# Thread: Least amount of materials problem

1. ## Least amount of materials problem

I've been given a problem and would like to see if I have thought correctly when solving it. A box without lid with a volume of 4dm3 should have as small area as possible for the 5 limiting sides. the base should be square.

My solution: The base area is b2 and the area for the 4 sides are b*h so total side area is 4*b*h. For the volume 4 we get that b2*h = 4 so h = 4/b2.
The total area is A = b2+4*b*(4/b2) -> A = b2 + 16/b.

Derivating and skissing the curve I found b=2 to be a minimum so with b=2 would therefor result in the least amount of materials used. Is this the correct steps for solving this type of problem?

2. ## Re: Least amount of materials problem

Originally Posted by dipsy34
I've been given a problem and would like to see if I have thought correctly when solving it. A box without lid with a volume of 4dm3 should have as small area as possible for the 5 limiting sides. the base should be square.

My solution: The base area is b2 and the area for the 4 sides are b*h so total side area is 4*b*h. For the volume 4 we get that b2*h = 4 so h = 4/b2.
The total area is A = b2+4*b*(4/b2) -> A = b2 + 16/b.

Derivating and skissing the curve I found b=2 to be a minimum so with b=2 would therefor result in the least amount of materials used. Is this the correct steps for solving this type of problem?
For a minimum require dA/db =0 Solve this and you will indeed get b=2

3. ## Re: Least amount of materials problem

Originally Posted by dipsy34
I've been given a problem and would like to see if I have thought correctly when solving it. A box without lid with a volume of 4dm3 should have as small area as possible for the 5 limiting sides. the base should be square.

My solution: The base area is b2 and the area for the 4 sides are b*h so total side area is 4*b*h. For the volume 4 we get that b2*h = 4 so h = 4/b2.
The total area is A = b2+4*b*(4/b2) -> A = b2 + 16/b.

Derivating and skissing the curve I found b=2 to be a minimum so with b=2 would therefor result in the least amount of materials used. Is this the correct steps for solving this type of problem?
I have no idea what "skissing the curve" could mean! b= 2 is correct- but you would still need to determine h from $4/b^2$. But we can't tell whether you used the "correct steps" until you tell us what steps you took.