Help with Minimum Area

[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!

 

[Wretchosoft]

You want to get your minimum as a function of one variable. Use substitution to make that happen.

 

[karisrou]

Alright cool. I've got b^2 + 128/b^2 = min

Should I take the derivative now?

 

[Dick]

You might want to make sure that's correct before you take the derivative. You've got h*b^2=32 and you want to minimize b^2+4*b*h. Are you sure that means b^2+128/b^2 should be minimized?

 

[karisrou]

But aren't I supposed to substitute h into b² + 4bh? and then solve for the critical point? Or would I just take the derivative of that? And then if that's what I'm supposed to do, do i take the derivative in relation to b or h?

 

[Dick]

But aren't I supposed to substitute h into b² + 4bh? and then solve for the critical point? Or would I just take the derivative of that? And then if that's what I'm supposed to do, do i take the derivative in relation to b or h?

You are supposed to do exactly that thing. But are you sure substituting h=32/b^2 into b^2+4hb gives you b^2+128/b^2?

 

[karisrou]

Right, so b² + 4b(32/b²)

Which simplifies to

b² + 128b/b²

Which simplifies to

b² + 128/b

 

[Dick]

Right, so b² + 4b(32/b²)

Which simplifies to

b² + 128b/b²

Which simplifies to

b² + 128/b

Right. Now take the derivative.

 

[karisrou]

so 2b - 128/b²

and then would you multiply everything by b² to clear the denominator? giving you 2b³ - 128 = 0
So b = 3rt64

So 4?

Then if b = 4

h must equal 2, per substitution into the original equation.

So then the areas of the sides are 8, and the base is 16, so the total area is 8 x 4 + 16 = 48

Is this right?

 

[Dick]

It looks good to me. But I've made mistakes before. Never hurts to double check. Seem ok to you?

 

[karisrou]

Yeah, everything looks right. Thanks very much for your help!

 

[Mark44]

Right, so b² + 4b(32/b²)

Which simplifies to

b² + 128b/b²

Which simplifies to

b² + 128/b

It would be a very good idea to work with equations, in part to help keep you focused on what you're doing.

You're trying to minimize the area as a function of b, or A(b) = b² + 4b(32/b²) = b² + 128/b

Then when you take the derivative, you get A'(b) = 2b - 128/b²

To find the critical point, you're going to set A'(b) to 0, so 2b - 128/b² = 0
You can multiply both sides of that equation by b² to get 2b³ - 128 = 0, and you can then solve that equation for b.