Ok the question is a rectangular storage container with an open top is to have a volume of 10 m^3. The length of the base is twice the width. Material for the sides cost 6$ per square meter and the cost of material for the base is 10$ per square meter. Find the cost of materials for the cheapest such container
I think i got the question I'm just not really 100% sure, any help would be appreciated.
Ok so i set up my volume of the container to be
V = 2W^2 * H
H = V / (2W^2)
H = 10 / (2W^2)
H = 5 / W^2
I set up my equation for surface area as follows:
P(w) = 10(2W^2) + 6 [ (2)(2W)(H) + 2(W)(H)]
= 20 W^2 + 36 (W)(H)
= 20 W^2 + 36(W)(5 / (W^2))
= 20 W^2 + 180 / W
P'(W) = 40 W - 180 / (W^2)
letting the above equation = 0 gave:
W = (cubedroot)(9/2)
so subbing into the original gave me:
20((cubedroot)(9/2)) ^2 + 180 / (cubedroot(9/2))
= 163.54$
Just kinda curious if i am going about this right.
Also i'm a litte confused on how to find the domain of the function in these type of problems, the way i looked at it was the W can approach zero but never equal 0 and it can approach infinite, but i really am not sure how to go about finding the domain.
I think i got the question I'm just not really 100% sure, any help would be appreciated.
Ok so i set up my volume of the container to be
V = 2W^2 * H
H = V / (2W^2)
H = 10 / (2W^2)
H = 5 / W^2
I set up my equation for surface area as follows:
P(w) = 10(2W^2) + 6 [ (2)(2W)(H) + 2(W)(H)]
= 20 W^2 + 36 (W)(H)
= 20 W^2 + 36(W)(5 / (W^2))
= 20 W^2 + 180 / W
P'(W) = 40 W - 180 / (W^2)
letting the above equation = 0 gave:
W = (cubedroot)(9/2)
so subbing into the original gave me:
20((cubedroot)(9/2)) ^2 + 180 / (cubedroot(9/2))
= 163.54$
Just kinda curious if i am going about this right.
Also i'm a litte confused on how to find the domain of the function in these type of problems, the way i looked at it was the W can approach zero but never equal 0 and it can approach infinite, but i really am not sure how to go about finding the domain.