# Thread: Help with Minimum Area

1. ## Help with Minimum Area

1) A cardboard box of 32in^3 volume with a square base and open top is to be constructed. Find the minimum area of cardboard needed.

Since it has a square base, length and width must be the same, variable b for base.

So volume could be written as: hb^2=32 and surface area as b^2+4bh=min.

Once I get here, I''m not quite sure what to do.
Any help is greatly appreciated!

2. Originally Posted by karisrou
1) A cardboard box of 32in^3 volume with a square base and open top is to be constructed. Find the minimum area of cardboard needed.

Since it has a square base, length and width must be the same, variable b for base.

So volume could be written as: hb^2=32 and surface area as b^2+4bh=min.

Once I get here, I''m not quite sure what to do.
Any help is greatly appreciated!
All you considerations and calculations are OK!

The value which should become a minimum is the area of the cardboard:

$\displaystyle a = b^2+4bh$

Use the side information(?) to eliminate one variable in a:

$\displaystyle hb^2 = 32 ~\implies~h=\dfrac{32}{b^2}$ . Plug in this term into a:

$\displaystyle a(b)= b^2+4b\cdot \dfrac{32}{b^2} = b^2 + \dfrac{128}b$

Solve the equation a'(b) = 0 for b:

$\displaystyle 2b - \dfrac{128}{b^2} = 0~\implies~2b^3 = 128~\implies~ b = 4$

Therefore h = 2