A metal can in the shape of a right circular cylinder needs to hold a volume of V cm3 . Throughout this problem V > 0 is a parameter that needs to be left as V . Suppose that the metal for the sides costs 5 cents per square cen- timeter to manufacture, whereas the top and bottom cost 10 cents per square centimeter to manufacture. Find the shape of the least expen- sive can. What is the cost of the least expensive can

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jtellezd jtellezd · Answer 1 · 2021-06-24T04:07:18+02:00

Answer:

C(min) = 0.5*V + √V/1.256 $

Step-by-step explanation:

The volume of a circular cylinder is: V(c) = π*r²*h where r is the radius of the circumference of the base and h is the height

The cost of the can is = the cost of (base and top) + lateral cost

Base surface = top surface = π*r²

Then cost of ( base + top ) is = (2* π*r² )*0,1

Lateral surface is = 2*π*r*h

Then cost of lateral surface is: (2*π*r*h)*0,5

Total cost C(t) = (2* π*r² )*0,1 + (2*π*r*h)*0,5

V = π*r²*h

Total cost as a function of (V >0 a parameter) and r then

h = V / π*r²

C(V,r) = (2* π*r² )*0,1 + π*r*(V / π*r²)

C(V,r) = 0.2*π*r² + V*/r

Taking derivatives on both sides of the equation we get:

C´(V,r) = 2*0.2*π*r - V/r²

C´(V,r) = 0 0.4*π*r - V/r = 0

Solving for r

0.4*π*r² - V = 0 ⇒ 1.256*r² = V r = √ V/ 1.256 cm

and h = V /π * (√ V/ 1.256)²

h = 1/ 1.256*π

h = 0.254 cm

C(V,r) = 0.2*π*r² + V*/r

C(min) = 0.2*π* (√ V/ 1.256)² + V/ √ V/ 1.256

C(min) = 0.2*π*V/1.256 + V/ √ V/ 1.256

C(min) = 0.5*V + √V/1.256 $