Answer:265.25
Step-by-step explanation:
Given
Volume of box[tex]=12 m^3[/tex]
height is 3 times the width
let height, length and breadth be H, L & B
[tex]H=3\times B[/tex]
[tex]V=12=L\times B\times H[/tex]
4=B^2\times L[/tex]
[tex]L=\frac{4}{B^2}[/tex]
Cost of side walls $ [tex]8/m^2[/tex]
Cost of base $ [tex]12/m^2[/tex]
Cost of side walls[tex]=(2LH+2BH)\cdot 8[/tex]
Cost of base [tex]=12LB[/tex]
Total cost [tex]C=16LH+16BH+12LB[/tex]
[tex]C=48LB+48B^2+12LB[/tex]
[tex]C=\frac{192}{B}+48B^2+\frac{48}{B}[/tex]
differentiate C w.r.t B to get minimum cost
[tex]\frac{\mathrm{d} C}{\mathrm{d} B}=-\frac{192}{B^2}+2\times 48 B-\frac{48}{B^2}[/tex]
[tex]B^3=\frac{240}{2\times 48}[/tex]
[tex]B=1.35 m[/tex]
[tex]H=3\times 1.35=4.07 m[/tex]
L=2.19 m
C=$ 265.25
