Question

A rectangular storage container with an open top is to have a volume of 10 m^3. The length of its base is twice the width. Material for the base costs $6 per m^2. Material for the sides costs $0.8 per m^2.

Required:
Find the dimensions of the container which will minimize cost and the minimum cost.

Answer 1

The dimension that minimizes the container is width = 1; base = 2 and height = 5

The minimum cost is $36

Step-by-step explanation:

Let the width be x

So:

`Width = x`

`Base = 2 * Width`

`Base = 2 * x`

`Base = 2x`

`Height = h`

Volume of the box is:

`Volume = 10m^3` ---- Given

Volume is calculated as:

`Volume = Base * Width * Height`

`Volume = 2x * x * h`

`Volume = 2x^2h`

Substitute 10 for Volume

`10 = 2x^2h`

Make h the subject

`h = (10)/(2x^2)`

`h = (5)/(x^2)`

Next, we calculate area of the sides.

`Area = Base + Sides`

Because it has an open top, the area is:

`Base\ Area = Base * Width`

`Sides\ Area = 2[(Width * Height) + (Base * Height)]`

`Base\ Area = 2x * x`

`Base\ Area = 2x^2`

`Sides\ Area = 2[(Width * Height) + (Base * Height)]`

`Side\ Area = 2[(x * h) + (2x * h)]`

`Side\ Area = 2[(xh) + (2xh)]`

`Side\ Area = 2[3xh]`

`Side\ Area = 6xh`

The base area costs $6 per m²

So, the cost of 2x² would be:

`Cost = 6 * 2x^2`

`Cost = 12x^2`

The side cost area costs $0.8 per m²

So, 6xh would cost

`Cost = 0.8 * 6xh`

`Cost = 4.8xh`

Total Cost (C) is:

`C = 12x^2 + 4.8xh`

Recall that
`h = (5)/(x^2)`

So:

`C = 12x^2 + 4.8x *(5)/(x^2)`

`C = 12x^2 + 4.8 *(5)/(x)`

`C = 12x^2 + (4.8 *5)/(x)`

`C = 12x^2 + (24)/(x)`

Take derivative of C

`C^(-1) = 24x - (24)/(x^2)`

Take LCM

`C^(-1) = (24x^3 - 24)/(x^2)`

Equate
`C^(-1)` to 0

`0 = (24x^3 - 24)/(x^2)`

Cross multiply

`24x^3 - 24 = 0 * x^2`

`24x^3 - 24 = 0`

Add 24 to both sides

`24x^3 = 24`

Divide through by 24

`x^3 = 1`

Take cube roots of both sides

`x = 1`

Recall that

`Width = x`

`Base = 2x`

`Height = h` and
`h = (5)/(x^2)`

Solve for these dimensions:

`Width = 1`

`Base = 2 * 1`

`Base = 2`

`h = (5)/(1^2)`

`h = (5)/(1)`

`h = 5`

i.e.

`Height = 5`

Hence, the dimension that minimizes the container is width = 1; base = 2 and height = 5

Recall that

`C = 12x^2 + (24)/(x)`

Substitute 1 for x

`C = 12(1^2) + (24)/(1)`

`C = 12 + 24`

`C = 36`

The minimum cost is $36