A square piece of tin of side 18 cm is to 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?
Solution and Explanation
Let the side of the square to be cut off be \( x \space cm\). Then, the length and the breadth of the box will be \((18 − 2x) \)cm each and the height of the box is \(x cm.\) Therefore, the volume\( V(x) \)of the box is given by,
\(V(x) = x(18 − 2x) ^{2}\)
\(v'(x)=(18-2x)^{2}-4x(18-2x)\)
\(=(18-2x)[18-2x-4x]\)
\(=(18-2x)(18-6x)\)
\(=6\times 2(9-x)(3-x)\)
Now,\(v'(x)=0=x=9 \space or\space x=3\)
If \(x = 9\), then the length and the breadth will become 0.
\(x ≠ 9.\)
\(x = 3.\)
\(v''(3)=-24(6-3)-72<0\)
Now,
By second derivative test,\( x = 3\) is the point of maxima of \(V\).
Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.
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