Let length of box =# x#, width of the box = #y# and the height of the box = #a# . [Note that the height of the box is also equal to length of the square cut out at the corners of the flat sheet from which it is constructed.]
So volume =#xya# . Considering the length of the box, it is seen that #[[16-x]]/2=a..........#[1]## and that #[[10 - y]]/2#=a..........#[2]#
Equating #[1]# and #[2] , # #[[16-x]]/2 #= #[[10-y]]/2#, and solving for #y# yields #y=[x-6]#........#[3]#
So volume of box #V# =#x[x-6]a# , but from .....#[1]# #a=[[16-x]]/2#
Thus volume of box in terms of #x#, #V =x[x-6][[16-x]]/2#,
=#[x^2-6x][8-x/2]# = [#8x^2- x^3/2-48x+6x^2/2##........#[4]#
Differentiating .....#[4]# wrt#x#, #[dV]/dx = 16x-3x^2/2-48+12x/2# = #0# for max/ min . Solving this expression, #[x-12][x-8/3]= 0# ,ie, #x= 12, x= 8/3#. Noting that the second derivative of #A# wrt#x# = #22-3x#, when #x= 12#, the second derivative is negative[ =#-14#] which indicates that this value of #x# ill maximise the volume of the box.
When #x= 12#, [from #[1]]# #a=[16-12]/2# = #2#. And from ...#[2]#, #[10-y]/2= a#, which when evaluated for #a=2# gives a value of #y# =#6#.
#V=[2][6][12]=144.# Hope this was helpful.
