Question

Suppose we want to build a rectangular storage container with open top whose volume is $$12 cubic meters. Assume that the cost of materials for the base is$$12 dollars per square meter, and the cost of materials for the sides is $$8 dollars per square meter. The height of the box is three times the width of the base. What’s the least amount of money we can spend to build such a container?

Answer 1

Answer:265.25

Step-by-step explanation:

Given

Volume of box$=12 m^3$

height is 3 times the width

let height, length and breadth be H, L & B

$H=3\times B$

$V=12=L\times B\times H$

4=B^2\times L[/tex]

$L=\frac{4}{B^2}$

Cost of side walls $ $8/m^2$

Cost of base $ $12/m^2$

Cost of side walls$=(2LH+2BH)\cdot 8$

Cost of base $=12LB$

Total cost $C=16LH+16BH+12LB$

$C=48LB+48B^2+12LB$

$C=\frac{192}{B}+48B^2+\frac{48}{B}$

differentiate C w.r.t B to get minimum cost

$\frac{\mathrm{d} C}{\mathrm{d} B}=-\frac{192}{B^2}+2\times 48 B-\frac{48}{B^2}$

$B^3=\frac{240}{2\times 48}$

$B=1.35 m$

$H=3\times 1.35=4.07 m$

L=2.19 m

C=$ 265.25