If an open box with a square base is to be made of a given quantity of cardboard of area $c^{2},$ then show that the maximum volume of the box is $\frac{c^{3}}{6 \sqrt{3}}$ cu units.

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Solution

Let the length of side of the square base of open box be x units and its height be y units.
$∴$ Area of the metal used $= x^{2} + 4 x y \Rightarrow x^{2} + 4 x y = c^{2} \Rightarrow y = \frac{c^{2} - x^{2}}{4 x}$
Now, volume of the box $(V) = x^{2} y \Rightarrow V = x^{2} . (\frac{c^{2} - x^{2}}{4 x}) = \frac{1}{4} x (c^{2} - x^{2}) = \frac{1}{4} (c^{2} x - x^{3})$

On differentiating both sides w.r.t., x , we get
$\frac{d V}{d x} = \frac{1}{4} (c^{2} - 3 x^{2}) N o w, \frac{d V}{d x} = 0 \Rightarrow c^{2} = 3 x^{2} \Rightarrow x^{2} = \frac{c^{2}}{3} \Rightarrow x = \frac{c}{\sqrt{3}}$
Again, differentiating Eq. (ii) w.r.t, x, we get
$\frac{d^{2} V}{d x^{2}} = \frac{1}{4} (- 6 x) = \frac{- 3}{2} x < 0 ∴ {(\frac{d^{2} v}{d x^{2}})}_{a t x = \frac{c}{\sqrt{3}}} = - \frac{3}{2} . (\frac{c}{\sqrt{3}}) < 0$
Thus, we see that volume (v) is maximum at $x = \frac{c}{\sqrt{3}} .$
$∴$ Maximum volume of the box, $(V_{x = \frac{c}{\sqrt{3}}}) = \frac{1}{4} (c^{2} . \frac{c}{\sqrt{3}} - \frac{c^{3}}{3 \sqrt{3}}) = \frac{1}{4} . \frac{(3 c^{2} - c^{3})}{3 \sqrt{3}} = \frac{1}{4} . \frac{2 c^{3}}{3 \sqrt{3}} = \frac{c^{3}}{6 \sqrt{3}}$

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If an open box with a square base is to be made of a given quantity of cardboard of area c2, then show that the maximum volume of the box is c36√3 cu units.

If an open box with a square base is to be made of a given quantity of cardboard of area $c^{2},$ then show that the maximum volume of the box is $\frac{c^{3}}{6 \sqrt{3}}$ cu units.