LetthedensionsofthebaseoftheopenboxbexcmandxcmandheightbeycmTheareaofthematerialoftheopenbox⇒(Areaofthebase)+4(areaofverticalface)⇒192=(x×x)+4(xy)⇒192=x2+4xy⇒192−x24x=yLetvbethevolumeofthebox,V=basearea×height=x2y=x2(192−x24x)∴V=x(192−x24)=192x−x34dvdx=192−3×x24dvdx=0x2=64since,xislengthoftheside,itcannotbe0−ve∴x≠−8and,d2vdx2=−32x(d2vdx2)x=8=−32(8)V=ismaximumwhenx=8cmwhenx=8,theny=192−824(8)=4cm∴Dimensionsare8cm&4cmlestpossiblevolumeofbox(cube)V=x2y=(8)2(4)=256sqcm