A rectangular storage container with an open top is to have a volume of 10 m3. The length of its base is twice the width. Material for the base costs $9 per m2. Material for the sides costs $150 per m2. Find the dimensions of the container which will minimize cost and the minimum cost. I need the minimum cost , I found the dimensions to length=10

Width=5 height=0.2. Thanks so much in advance!

L = 2 w

V = L w h = 2 w^2 h = 10
so w^2 h = 5

base area = L w =2 w^2
base cost = 18 w^2

ends area = 2 w h
ends cost = 300 w h

sides area = 2 L h = 4 w h
sides cost = 600 w h

total cost = 900 w h + 18 w^2
but h = 5/w^2

c = 900 w (5/w^2) + 18 w^2
c = 4500/w + 18 w^2
dc/dw = -4500/w^2 + 36 w
= 0 for min
36 w^3 = 4500
w = 5 sure enough

$4500 comes up as incorrect :(

Where did you get 4500 from ?

Damon correctly found w = 5

and his cost equation was
c = 4500/w + 18 w^2
= 4500/5 + 18(25) = 1350

So sorry for that!!

To find the dimensions of the container that will minimize the cost, we need to determine the cost function and then minimize it using calculus.

Let's start by defining the variables:
- Length of the base: L
- Width of the base: W
- Height of the container: H

Given that the volume of the container is 10 m^3, we have:

L × W × H = 10

Also, the length of the base is twice the width, so we can write:

L = 2W

Now, let's determine the cost function. The cost of the base is given by:

Cost_base = Area_base × Cost_per_m2 = (L × W) × 9

The cost of the four sides is given by:

Cost_sides = Area_sides × Cost_per_m2 = 2LH + 2WH × 150

The total cost function is the sum of the base cost and the side cost:

Cost = Cost_base + Cost_sides = (L × W) × 9 + (2LH + 2WH) × 150

Using the volume constraint, we can express the cost function in terms of a single variable. Substituting L = 2W and H = 10/(2W^2), we have:

Cost = (2W × W) × 9 + (2(2W)(10/(2W^2)) + 2W(10/(2W^2))) × 150
= 18W^2 + (40/W + 10/W) × 150
= 18W^2 + (50/W) × 150

To find the dimensions that minimize the cost, we need to find the critical points of the cost function. The critical points occur where the derivative of the cost function with respect to W is equal to zero.

Taking the derivative of the cost function with respect to W, we get:

dCost/dW = 36W - (50/W^2) × 150

Setting the derivative equal to zero and solving for W:

36W - (50/W^2) × 150 = 0
36W^3 = 50 × 150
W^3 = (50 × 150)/36
W^3 = 625/3
W ≈ 7.34

Plugging this value of W back into the L = 2W equation, we have:

L = 2 × 7.34 ≈ 14.68

Lastly, substituting the values of W and L into the volume equation, we can find the value of H:

(14.68) × (7.34) × H = 10
H ≈ 0.192

So, the dimensions that minimize the cost are approximately:
Length: 14.68 m
Width: 7.34 m
Height: 0.192 m

To find the minimum cost, substitute these values back into the cost function:

Cost = (14.68 × 7.34) × 9 + (2 × 14.68 × 0.192 + 2 × 7.34 × 0.192) × 150

Calculating this value, we find that the minimum cost is approximately $2738.41.

Therefore, the minimum cost is approximately $2738.41, with the dimensions Length: 14.68 m, Width: 7.34 m, and Height: 0.192 m.