Retire zero la bour li dan trou fess to pu ggnr bon
5c + 3 - 2c + 5
5c - 2c + 3 + 5
3c + 8 ~~~
X^2 - 3x + 5
2^2 - 3(2) + 5
4 - 6 + 5
3
Answer:265.25
Step-by-step explanation:
Given
Volume of box
height is 3 times the width
let height, length and breadth be H, L & B


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

Cost of side walls $ 
Cost of base $ 
Cost of side walls
Cost of base 
Total cost 


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




L=2.19 m
C=$ 265.25