Question

An open box with square base is to be made out of a given quantity of cardboard of area \(P^2\) sq. units. The maximum volume of the box is :

• A. \(\frac{P^3}{6}\)
• B. \(\frac{P}{6\sqrt{3}}\)
• C. \(\frac{P^2}{6\sqrt{3}}\)
• D. \(\frac{P^3}{6\sqrt{3}}\)

Answer 1

The Correct Option is D

To determine the maximum volume of an open box with a square base made from a cardboard of area \(P^2\), let's follow these steps:

1. Define the variables: Let the side of the square base be \(x\) and the height of the box be \(h\). The area of the cardboard forms the base and four sides, so:

\[x^2+4xh=P^2\]

2. Solve for \(h\):

\[h=\frac{P^2-x^2}{4x}\]

3. The volume \(V\) of the box is given by:

\[V=x^2h=x^2\left(\frac{P^2-x^2}{4x}\right)=\frac{x(P^2-x^2)}{4}\]

4. Differentiating \(V\) with respect to \(x\) and setting the derivative equal to zero to find the critical points:

\[\frac{dV}{dx}=\frac{P^2}{4}-\frac{3x^2}{4}\]

Setting \(\frac{dV}{dx}=0\):

\[\frac{P^2}{4}=\frac{3x^2}{4}\Rightarrow x^2=\frac{P^2}{3}\Rightarrow x=\frac{P}{\sqrt{3}}\]

5. Substitute \(x\) back into the equation for \(h\):

\[h=\frac{P^2-\left(\frac{P^2}{3}\right)}{4\left(\frac{P}{\sqrt{3}}\right)}=\frac{2P}{\sqrt{3}\cdot4}=\frac{P\sqrt{3}}{6}\]

6. Now substitute \(x\) and \(h\) back into the volume equation:

\[V=\frac{\left(\frac{P}{\sqrt{3}}\right)^2\cdot\left(\frac{P\sqrt{3}}{6}\right)}{4}=\frac{P^3\cdot\sqrt{3}}{6\cdot 3\sqrt{3}}=\frac{P^3}{6\sqrt{3}}\]

Thus, the maximum volume of the box is \(\frac{P^3}{6\sqrt{3}}\).