Respuesta :
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= [tex]\pi r^2h[/tex]
According to the problem,
[tex]\pi r^2 h=25[/tex]
[tex]\Rightarrow h=\frac{25}{\pi r^2}[/tex]
The surface area of the base of the can is = [tex]\pi r^2[/tex]
The metal for the bottom will cost $2.00 per cm²
The metal cost for the base is =$(2.00× [tex]\pi r^2[/tex])
The lateral surface area of the can is = [tex]2\pi rh[/tex]
The metal for the side will cost $1.25 per cm²
The metal cost for the base is =$(1.25× [tex]2\pi rh[/tex])
[tex]=\$2.5 \pi r h[/tex]
Total cost of metal is C= 2.00 [tex]\pi r^2[/tex]+[tex]2.5 \pi r h[/tex]
Putting [tex]h=\frac{25}{\pi r^2}[/tex]
[tex]\therefore C=2\pi r^2+2.5 \pi r \times \frac{25}{\pi r^2}[/tex]
[tex]\Rightarrow C=2\pi r^2+ \frac{62.5}{ r}[/tex]
Differentiating with respect to r
[tex]C'=4\pi r- \frac{62.5}{ r^2}[/tex]
Again differentiating with respect to r
[tex]C''=4\pi + \frac{125}{ r^3}[/tex]
To find the minimize cost, we set C'=0
[tex]4\pi r- \frac{62.5}{ r^2}=0[/tex]
[tex]\Rightarrow 4\pi r=\frac{62.5}{ r^2}[/tex]
[tex]\Rightarrow r^3=\frac{62.5}{ 4\pi}[/tex]
⇒r=1.71
Now,
[tex]\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0[/tex]
When r=1.71 cm, the metal cost will be minimum.
Therefore,
[tex]h=\frac{25}{\pi\times 1.71^2}[/tex]
⇒h=2.72 cm
Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.
