# Question about the minimum area

1. Apr 27, 2009

### Lord Dark

1. The problem statement, all variables and given/known data
hi guys, how are you all,,
container has a square base and the volume of it V= 1000 , find the minimum area of the surface that the container can have ...

2. Relevant equations
A=x^2 , V=x^2 * h

3. The attempt at a solution
i started by saying: 1000=x^2 * h then i can get h =1000/x^2 and A=x^2 then i differentiate 0=2x*h+x^2*h` ... i get at the end (2000/x)-(2000/x)=0 and this doesn't make any sense ,,i tried another way ,, i assumed it should be a cube for the minimum area so i get 6x^2=1000 then i get (x=13) is the answer right?? ,, is the question even right ?? anyone knows how it should be at least ??

Last edited: Apr 27, 2009
2. Apr 27, 2009

### DorianG

By minimum area, do you mean the sum of the area of the six surfaces? Are there six, is the top included?
Often these questions are asked to minimise material for packaging to hold a certain volume which I'd say this question is about. Not sure though...

3. Apr 27, 2009

### Lord Dark

i really don't know ,, he didn't remember ,, although, if it was the minimum area of the surface of the container ,, it should be cube right ?? so can i apply this equation ?
(6x^2=1000) ??

4. Apr 27, 2009

### Lord Dark

i think i got it ,, can anyone check my answer now ,,

1000=x^2*y , y=1000/x^2 , surface area = 2x^2+4x*(1000/x^2) then differentiate i get 4x-(4000/x^2)=0 then i get x=10 and y=10 so it's cube alright :D