Question

A rectangular sheet of tin 45cm by 24cm is to be made into a box without top,by cutting off square from each corner and folding up the flaps.What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

Answer 1

Let the side of the square to be cut off be \(xcm\).Then,the height of the box is \(x\),the

length is \(45−2x\),and the breadth is \(24−2x\).

Therefore,the volume \(V(x)\)of the box is given by,

\(V(x)=x(45-2x)(24-2x)\)

\(=x(1080-90x-48xx+x^{2})\)

\(=4x^{3}-138x^{2}+1080x\)

\(∴V'(x)=12x^{2}-276x+1080\)

\(=12(x^{2}-23x+90)\)

\(=12(x-18)(x-5)\)

\(V''(x)=24x-276=12(2x-23)\)

Now,\(V'(x)=0⇒x=18 \space and\space  x=5\)

It is not possible to cut off a square of side 18cm from each corner of the rectangular

sheet.Thus,x cannot be equal to 18.

\(so\space  x=5\)

Now,\(V''(5)=12(10-23)=12(-13)=-156<0\)

∴By second derivative test,\(x=5\) is the point of maxima.

Hence,the side of the square to be cut off to make the volume of the box maximum

possible is \(5cm.\)