A rectangular storage container with an open top is to have a volume of k cubic meters. The length of its base is twice its width. The material for the base costs $6 per square meters and the material for the sides costs $10 per square meter.

Find the dimensions of the container having the least cost.

area of base = 2w^2

height = k/2w^2
area of sides = 4*k/2w^2 = 2k/w^2

cost = 6(2w^2) + 10(2k/w^2)
= 12w^2 + 20k/w^2

dc/dw = 24w - 40k/w
dc/dw = 0 when

24w = 40k/w
24w^2 = 40k
w^2 = 5/3 k
w = √(5k/3)

least cost box is thus √(5k/3) x √(5k/3) x 3/10

error starts here:

dc/dw = 24w - 40k/w

It should be:

dc/dw = 24w - 40k/w^3
24w = 40k/w^3
24w^4 = 40k
w = ∜(5k/3)

least cost box is thus ∜(5k/3) x ∜(5k/3) x √(3k)/(2√5)

= 3k/w

To find the dimensions of the container with the least cost, we need to minimize the cost function. Let's first determine the dimensions of the storage container.

Let's assume the width of the base is x meters. Then, the length of the base would be 2x meters, according to the problem statement.

The height of the container is unknown at this point, so let's call it h meters.

The volume of the container is given as k cubic meters. We can write the equation for volume as:

Volume = Length × Width × Height

Since the length is 2x, the width is x, and the height is h, we can write the volume equation as:

k = 2x × x × h
k = 2x^2h

Now, we need to express the surface area of the container in terms of x and h.

The surface area consists of the base area, which is the length times the width, and the area of the four sides, which is the perimeter of the base times the height.

Base Area = Length × Width
Base Area = 2x × x
Base Area = 2x^2

Perimeter of the Base = 2 × (Length + Width)
Perimeter of the Base = 2 × (2x + x)
Perimeter of the Base = 2 × (3x)
Perimeter of the Base = 6x

Area of the Four Sides = Perimeter of the Base × Height
Area of the Four Sides = 6x × h
Area of the Four Sides = 6xh

The total surface area is the sum of the base area and the area of the four sides:

Total Surface Area = Base Area + Area of Four Sides
Total Surface Area = 2x^2 + 6xh

Now, let's express the cost function in terms of x and h.

Cost = Cost of Base + Cost of Sides
Cost = (Area of Base × Cost per square meter) + (Area of Sides × Cost per square meter)
Cost = (2x^2 × 6) + (6xh × 10)
Cost = 12x^2 + 60xh

We want to find the dimensions (x and h) that minimize the cost function while satisfying the volume equation k = 2x^2h.

Now, let's differentiate the cost function with respect to x (holding h constant) and set it to zero to find the critical points:

dCost/dx = 24x + 60h = 0

To find the dimensions that minimize the cost, we need to solve this equation for x.

Umm, I thought we were computing a "Rectangular" box? Shouldn't that 2k/w^2 be a 3k/w?

Area of sides =2 (2wh + wh)

= 2 (3wh)

= 6wh
= 6w (k/2w^2)
= 6wk/2w^2
= 3k/2w