Answer:
Radius=2.09 cm
Height,h=14.57 cm
Step-by-step explanation:
We are given that
Volume of cylinderical shaped can=200 cubic cm.
Cost of sides of can=0.02 cents per square cm
Cost of top and bottom of the can =0.07 cents per square cm
Curved surface area of cylinder=[tex]2\pi rh[/tex]
Area of circular base=Area of circular top=[tex]\pi r^2[/tex]
Total cost,C(r)=[tex]0.02\times 2\pi rh+2\pi r^2\times 0.07[/tex]
Volume of cylinder,[tex]V=\pi r^2 h[/tex]
[tex]200=\pi r^2 h[/tex]
[tex]h=\frac{200}{\pi r^2}[/tex]
Substitute the value of h
[tex]C(r)=0.02\times 2\pi r\times \frac{200}{\pi r^2}+2\pi r^2\times 0.07[/tex]
[tex]C(r)=\frac{8}{r}+0.14\pi r^2[/tex]
Differentiate w.r.t r
[tex]C'(r)=-\frac{8}{r^2}+0.28\pi r[/tex]
[tex]C'(r)=0[/tex]
[tex]-\frac{8}{r^2}+0.28\pi r=0[/tex]
[tex]0.28\pi r=\frac{8}{r^2}[/tex]
[tex]r^3=\frac{8}{0.28\pi}=9.095[/tex]
[tex]r=(9.095)^{\frac{1}{3}}=2.09[/tex]
Again, differentiate w.r.t r
[tex]C''(r)=\frac{16}{r^3}+0.28\pi[/tex]
Substitute the value of r
[tex]C''(2.09)=\frac{16}{(2.09)^3}+0.28\pi=2.63>0[/tex]
Therefore,the product cost is minimum at r=2.09
h=[tex]\frac{200}{\pi (2.09)^2}=14.57[/tex]
Radius of can,r=2.09 cm
Height of cone,h=14.57 cm
