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 $\displaystyle V_0 = 27 = a b c$
The surface area is $\displaystyle A(a,b,c) = 2ab + 2bc + 2ac = 2\left(ab + \frac{V_0}{a} + \frac{V_0}{b}\right)$
The extremum of A is obtained when $\displaystyle \frac{\partial A}{\partial a} = 0$ and $\displaystyle \frac{\partial A}{\partial b} = 0$
Spoiler:
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
I prefer to use $\displaystyle V_0$, perform all the calculations and only at the end substitute the value because :
- it can be applied to any value of $\displaystyle V_0$
- it allows to find possible mistakes : when I find $\displaystyle b - \frac{V_0}{a^2} = 0$, I know that this is a homogeneous equation since b is expressed in cm, V0 in cm^3 and aČ in cm. If I use the value of V0, I am losing this possibility.