# optimization

• November 1st 2012, 12:30 PM
pnfuller
optimization
a rectangular storage container with an open top is to have a volume of 10 cubic meters. the length of its base is twice the width. material for the base costs $10 per square meter. material for the sides costs$6 per square meter. find the cost of materials for the cheapest such container.

how do you find the length, width and height for volume?
• November 1st 2012, 12:44 PM
MarkFL
Re: optimization
What is the area of the rectangular base in terms of the width?
• November 1st 2012, 12:47 PM
pnfuller
Re: optimization
the area would be 2w^2 but i need volume with height in it how do i do that?
• November 1st 2012, 12:51 PM
MarkFL
Re: optimization
Good. Now we know the volume in cubic meters is:

$V=2w^2h=10$

So, we have:

$h=\frac{5}{w^2}$

Now, what is the combined surface area of the 4 sides?
• November 1st 2012, 01:04 PM
pnfuller
Re: optimization
20/w^2? is that right? im not really understanding that part
• November 1st 2012, 01:07 PM
MarkFL
Re: optimization
The total surface area of the sides is:

$2(2wh)+2(wh)=6wh$

Do you see how I found this?

Now, substitute for $h$ to get the area of the sides as a function of $w$.
• November 1st 2012, 01:18 PM
pnfuller
Re: optimization
but the box is suppose to have an open top so should i do it with 5 sides?
• November 1st 2012, 01:24 PM
MarkFL
Re: optimization
Yes, there will be the base, whose area you have already found is $A_b=2w^2$ and the 4 sides. Now, you should have the area of the sides is:

$A_s=\frac{30}{w}$

Both of these are in square meters. So, now you want to construct the cost function. Take the cost of the material per square meter times the area in square meters to get the cost, then sum these together to get the total cost in dollars:

$C(w)=10A_b+6A_s$

Now, minimize this function.
• November 1st 2012, 01:38 PM
pnfuller
Re: optimization
i got 3.5569 as the minimum value but where do i plug that back into to get the cost of the materials?
Quote:

Originally Posted by MarkFL2
Yes, there will be the base, whose area you have already found is $A_b=2w^2$ and the 4 sides. Now, you should have the area of the sides is:

$A_s=\frac{30}{w}$

Both of these are in square meters. So, now you want to construct the cost function. Take the cost of the material per square meter times the area in square meters to get the cost, then sum these together to get the total cost in dollars:

$C(w)=10A_b+6A_s$

Now, minimize this function.

• November 1st 2012, 03:15 PM
MarkFL
Re: optimization
You should have:

$C(w)=20(w^2+9w^{-1})$ and so:

$C'(w)=20(2w-9w^{-2})=\frac{20(2w^3-9)}{w^2}=0$

Since we want $0, we take the critical value from the numerator:

$w=\left(\frac{9}{2} \right)^{\frac{1}{3}$

Using the first derivative test, we find this critical number is at a minimum. Hence:

$C_{\text{min}}=C\left(\left(\frac{9}{2} \right)^{\frac{1}{3} \right)$

Now evaluate the cost function at this critical value to find the minimum cost in dollars.

Please show how you arrived at 3.5569 as your critical number, so we can address the error in the method/computation.
• November 1st 2012, 03:34 PM
pnfuller
Re: optimization
i thought the c(w)=10(2w^2)+6(30/w)

how did you get a different c(w) ?
• November 1st 2012, 03:42 PM
MarkFL
Re: optimization
My apologies, a post made in haste. I have edited it.
• November 1st 2012, 03:53 PM
pnfuller
Re: optimization
would the cheapest cost then be \$163.54?
Quote:

Originally Posted by MarkFL2
My apologies, a post made in haste. I have edited it.

• November 1st 2012, 03:55 PM
MarkFL
Re: optimization
Yes, that's what I get too.
• November 2nd 2012, 01:42 AM
rachit
Re: optimization
Yes this is a exactly easy way for solve this problem.