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)