A rectangular storage container with an open top is to have a volume of 10 . The length of its base is twice the width. Material for the base costs $12 per square meter. Material for the sides costs $5 per square meter. Find the cost of materials for the cheapest such container.

If the base has dimensions x and 2x, then we have

2x^2y = 10
so, y = 5/x^2

So, the cost of the container is

c(x) = 12*2x^2 + 5*2xy + 5*xy = 24x^2 + 15xy = 24x^2 + 75/x

so, find c(x) when dc/dx=0

Would dc/dx=0 when x is 1.16?

yes

To find the cost of materials for the cheapest container, we need to determine the dimensions of the container that minimize the cost. Let's break down the problem step by step:

1. Define the variables:
- Let the width of the base of the container be 'w' meters.
- Since the length of the base is twice the width, the length will be '2w' meters.
- The height of the container is 'h' meters.

2. Determine the volume of the container:
- The volume of a rectangular container is given by the formula V = length x width x height.
- In this case, V = (2w)w(h) = 2w^2h.
- We know the volume is 10, so 2w^2h = 10.

3. Express the cost of materials in terms of the variables:
- The cost of the base material is $12 per square meter, which is equal to the area of the base, 2w^2.
- The cost of the side material is $5 per square meter, which is equal to the area of the four sides, 2lw + 2lh = 2(2w)h + 2w(h) = 6wh.

4. Express the total cost in terms of the variables:
- The total cost is the cost of the base plus the cost of the sides, C = 12(2w^2) + 5(6wh) = 24w^2 + 30wh.

5. Use the volume constraint to express one variable in terms of the others:
- From step 2, we have the equation 2w^2h = 10.
- Rearrange this equation to solve for h: h = 10/(2w^2) = 5/w^2.

6. Substitute the expression for h from step 5 into the total cost equation from step 4:
- C = 24w^2 + 30w * (5/w^2) = 24w^2 + 150/w.

7. Determine the derivative of the total cost equation with respect to w:
- dC/dw = 48w - 150/w^2.

8. Set the derivative equal to zero and solve for w:
- 48w - 150/w^2 = 0.
- Multiply through by w^2: 48w^3 - 150 = 0.
- Solve for w: w^3 = 150/48 = 25/8.
- Take the cube root: w = (25/8)^(1/3) β‰ˆ 1.857.

9. Plug the value of w back into the expression for h:
- h = 5/w^2 = 5/(1.857^2) β‰ˆ 1.861.

10. Calculate the cost of materials using the optimal values of w and h:
- C = 24w^2 + 30wh = 24(1.857^2) + 30(1.857)(1.861).
- The cost of materials for the cheapest container is the calculated value of C.