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Mumz [18]
3 years ago
12

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. mate

rial 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.
Mathematics
1 answer:
katrin [286]3 years ago
7 0

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.

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