What are the dimensions of a rectangular storage container with an open top that has a volume of 10 m³ and a base length that is twice the width, considering material costs?

To find the dimensions of the rectangular storage container, let’s denote the width of the base as w meters. Since the length of the base is twice the width, we can express the length as l = 2w meters. The volume V of the container is given by the formula:

V = length × width × height

Given that the volume is 10 m³, we can write:

10 = (2w) × w × h

From this, we can simplify it:

10 = 2w²h

Now, we can express the height in terms of width:

h = rac{10}{2w^2} = rac{5}{w^2}

The next step is to calculate the cost of the material for the base. The area of the base is:

Area = length × width = 2w × w = 2w²

If the cost of the material for the base is $15 per square meter, then the total cost C of the material for the base can be calculated as:

C = Area × cost per square meter = 2w² × 15 = 30w²

We need to minimize this cost while maintaining the volume constraint. Substituting the expression for h back into the volume equation gives us:

10 = 2w² × rac{5}{w^2}

Through solving the equation, we find that:

w = 1

So, Length: l = 2w = 2

Height: h = rac{5}{1^2} = 5

Finally, the dimensions of the container are:

Width: 1 m

Length: 2 m

Height: 5 m

In conclusion, the dimensions of the rectangular storage container with a volume of 10 m³, where the length of the base is twice the width, are a width of 1 meter, a length of 2 meters, and a height of 5 meters, resulting in a material cost of $30.

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