A rectangular storage container with an open top is to have a volume of 10m^3. The length of its base is twice the width. Material for the base costs $3 per m^2. Material for the sides costs $10.8 per m^2. Find the dimensions of the container which will minimize cost and the minimum cost.

Base length=?m
Base width=?m
height=?
minimum cost=$?

if the base width is w,

base is w*2w
height is 10/2w^2 = 5/w^2

area is thus
a = w*2w + 2(w*h) + 2(2w*h)
= 2w^2 + 6wh
cost is thus

c = 3*2w^2 + 10.8*6wh
= 6w^2 + 64.8w*5/w^2
= 6w^2 + 324/w

dc/dw = 12w - 324/w^2
= 12(w^3 - 27)/w^2
since w≠0,
dc/dw is minimum when w^3=27, or w=3

So, the box is 3 x 6 x 5/9
and the minimum cost is $162

To find the dimensions of the container that will minimize cost and the minimum cost, we need to set up an equation that represents the cost function.

Let's assume the width of the base is x meters. Therefore, the length of the base is 2x meters.

The height of the container can be found by dividing the volume by the area of the base:

height = volume / base area

Given that the volume is 10m^3, and the base area is length x width, we have:

height = 10 / (2x * x)
height = 10 / (2x^2)

Now, we can express the cost function in terms of x:

cost = 3 * base area + 2 * (height * (base length + base width)

Given that the cost for the base area is $3 per m^2 and the cost for the sides is $10.8 per m^2, we have:

cost = 3 * (2x * x) + 2 * (10.8 * (2x + x))

Simplifying this equation, we get:

cost = 6x^2 + 64.8x

To minimize the cost, we need to find the value of x that minimizes the cost function. To do this, we can take the derivative of the cost function with respect to x and set it equal to zero:

d(cost)/dx = 12x + 64.8 = 0

Solving for x, we have:

12x = -64.8
x = -64.8 / 12
x = -5.4

Since negative width doesn't make sense in this context, we ignore this solution.

Therefore, the width of the base is approximately 5.4 meters.

The length of the base is twice the width, so the length is approximately 10.8 meters.

Now, let's find the height:

height = 10 / (2 * 5.4 * 5.4)
height ≈ 10 / 58.32
height ≈ 0.1714 meters

So, the dimensions of the container that will minimize cost are:

Base length ≈ 10.8 meters
Base width ≈ 5.4 meters
Height ≈ 0.1714 meters

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

cost = 6*(5.4)^2 + 64.8*(5.4)
cost ≈ 174.24 + 350.88
cost ≈ $525.12

Therefore, the minimum cost is approximately $525.12.

To find the dimensions of the container that will minimize cost and the minimum cost itself, we need to express the cost of materials as a function of the dimensions and then apply calculus to find the critical points and determine the minimum.

Let's start by introducing variables for the dimensions of the rectangle: the width (W), the length (L), and the height (H).

Given that the volume of the container is 10m^3, we can write the equation for volume as:

V = L * W * H

Since the length of the base is twice the width, we have L = 2W.

Substituting this into the volume equation, we get:

10 = 2W * W * H
10 = 2W^2 * H
5 = W^2 * H

Now, let's express the cost of materials as a function of the dimensions. The cost of the base is given as $3 per m^2, so the cost of the base, C_base, is:

C_base = 3 * (L * W)

The cost of the sides is given as $10.8 per m^2, and since there are four sides, the total cost of the sides, C_sides, is:

C_sides = 4 * (L * H) * 10.8

The total cost, C_total, is the sum of the cost of the base and the cost of the sides:

C_total = C_base + C_sides
C_total = 3 * (L * W) + 4 * (L * H) * 10.8
C_total = 3 * (2W * W) + 4 * (2W * H) * 10.8
C_total = 6W^2 + 86.4W * H

Now that we have expressed the total cost as a function of the dimensions, we can differentiate it with respect to W and H to find the critical points.

dC_total/dW = 12W + 86.4H = 0 (equation 1)
dC_total/dH = 86.4W = 0 (equation 2)

From equation 2, we have W = 0, but since we are dealing with physical dimensions, let's ignore this solution.

From equation 1, we get:
12W = -86.4H
W = -7.2H

Substituting this value of W back into the equation 5 = W^2 * H, we have:
5 = (-7.2H)^2 * H
5 = 51.84H^3

Solving for H, we get:
H^3 = 5/51.84
H = (5/51.84)^(1/3)

Substituting this value of H back into W = -7.2H, we can find W and then calculate L (L = 2W). Finally, we can find the minimum cost by substituting these dimensions into C_total.

Once we have the numerical values for W, L, H, and C_total, we can find the minimum.

Please note that the specific numerical results will depend on the calculations, and the final values need to be rounded to the appropriate decimal places.