1. ## Parallelepiped

I'm stuck with my homework in the following task:

a rectangular parallelepiped ( b*b*h ) holds K dm^3.
Show that the total surface area of the side faces and top and bottom surface is minimal if h/b = 1

see picture in the link: https://myalbum.com/album/k1KaEgC1WieE

I must first describe a function / formula with one variable , who can help me thus on the way?
I fail to derive it by myself.

2. ## Re: Parallelepiped

Hey westerwolde.

Can you show us the total surface area? [Hint - you have six sides to consider].

3. ## Re: Parallelepiped

I think it is: 2*b + 4*h

4. ## Re: Parallelepiped

Hint - The surface area of a b x b block is b*b and for b x h it is b*h.

Take a closer look at the box and write out the terms for all sides you need to include.

5. ## Re: Parallelepiped

Originally Posted by westerwolde
I think it is: 2*b + 4*h
This is a length, not an area.

6. ## Re: Parallelepiped

I looked at the picture again, i think it should be : 4*bh + 2*bb

7. ## Re: Parallelepiped

Originally Posted by westerwolde
I looked at the picture again, i think it should be : 4*bh + 2*bb
correct ...

$S= 2(2bh+b^2)$

what method will you use to determine the minimum surface area, $S$ ?

8. ## Re: Parallelepiped

So you want to minimize S= 4bh+ 2b^2 subject to the constraint V= b^2h= K.

The simpler in concept is to solve the constraint for h= K/b^2 and replace h in the formula for S so that you have a function in the single variable b to minimize. Do you know how to do that? Set the derivative of S, with respect to b, equal to 0 and solve for b.

9. ## Re: Parallelepiped

I was going to do it as described by HallsofIvy, in another task has already succeeded but ther was given the formula.

10. ## Re: Parallelepiped

I think there is something wrong with my development, can you see what goes wrong?

h= k/b^2
s= 4bh+ 2b^2

I replace h in s

=> s= 4k/b^2 + 2b^2

ds/db=0 => -4k/b^2 + 4b=0

and when I now solve for b,I get specific outcomes ..

11. ## Re: Parallelepiped

$S = 2(2bh + b^2)$

$b^2h = k \implies \color{red}{h = \dfrac{k}{b^2}} \implies S = 2\left(2b \cdot \dfrac{k}{b^2} + b^2\right) = 2\left(\dfrac{2k}{b} + b^2\right)$

$\dfrac{dS}{db} = 2\left(-\dfrac{2k}{b^2} + 2b\right)$

$2\left(-\dfrac{2k}{b^2} + 2b\right) = 0 \implies b = \dfrac{k}{b^2}$

what else equals $\dfrac{k}{b^2}$ ?

12. ## Re: Parallelepiped

h is equal to k/b^2 ?

How do you get b= k/b^2 ?

13. ## Re: Parallelepiped

How do you get b= k/b^2 ?
Originally Posted by skeeter
$\dfrac{dS}{db} = 2\left(-\dfrac{2k}{b^2} + 2b\right)$

$\color{red}{2\left(-\dfrac{2k}{b^2} + 2b\right) = 0}$
I solved the above equation ...

14. ## Re: Parallelepiped

okay, I had solved the equation as b=^3√ k , but that's wrong.

what do you mean with ; what else equals k/b^2 ? How do I go from here?

15. ## Re: Parallelepiped

I had solved the equation as b=^3√ k , but that's wrong.
$2\left(-\dfrac{2k}{b^2}+2b\right) = 0 \implies 4\left(-\dfrac{k}{b^2}+b\right) = 0 \implies \color{red}{b = \dfrac{k}{b^2}}$

... this is the value of $b$ that minimizes the volume.

what do you mean with ; what else equals k/b^2 ? How do I go from here?
from the original equation for volume ...

$k = b^2 h \implies \color{red}{h = \dfrac{k}{b^2}}$

take a look again at your first post ... what you were supposed to show?

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