Question

A rectangular storage container with an open top is to have a volume of 10 m3 . then length of its base is twice the width. material for the base costs $10 per square meter. material for the sides cost $6 per square meter. find the cost of materials for the cheapest such container.

Answer 1

Answer:

The cost of materials for the cheapest such container is $163.54.

Step-by-step explanation:

A rectangular storage container with an open top is to have a volume of 10 m³.

The volume of the rectangle is

$\text{Volume} =\text{Length} \times \text{Width} \times \text{Height}$

Length of its base is twice the width.

Let Width be 'w'.

Length is l=2w.

Height be 'h'.

$10 =2w\times w\times h$

$10=2w^2h$

The height in terms of width is represented as,

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

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

According to question,

The cost is 10 times the area of the base and 6 times the total area of the sides.

i.e. Cost is given by,

$C=10(L\times W)+6(2\times L\times H+2\times W\times H)$

$C=10(2w\times w)+6(2\times 2w\times \frac{5}{w^2}+2\times w\times \frac{5}{w^2})$

$C=20w^2+\frac{120}{w}+\frac{60}{w}$

$C(w)=20w^2+\frac{180}{w}$

To get the minimum value,

Differentiate the cost w.r.t 'w',

$C'(w)=20\frac{d(w^2)}{dw}+180\frac{d(w^{-1})}{dw}$

$C'(w)=20\times 2w-180 w^{-2}$

$C'(w)=40w-\frac{180}{w^2}$

To find critical points put derivate =0,

$40w-\frac{180}{w^2}=0$

$40w=\frac{180}{w^2}$

$w^3=\frac{180}{40}$

$w=\sqrt[3]{4.5}$

$w=1.65$

We find the second derivative to minimize,

$C''(w)=40\frac{d(w)}{dw}-180\frac{d(w^{-2})}{dw}$

$C''(w)=40+360(w^{-3})$

$C''(w)>0$

As $C''(w)>0$ it is the minimum cost.

The cost is minimum at w=1.65.

Substitute the values in the cost function,

$C(1.65)=20(1.65)^2+\frac{180}{1.65}$

$C(1.65)=54.45+109.09$

$C(1.65)=163.54$

Therefore, the cost of materials for the cheapest such container is $163.54.

Answer 2

The volume of a box is the amount of space in it.

The cheapest cost of such container is $163.54

The volume is given as:

$\mathbf{V = 10}$

The relationship between the dimension is:

$\mathbf{L = 2W}$

So, the volume is:

$\mathbf{V = LWH}$

Substitute 2W for L

$\mathbf{V = 2W^2H}$

Substitute 10 for V

$\mathbf{10 = 2W^2H}$

Divide both sides by 2

$\mathbf{5 = W^2H}$

Make H the subject

$\mathbf{H = \frac{5}{W^2}}$

The surface area is calculated as:

$\mathbf{A = Base + 2 \times Sides_1 + 2 \times Sides_2}$

The cost function is then calculated as:

$\mathbf{C =10 \times Base +6 \times ( 2 \times Sides_1 + 2 \times Sides_2)}$

So, we have:

$\mathbf{C =10 \times LW +6 \times (2 \times WH + 2 \times LH)}$

Substitute $\mathbf{L = 2W}$

$\mathbf{C =10 \times 2W^2 +6 \times (2 \times WH + 2 \times 2WH)}$

$\mathbf{C =10 \times 2W^2 +6 \times (2WH + 4WH)}$

$\mathbf{C =10 \times 2W^2 +6 \times (6WH)}$

$\mathbf{C =20W^2 +36WH}$

Substitute $\mathbf{H = \frac{5}{W^2}}$

$\mathbf{C =20W^2 +36W\times \frac{5}{W^2}}$

$\mathbf{C =20W^2 + \frac{180}{W}}$

Differentiate

$\mathbf{C' =40W - \frac{180}{W^2}}$

Set to 0

$\mathbf{40W - \frac{180}{W^2} = 0}$

Divide through by W

$\mathbf{40 - \frac{180}{W^3} = 0}$

Rewrite as:

$\mathbf{\frac{180}{W^3} = 40}$

Solve for W

$\mathbf{40W^3 = 180}$

Divide both sides by 40

$\mathbf{W^3 = 4.5}$

Take cube roots

$\mathbf{W = 1.65}$

Recall that: $\mathbf{C =20W^2 + \frac{180}{W}}$

So, we have:

$\mathbf{C = 20 \times 1.65^2 + \frac{180}{1.65^2}}$

$\mathbf{C = 163.54}$

Hence, the cheapest cost of such container is $163.54

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