Suppose the base-length, base-width and height of the box
are #l,w, and h,# resp.
#:." The volume v of the box"=lwh=200.................."[Given]."#
But, as, #w=2l, :. v=l(2l)h=200, or, 2l^2h=200.......(star).#
Now, to minimize the material cost of the box, the surface
area of the box must be minimum.
Knowing that the box is open at the top, its surface area #s# is,
#s=2hl+2wh+wl=2hl+2(2l)h+(2l)l,.......[because, w=2l],#
#=2hl+4hl+2l^2=6hl+2l^2,#
#=6(200/(2l^2))l+2l^2.............................................[because, (star)],#
#=2l^2+600/l,# which is a function of #l,# so let us write,
#s(l)=2l^2+600/l.#
For #s_(min), s'(l)=0. and, s''(l) gt 0.#
#s(l)=2l^2+600/l rArr s'(l)=2(2l)+600(-1/l^2),#
#=4l-600/l^2, and,#
#s''(l)=4-600(-2l^(-2-1))=4+1200/l^3.#
#:. s'(l)=0 rArr 4l-600/l^2=0 rArr l^3=150, or, l=150^(1/3).#
Also, for #s=150^(1/3), s''(l)=4+1200/150 gt 0.#
Hence, #l=150^(1/3)# gives #s_(min).#
Thus, the reqd. dimension of the box, in meters, are,
#l=150^(1/3), w=2(150)^(1/3) and h=100/((150)^(2/3)).#
Enjoy Maths.!
