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}](https://tex.z-dn.net/?f=%5Ctext%7BVolume%7D%20%3D%5Ctext%7BLength%7D%20%5Ctimes%20%5Ctext%7BWidth%7D%20%5Ctimes%20%5Ctext%7BHeight%7D)
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](https://tex.z-dn.net/?f=10%20%3D2w%5Ctimes%20w%5Ctimes%20h)
![10=2w^2h](https://tex.z-dn.net/?f=10%3D2w%5E2h)
The height in terms of width is represented as,
![h=\frac{10}{2w^2}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B10%7D%7B2w%5E2%7D)
![h=\frac{5}{w^2}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B5%7D%7Bw%5E2%7D)
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)](https://tex.z-dn.net/?f=C%3D10%28L%5Ctimes%20W%29%2B6%282%5Ctimes%20L%5Ctimes%20H%2B2%5Ctimes%20W%5Ctimes%20H%29)
![C=10(2w\times w)+6(2\times 2w\times \frac{5}{w^2}+2\times w\times \frac{5}{w^2})](https://tex.z-dn.net/?f=C%3D10%282w%5Ctimes%20w%29%2B6%282%5Ctimes%202w%5Ctimes%20%5Cfrac%7B5%7D%7Bw%5E2%7D%2B2%5Ctimes%20w%5Ctimes%20%5Cfrac%7B5%7D%7Bw%5E2%7D%29)
![C=20w^2+\frac{120}{w}+\frac{60}{w}](https://tex.z-dn.net/?f=C%3D20w%5E2%2B%5Cfrac%7B120%7D%7Bw%7D%2B%5Cfrac%7B60%7D%7Bw%7D)
![C(w)=20w^2+\frac{180}{w}](https://tex.z-dn.net/?f=C%28w%29%3D20w%5E2%2B%5Cfrac%7B180%7D%7Bw%7D)
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}](https://tex.z-dn.net/?f=C%27%28w%29%3D20%5Cfrac%7Bd%28w%5E2%29%7D%7Bdw%7D%2B180%5Cfrac%7Bd%28w%5E%7B-1%7D%29%7D%7Bdw%7D)
![C'(w)=20\times 2w-180 w^{-2}](https://tex.z-dn.net/?f=C%27%28w%29%3D20%5Ctimes%202w-180%20w%5E%7B-2%7D)
![C'(w)=40w-\frac{180}{w^2}](https://tex.z-dn.net/?f=C%27%28w%29%3D40w-%5Cfrac%7B180%7D%7Bw%5E2%7D)
To find critical points put derivate =0,
![40w-\frac{180}{w^2}=0](https://tex.z-dn.net/?f=40w-%5Cfrac%7B180%7D%7Bw%5E2%7D%3D0)
![40w=\frac{180}{w^2}](https://tex.z-dn.net/?f=40w%3D%5Cfrac%7B180%7D%7Bw%5E2%7D)
![w^3=\frac{180}{40}](https://tex.z-dn.net/?f=w%5E3%3D%5Cfrac%7B180%7D%7B40%7D)
![w=\sqrt[3]{4.5}](https://tex.z-dn.net/?f=w%3D%5Csqrt%5B3%5D%7B4.5%7D)
![w=1.65](https://tex.z-dn.net/?f=w%3D1.65)
We find the second derivative to minimize,
![C''(w)=40\frac{d(w)}{dw}-180\frac{d(w^{-2})}{dw}](https://tex.z-dn.net/?f=C%27%27%28w%29%3D40%5Cfrac%7Bd%28w%29%7D%7Bdw%7D-180%5Cfrac%7Bd%28w%5E%7B-2%7D%29%7D%7Bdw%7D)
![C''(w)=40+360(w^{-3})](https://tex.z-dn.net/?f=C%27%27%28w%29%3D40%2B360%28w%5E%7B-3%7D%29)
![C''(w)>0](https://tex.z-dn.net/?f=C%27%27%28w%29%3E0)
As
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}](https://tex.z-dn.net/?f=C%281.65%29%3D20%281.65%29%5E2%2B%5Cfrac%7B180%7D%7B1.65%7D)
![C(1.65)=54.45+109.09](https://tex.z-dn.net/?f=C%281.65%29%3D54.45%2B109.09)
![C(1.65)=163.54](https://tex.z-dn.net/?f=C%281.65%29%3D163.54)
Therefore, the cost of materials for the cheapest such container is $163.54.