Question

A rectangular storage container with an open top is to have a volume of 8 m^3. The length of its base is twice the width. Material for the base costs 10 dollars per square meter. Material for the sides costs 4 dollars per square meter. Find the cost of materials for the cheapest such container.

Let ell denote the length of the base, w the width of the base, and let h denote the height of the container. First, write an equation for the cost of the materials in terms of ell,w, and h.
Cost =

Answer 1

Now use the information given to solve for ell in terms of w, and to solve for h in terms of w. Using these, we may express the cost as a function of w alone:
C(w) =

Answer 2

Using this, we find the width of the cheapest such container is:

and the cost of materials for the cheapest such container is:

Answer 3

L = Length of container
W = Width of container
H = Height of container

There. You're half way there! What's next?

Answer 4

hmmmm

Answer 5

V= L * W * h

Answer 6

Excellent. V = L*W*H = 8 m^3

Now this part: "The length of its base is twice the width"

Answer 7

2w*w*h?

Answer 8

h=(8m^3)/(2w^2)

Answer 9

Well, I would be happier if you:

1) Explicitly showed L = 2W, and
2) Substituted without changing varible names: V = 2W*W*H = 8 m^3
3) And even happier if you then simplified: V = 2H*W^2 = 8 m^3

Answer 10

I see, i have no idea what the next step is

Answer 11

Now, we're down to the cost of the container.

Area of Base?
Lateral Surface Area?

Answer 12

how do i find the cost of the container?

Answer 13

Answer the two questions I just posed. Express both areas in terms of W.

Answer 14

i don't know what i need to don :(

Answer 15

Yes you do.

Base = L*W
Lateral Surface Area = 2*L*H + 2*W*H

What about that do you know know how to do?

Answer 16

oh i see

Answer 17

Now, write them in terms of W.

Answer 18

L = 2W
H=(8m^3)/(2W^2)

Answer 19

W=Base/H
W=-L

Answer 20

Can you just give me the answer then i can just look through it please?

Answer 21

then i can find out how i can do this problem

Answer 22

No. You are so close. Why are you giving up?

Base = L*W

Given L = 2W, we have

Base = (2W)*W = 2W^2

Are you SURE that's hard?

You do the Lateral Surface Area.

Lateral Surface Area = 2*L*H + 2*W*H

Given L = 2W, and H=(8m^3)/(2W^2)

Lateral Surface Area = 2*(_____)*(_____) + 2*W*(_____)

Answer 23

=2(2w)*((8m^3)/(2W^2))+2*W((8m^3)/(2W^2))

Answer 24

=16m^3/w+8m^3/2W

Answer 25

Excellent. One more thing. That's just the area. We still don't have the Cost.

Here's the expression to minimize: 10*(Base Area) + 4*(Lateral Surface Area)

Answer 26

minimize=20w^2+(96m^3)/w

Answer 27

W=(-96m^3)/(2w^2)

Answer 28

96?

Answer 29

=-48m^3/w^2

Answer 30

64+32

Answer 31

10(2W^2)+4(16m^3/w+8m^3/w)

Answer 32

20W^2+(64m^3/w+32m^3/w)

Answer 33

Lateral Surface Area
16m^3/W+8m^3/2W = 32m^3/2W+8m^3/2W = 40m^3/2W = 20m^3/W

Answer 34

it is not 2W, it is just W

Answer 35

i wrote it wrong

Answer 36

24m^3/w

Answer 37

Given L = 2W, and H=(8m^3)/(2W^2)

Lateral Surface Area = 16/W + 8/W = 24/W

Okay, I'll bite. Be neater and more careful.

Answer 38

now what is the next step i need to do?

Answer 39

You already had it: minimize=20w^2+(96m^3)/w

Answer 40

do i add them together?

Answer 41

Do you want the total cost? You can build a container with only a bottom, can you? Only sides? Yes, add them.