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

Answer:

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

The minimum cost is $36

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

$h = \frac{10}{2x^2}$

$h = \frac{5}{x^2}$

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 $h = \frac{5}{x^2}$

So:

$C = 12x^2 + 4.8x *\frac{5}{x^2}$

$C = 12x^2 + 4.8 *\frac{5}{x}$

$C = 12x^2 + \frac{4.8 *5}{x}$

$C = 12x^2 + \frac{24}{x}$

Take derivative of C

$C^{-1} = 24x - \frac{24}{x^2}$

Take LCM

$C^{-1} = \frac{24x^3 - 24}{x^2}$

Equate $C^{-1}$ to 0

$0 = \frac{24x^3 - 24}{x^2}$

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 $h = \frac{5}{x^2}$

Solve for these dimensions:

$Width = 1$

$Base = 2 * 1$

$Base = 2$

$h = \frac{5}{1^2}$

$h = \frac{5}{1}$

$h = 5$

i.e.

$Height = 5$

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

Recall that

$C = 12x^2 + \frac{24}{x}$

Substitute 1 for x

$C = 12(1^2) + \frac{24}{1}$

$C = 12 + 24$

$C = 36$

The minimum cost is $36