# Thread: Word Problem continued Finding minimum.

1. ## Word Problem continued Finding minimum.

I have no clue who closed the original topic or why

romsek sent me a PM trying to help but the format he used didn't work in there so maybe it will work here.

You need to make a closed box with a square base and volume of 32,000 inches cubed . The material for the top and bottom of the box costs $1 per square inch, and the materials for the sides cost 0.25 dollars per square inch, Find the dimensions of the box that minimize the cost. Originally Posted by romsek Ok the box has dimensions$s \times s \times hV= s^2 h = 32000~in^3h = \dfrac{V}{s^2}C = s^2 + 4 (0.25)s h = s^2 + sh~\

$C = s^2 + s \dfrac{V}{s^2} = s^2 + \dfrac{V}{s}$

$\dfrac{dC}{ds} = 2s -\dfrac{V}{s^2}$

Now we set $\dfrac{dC}{ds}=0$ and solve for $s$

$2s - \dfrac{V}{s^2} = 0$

$2s = \dfrac{V}{s^2}$

$s^3 = \dfrac{V}{2}$

$s = \left(\dfrac{V}{2}\right)^{1/3} = \left(16000\right)^{1/3} = 20 \sqrt[3]{2} \approx 25.198~in$

$h = \dfrac{32000}{s^2} = 800 \times 2^{2/3} \approx 1269.92~in$

you'll need to verify this is a minimum by checking the $\dfrac{d^2C}{ds^2}$ at $s=20\sqrt[3]{2}$
and ensuring it's positive.

2. ## Re: Word Problem continued Finding minimum.

Hey XxXScorpionXxX.

What are you having problem with? Understanding the solution romsek gave you or something else?

3. ## Re: Word Problem continued Finding minimum.

Originally Posted by chiro
Hey XxXScorpionXxX.

What are you having problem with? Understanding the solution romsek gave you or something else?
well i had to post his solution here so i could even see it.

He also explain that once I find the minimum I plug it into the cost function and I get the minimum cost. Looking at his solution though it seems off for a question I'm supposed to be able to solve with out a calculator. How do I find out the cubed root of 2 long hand?

4. ## Re: Word Problem continued Finding minimum.

I get a different result from romsek because I read the problem differently. Obviously

$32000 = V = s^2h \implies h = \dfrac{32000}{s^2}.$

But I read the OP as requiring the box to have a top and bottom, which changes the cost equation.

$cost\ of\ bottom = s^2 * 1 = s^2 \implies cost\ of\ top\ and\ bottom = 2s^2.$

$cost\ of\ a\ side = s * h * 0.25 = s * \dfrac{32000 * 0.25}{s^2} = \dfrac{8000}{s} \implies$

$cost\ of\ 4\ sides = \dfrac{32000}{s}.$

$total\ cost = C = 2s^2 + \dfrac{32000}{s} \implies \dfrac{dC}{ds} = 4s - \dfrac{32000}{s^2} \implies$

$\dfrac{dC}{ds} = 0 \implies 4s - \dfrac{32000}{s^2} = 0 \implies 4s^3 - 32000 = 0 \implies$

$s^3 = 8000 \implies s = 20 \implies h = \dfrac{32000}{20^2} = 80.$

5. ## Re: Word Problem continued Finding minimum.

I think thats the right answer thanks to both of you guys though.

6. ## Re: Word Problem continued Finding minimum.

Originally Posted by JeffM
I get a different result from romsek because I read the problem differently. Obviously

$32000 = V = s^2h \implies h = \dfrac{32000}{s^2}.$

But I read the OP as requiring the box to have a top and bottom, which changes the cost equation.

$cost\ of\ bottom = s^2 * 1 = s^2 \implies cost\ of\ top\ and\ bottom = 2s^2.$

$cost\ of\ a\ side = s * h * 0.25 = s * \dfrac{32000 * 0.25}{s^2} = \dfrac{8000}{s} \implies$

$cost\ of\ 4\ sides = \dfrac{32000}{s}.$

$total\ cost = C = 2s^2 + \dfrac{32000}{s} \implies \dfrac{dC}{ds} = 4s - \dfrac{32000}{s^2} \implies$

$\dfrac{dC}{ds} = 0 \implies 4s - \dfrac{32000}{s^2} = 0 \implies 4s^3 - 32000 = 0 \implies$

$s^3 = 8000 \implies s = 20 \implies h = \dfrac{32000}{20^2} = 80.$
sigh, Jeff is correct, I screwed up again.