# Math Help - calc optimization help

1. ## calc optimization help

Alright! My final question on here till next semester lol. Doing some last minute review for my final tonight and came across this problem having a little trouble with it.

A rectanglular storage container with an open top is to have a volume of 10 cubic-meters. Material for the sides costs $6 per square-meter. Material for the base costs$10 per square-meter. Find the cost of materials for the cheapest container.

Thanks

{edit} ah yes sorry. Length of its base is twice the width.

2. I would say that you are missing a piece of information. Are you sure that you typed the entire problem?

3. Originally Posted by UMStudent
Alright! My final question on here till next semester lol. Doing some last minute review for my final tonight and came across this problem having a little trouble with it.

A rectanglular storage container with an open top is to have a volume of 10 cubic-meters. Material for the sides costs $6 per square-meter. Material for the base costs$10 per square-meter. Find the cost of materials for the cheapest container.

Thanks

{edit} ah yes sorry. Length of its base is twice the width.
L = 2x
W = x
H = h

So Volume = 2x^2h = 10

Exterior Area = 2(L + B)H + (L x B) [Because you do not have a lid]
= 2(2x + x)h + (2x . x)
= 4x^2h + 2x^2
= 6xh + 2x^2

But we know that h = (10)/(2x^2)

= 6x((10)/(2x^2)) + 2x^2
= 60x/2x^2 + 2x^2
= 30/x + 2x^2

dA/dx = -30/x^2 + 4x = 0
-30 = -4x^3
7,5 = x^3
x = 1,957433821

Now you can calculate the rest.

And this is your bottom lid:2x^2

4. Originally Posted by UMStudent
Alright! My final question on here till next semester lol. Doing some last minute review for my final tonight and came across this problem having a little trouble with it.

A rectanglular storage container with an open top is to have a volume of 10 cubic-meters. Material for the sides costs $6 per square-meter. Material for the base costs$10 per square-meter. Find the cost of materials for the cheapest container.

Thanks

{edit} ah yes sorry. Length of its base is twice the width.
Let the width be w metres, then the length is 2w metres, and as the volume
is 10 cubic metres the height is: 5/w^2 metres.

Total area of the sides is:

As = 2*w*5/w^2 + 4*w*5/w^2

... = 30/w

The area of the base is:

Ab = 2*w^2

So the cost is:

C = 180/w + 20*w^2

Now you need to find the value of w that minimises C, and that minimum

RonL

5. Originally Posted by CaptainBlack
Let the width be w metres, then the length is 2w metres, and as the volume
is 10 cubic metres the height is: 5/w^2 metres.

Total area of the sides is:

As = 2*w*5/w^2 + 4*w*5/w^2

... = 30/w

The area of the base is:

Ab = 2*w^2

So the cost is:

C = 180/w + 20*w^2

Now you need to find the value of w that minimises C, and that minimum

RonL
Is my solution wrong Captain?

6. Ok
let x, y, z be the dimensions of the box.
Since volum should be 10, (1) xyz=10
Length of its base is twice the width: (2) y=2z
(3) Cost=6(2xy+2xz)+10yz=12xy+12xz+10yz
You need a function cost having just one variable, so let's use (1) and (2) to simplify (3)
(2) into (1) gives 2xz^2=10 or x=5/z^2
Using this relation and (2) into (3) gives:
cost=20z^2+180/z
Solve the equation Cost'=0 to find z which gives minimum cost.
40z-180/z^2=0
40z^3-180=0
z^3=4.5
z=1.65
x=1.84
y=3.3

7. Originally Posted by janvdl
Is my solution wrong Captain?
Your solution is wrong since you did not minimize the cost, but the area.

8. Originally Posted by janvdl
Is my solution wrong Captain?
I havn't checked it, you posted while I was writting mine, so I did not see
yours. If one is wrong its just as likely to be mine.

RonL

9. Oh of course! He pays diff prices for diff parts of the box.