In exercise 23, suppose that the piece of metal has length twice the width and 4-in squares are cut from the corners. if the volume of the box is 1536in^3 what were the original dimensions of the piece of metal?

L=2w

l x w=4 in²

V=lx wx H=1536in^3, so H = 1536 / lxw=384

and l x (l/2) xH = 1536

l²/2 x384 = 1536, implies I=sqrt (1536/192) so I = 2.82

and when l = 2.82=2w we can get w=I/2=2.82/2=1.41

finally the original dimensions of the piece of metal were

I=2.82, w=1.41 and H = 384 (the uniti is inch)