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

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  $

last one.

step-by-step explanation:

last one.

generally speaking, the others follow some sort of pattern that correlates to either linear (looks like a straight line) or sinusoidal (looks like an s).

the last one is just a random cloud of points that doesn't correlate to any particular known order.

a. she will receive a refund of $278.

step-by-step explanation: