maximum volume is \(\frac{p^3}{6\sqrt{3}}\) cubic units.
NCERT Exemplar:
Application of Derivatives
