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A square piece of tin of side $18 \mathrm{~cm}$ is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.
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Verified Answer
Let each side of the square to be cut off be $\mathrm{x} \mathrm{cm}$.
$\therefore$ for the box length $=18-2 \mathrm{x}:$ breadth $=18-2 \mathrm{x}$ and height $=\mathrm{x}$
$\therefore$ Volume $=\mathrm{x}(18-2 \mathrm{x})^2 ; \frac{\mathrm{dv}}{\mathrm{dx}}=(18-2 \mathrm{x})(18-6 \mathrm{x})$
For maxima and minima,
$\frac{\mathrm{dv}}{\mathrm{dx}}=0 \Rightarrow(18-2 \mathrm{x})(18-6 \mathrm{x})=0 \Rightarrow \mathrm{x}=3,9$
But $x=9 \mathrm{~cm}$ is not possible
Also $\frac{\mathrm{d}^2 \mathrm{v}}{\mathrm{dx}^2}=(18-2 \mathrm{x})(-6)+(18-6 \mathrm{x})(-2)$
At $x=3, \frac{\mathrm{d}^2 \mathrm{v}}{\mathrm{dx}^2}=(18-6)(-6)+(18-18)(-2)=-72 < 0 \quad \therefore \mathrm{x}=3$ attains for $\max ^{\mathrm{m}}$ volume
i.e. Square of side $=3 \mathrm{~cm}$ is cut from each corner.
$\therefore$ for the box length $=18-2 \mathrm{x}:$ breadth $=18-2 \mathrm{x}$ and height $=\mathrm{x}$
$\therefore$ Volume $=\mathrm{x}(18-2 \mathrm{x})^2 ; \frac{\mathrm{dv}}{\mathrm{dx}}=(18-2 \mathrm{x})(18-6 \mathrm{x})$
For maxima and minima,
$\frac{\mathrm{dv}}{\mathrm{dx}}=0 \Rightarrow(18-2 \mathrm{x})(18-6 \mathrm{x})=0 \Rightarrow \mathrm{x}=3,9$
But $x=9 \mathrm{~cm}$ is not possible
Also $\frac{\mathrm{d}^2 \mathrm{v}}{\mathrm{dx}^2}=(18-2 \mathrm{x})(-6)+(18-6 \mathrm{x})(-2)$
At $x=3, \frac{\mathrm{d}^2 \mathrm{v}}{\mathrm{dx}^2}=(18-6)(-6)+(18-18)(-2)=-72 < 0 \quad \therefore \mathrm{x}=3$ attains for $\max ^{\mathrm{m}}$ volume
i.e. Square of side $=3 \mathrm{~cm}$ is cut from each corner.
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