# Thread: strange small story problem-closed box area

1. ## strange small story problem-closed box area

If a closed rectangular box should have volume 27ccm, find the dimensions
that will make the surface area of closed box minimum possible.

Nothing else is said about box...

2. Originally Posted by fiksi
If a closed rectangular box should have volume 27ccm, find the dimensions
that will make the surface area of closed box minimum possible.

Nothing else is said about box...
Hi

Let a, b and c the length of the box

The volume of the box is $V_0 = 27 = a b c$

The surface area is $A(a,b,c) = 2ab + 2bc + 2ac = 2\left(ab + \frac{V_0}{a} + \frac{V_0}{b}\right)$

The extremum of A is obtained when $\frac{\partial A}{\partial a} = 0$ and $\frac{\partial A}{\partial b} = 0$

Spoiler:

$b - \frac{V_0}{a^2} = 0$

$a - \frac{V_0}{b^2} = 0$

$V_0 = a^2b = ab^2 = abc \implies a = b = c$

The box must be a cube

3. Originally Posted by running-gag
Hi

Let a, b and c the length of the box

The volume of the box is $V_0 = 27 = a b c$

The surface area is $A(a,b,c) = 2ab + 2bc + 2ac = 2\left(ab + \frac{V_0}{a} + \frac{V_0}{b}\right)$

The extremum of A is obtained when $\frac{\partial A}{\partial a} = 0$ and $\frac{\partial A}{\partial b} = 0$

Spoiler:

$b - \frac{V_0}{a^2} = 0$

$a - \frac{V_0}{b^2} = 0$

$V_0 = a^2b = ab^2 = abc \implies a = b = c$

The box must be a cube
So i don't even need 27ccm data, aye? I thought in the start that it should be 3x3x3, for the area to be min possible.

4. I have considered the generic case but you need the data to compute a,b,c which are 3 cm

5. Originally Posted by running-gag
I have considered the generic case but you need the data to compute a,b,c which are 3 cm
Of course I need it to know a,b,c. Your generic solution was nice as it applies all-round. I meant, I don't need any data about the box itself(that's what i meant), as it must be a cube. From 27ccm is 3,3,3.

thx again. I guess I can apply this to all similar problems

6. I prefer to use $V_0$, perform all the calculations and only at the end substitute the value because :
- it can be applied to any value of $V_0$
- it allows to find possible mistakes : when I find $b - \frac{V_0}{a^2} = 0$, I know that this is a homogeneous equation since b is expressed in cm, V0 in cm^3 and a² in cm. If I use the value of V0, I am losing this possibility.