Answer:
Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.
Step-by-step explanation:
Given that, the volume of cylindrical can with out top is 25 cm³.
Consider the height of the can be h and radius be r.
The volume of the can is V= ![\pi r^2h](https://tex.z-dn.net/?f=%5Cpi%20r%5E2h)
According to the problem,
![\pi r^2 h=25](https://tex.z-dn.net/?f=%5Cpi%20r%5E2%20h%3D25)
![\Rightarrow h=\frac{25}{\pi r^2}](https://tex.z-dn.net/?f=%5CRightarrow%20h%3D%5Cfrac%7B25%7D%7B%5Cpi%20r%5E2%7D)
The surface area of the base of the can is = ![\pi r^2](https://tex.z-dn.net/?f=%5Cpi%20r%5E2)
The metal for the bottom will cost $2.00 per cm²
The metal cost for the base is =$(2.00×
)
The lateral surface area of the can is = ![2\pi rh](https://tex.z-dn.net/?f=2%5Cpi%20rh)
The metal for the side will cost $1.25 per cm²
The metal cost for the base is =$(1.25×
)
![=\$2.5 \pi r h](https://tex.z-dn.net/?f=%3D%5C%242.5%20%5Cpi%20r%20h)
Total cost of metal is C= 2.00
+![2.5 \pi r h](https://tex.z-dn.net/?f=2.5%20%5Cpi%20r%20h)
Putting ![h=\frac{25}{\pi r^2}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B25%7D%7B%5Cpi%20r%5E2%7D)
![\therefore C=2\pi r^2+2.5 \pi r \times \frac{25}{\pi r^2}](https://tex.z-dn.net/?f=%5Ctherefore%20C%3D2%5Cpi%20r%5E2%2B2.5%20%5Cpi%20r%20%5Ctimes%20%5Cfrac%7B25%7D%7B%5Cpi%20r%5E2%7D)
![\Rightarrow C=2\pi r^2+ \frac{62.5}{ r}](https://tex.z-dn.net/?f=%5CRightarrow%20C%3D2%5Cpi%20r%5E2%2B%20%5Cfrac%7B62.5%7D%7B%20r%7D)
Differentiating with respect to r
![C'=4\pi r- \frac{62.5}{ r^2}](https://tex.z-dn.net/?f=C%27%3D4%5Cpi%20r-%20%5Cfrac%7B62.5%7D%7B%20r%5E2%7D)
Again differentiating with respect to r
![C''=4\pi + \frac{125}{ r^3}](https://tex.z-dn.net/?f=C%27%27%3D4%5Cpi%20%2B%20%5Cfrac%7B125%7D%7B%20r%5E3%7D)
To find the minimize cost, we set C'=0
![4\pi r- \frac{62.5}{ r^2}=0](https://tex.z-dn.net/?f=4%5Cpi%20r-%20%5Cfrac%7B62.5%7D%7B%20r%5E2%7D%3D0)
![\Rightarrow 4\pi r=\frac{62.5}{ r^2}](https://tex.z-dn.net/?f=%5CRightarrow%204%5Cpi%20r%3D%5Cfrac%7B62.5%7D%7B%20r%5E2%7D)
![\Rightarrow r^3=\frac{62.5}{ 4\pi}](https://tex.z-dn.net/?f=%5CRightarrow%20%20r%5E3%3D%5Cfrac%7B62.5%7D%7B%204%5Cpi%7D)
⇒r=1.71
Now,
![\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0](https://tex.z-dn.net/?f=%5Cleft%20C%27%27%5Cright%7C_%7Bx%3D1.71%7D%3D4%5Cpi%20%2B%5Cfrac%7B125%7D%7B1.71%5E3%7D%3E0)
When r=1.71 cm, the metal cost will be minimum.
Therefore,
![h=\frac{25}{\pi\times 1.71^2}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B25%7D%7B%5Cpi%5Ctimes%201.71%5E2%7D)
⇒h=2.72 cm
Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.