A cylindrical can is to be made to hold 50 cm3 of oil. Determine the dimensions of the can that will minimize its surface area. What is the minimum surface area

S(min) = 59,66 cm²

Step-by-step explanation:

The volume of a cylindrical can is:

V(c) = π*x²*h         where   x is radius of the base and h the height

V(c) = 50 cm³

50 =  π*x²*h      (1)

The surface area of the can (Sc) is  Surface area of the base (Sb) plus surface lateral area (Sl)

S(b) =  π*x²  

And  S(l) = 2*π*x*h

Then

S(c) =  π*x²  +  2*π*x*h

And  surface area as a function of x is

From equation (1)

h = 50 /π*x²   and plugging this value in the previous expression

S(x) =  π*x²  +  2*π*x*(50/π*x²)

S(x) =  π*x²  + 100/x

Taking derivatives on both sides of the equation

S´(x) = 2*π*x - 100/x²

S´(x) = 0     means       2*π*x - 100/x² = 0

π*x - 50/x²  = 0

π*x³ - 50  = 0

π*x³  = 50

x³ = 50 / 3,14

x³ = 15,92

x  = 2,51 cm

And  h =  50 / π* (2,51)²

h = 2,53 cm

Then minimum surface area of the can is:

S(min) = 19,78 + 39,88

S(min) = 59,66 cm²

168

step-by-step explanation:

i am pretty sure that the answer is 168

4932,9180,7625 are divisible by 9