Let #{(l" ... the length of the base"),(w" ... the width of the base"),(h" ... the height of the box"):}#
Since the length is twice the width, we have
#l=2w#
Since the volume #V# of the box is 10 #m^3#, we have
#V=l cdot w cdot h=2w cdot w cdot h=2w^2h=10#
By dividing by #2w^2#,
#h=10/{2w^2}=5/w^2#
Since the area of the base is
#l cdot w=2w cdot w=2w^2#,
and the combined area of the four walls is
#(2l+2w)cdot h=(2cdot2w+2w)cdot 5/w^2=30/w#,
the total cost #C(w)# of the material can be expressed by
#C(w)=5 cdot 2w^2+3 cdot 30/w=10w^2+90/w#
Now, we want to minimize #C(w)# on #(0, infty)#.
Let us find its critical numbers.
#C'(w)=20w-90/w^2={10(2w^3-9)}/w^2=0 => w=(4.5)^{1/3}#
Since #C'(w)<0# on #(0,(4.5)^{1/3})# and #C'(w)>0# on #((4.5)^{1/3}.infty)#, the critical number yields the minimum cost.
Hence, the minimum total cost is
#C((4.5)^{1/3})=10(4.5)^{2/3}+90/(4.5)^{1/3}#
I hope that this was helpful.
