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please help with calculus optimization problem.?
a box with a square base and a closed top has a volume of 3000 cubic feet. the top costs $4 sq. ft, the bottom costs $8 sq. ft, and the sides cost $2 sq ft. Find the dimensions of the box that will minimize its total cost of construction. Im not so sure on how to do this and my final exam is tomorrow.
1 Answer
- 1 decade agoFavorite Answer
let the sides of square be a ft. (breath= a ft) and height of box be h ft.
then total area of bottom=a^2
cost of bottom=8a^2
area of sides= 2ah
cost of sides= 2*(2ah)= 4ah
area of top= a^2
cost of top= 4a^2
total cost= (4a^2)+(8a^2)+(4ah)
=(12a^2)+4ah
now it is given that total volume=3000 cubic ft
therefore, (area of base*height)=3000
(a^2)*h=3000
use h=[(3000)/(a^2)]
expression for cost becomes
12(a^2)+{4a*(3000/(a^2))}
=12(a^2)+12000/a
simply differentiate and get the answer