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MrRa [10]
3 years ago
12

Suppose we want to build a rectangular storage container with open top whose volume is $$12 cubic meters. Assume that the cost o

f 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?
Mathematics
1 answer:
Snezhnost [94]3 years ago
3 0

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

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