Answer:
$ 327.08
Step-by-step explanation:
Let w be the width ( in meters ) of the container,
⇒ Length of the container = 2w,
If h be the height of the container,
So, the volume of the container = length × width × height
= 2w × w × h
= 2w² h
According to the question,
![2w^2h=10](https://tex.z-dn.net/?f=2w%5E2h%3D10)
![\implies h=\frac{10}{2w^2}=\frac{5}{w^2}](https://tex.z-dn.net/?f=%5Cimplies%20h%3D%5Cfrac%7B10%7D%7B2w%5E2%7D%3D%5Cfrac%7B5%7D%7Bw%5E2%7D)
Now, the area of the base = length × width
![=2w^2](https://tex.z-dn.net/?f=%3D2w%5E2)
Area of sides = 2 × length × height + 2 × width × height
![=2\times 2w\times h+2\times w\times h](https://tex.z-dn.net/?f=%3D2%5Ctimes%202w%5Ctimes%20h%2B2%5Ctimes%20w%5Ctimes%20h)
![=4w\times \frac{5}{w^2}+2w\times \frac{5}{w^2}](https://tex.z-dn.net/?f=%3D4w%5Ctimes%20%5Cfrac%7B5%7D%7Bw%5E2%7D%2B2w%5Ctimes%20%5Cfrac%7B5%7D%7Bw%5E2%7D)
![=\frac{20}{w}+\frac{10}{w}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B20%7D%7Bw%7D%2B%5Cfrac%7B10%7D%7Bw%7D)
![=\frac{30}{w}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B30%7D%7Bw%7D)
Since, material for the base costs $20 per square meter and material for the sides costs $12 per square meter,
Hence, total cost,
![C(w) = 2w^2\times 20 +\frac{30}{w}\times 12](https://tex.z-dn.net/?f=C%28w%29%20%3D%202w%5E2%5Ctimes%2020%20%2B%5Cfrac%7B30%7D%7Bw%7D%5Ctimes%2012)
Differentiating with respect to w,
![C'(w) = 80w - \frac{360}{w^2}](https://tex.z-dn.net/?f=C%27%28w%29%20%3D%2080w%20-%20%5Cfrac%7B360%7D%7Bw%5E2%7D)
Again differentiating with respect to w,
![C''(w) = 80 +\frac{720}{w^3}](https://tex.z-dn.net/?f=C%27%27%28w%29%20%3D%2080%20%2B%5Cfrac%7B720%7D%7Bw%5E3%7D)
For maxima or minima,
C'(w) = 0
![80w - \frac{360}{w^2}=0](https://tex.z-dn.net/?f=80w%20-%20%5Cfrac%7B360%7D%7Bw%5E2%7D%3D0)
![80w^3-360=0](https://tex.z-dn.net/?f=80w%5E3-360%3D0)
![80w^3=360](https://tex.z-dn.net/?f=80w%5E3%3D360)
![\implies w=\sqrt[3]{\frac{360}{80}}=1.65096362445\approx 1.651](https://tex.z-dn.net/?f=%5Cimplies%20w%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B360%7D%7B80%7D%7D%3D1.65096362445%5Capprox%201.651)
For w = 1.651, C''(w) = positive,
Thus, cost is minimum for width 1.651 meters,
And, the minimum cost = C(1.651) = ![40(1.651)^2+\frac{360}{1.651}=\$327.081706869\approx \$ 327.08](https://tex.z-dn.net/?f=40%281.651%29%5E2%2B%5Cfrac%7B360%7D%7B1.651%7D%3D%5C%24327.081706869%5Capprox%20%5C%24%20327.08)