A cylindrical drum is made to hold exactly 1m^3 in its interior. Assume that the material for the top and the bottom costs $20 per m^2, while that for the side costs $10 per m^2. Determine the radius of the drum that minimizes the cost of the material used.
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I posted this question before but I wasn't sure about one part.
Eqn's:
(1) Area = 2PI*r^2 + 2PI*rh
(2) Cost = 20*2PI*r^2 + 2PI*rh
Solving for "h" for eqn (1)
h = -Area(2PI*r^2/(2PI*r)
did I do this step right cuz after i solve for h I am able to sub that into eqn 2, and find DCost/dt but I feel Im missing a step. Any help will be appreciated! :)
Use the volume equation to solve for h.
Volume= PI r^2 h
h= volume/PIr^2
Then put that into cost.
Cost = 20*2PI*r^2 + 2PI*r(volume/PIr^2)
Now simplify, and take the derivative of cost with respect to r.
To find the radius of the drum that minimizes the cost of the material used, we need to find the derivative of the cost function with respect to the radius, and then set it equal to zero. This will give us the critical points where the cost function is minimized.
First, let's simplify the cost equation:
Cost = 20*2π*r^2 + 2π*r(volume/πr^2)
Simplifying further:
Cost = 40πr^2 + 2π(volume/r)
Now let's differentiate the cost function with respect to r:
dCost/dr = 80πr - 2π(volume/r^2)
Setting the derivative equal to zero:
80πr - 2π(volume/r^2) = 0
Dividing through by 2π:
40r - (volume/r^2) = 0
Multiplying through by r^2:
40r^3 - volume = 0
Now, substitute the volume value of 1m^3:
40r^3 - 1 = 0
Solving for r^3:
40r^3 = 1
r^3 = 1/40
Taking the cube root:
r = (1/40)^(1/3)
Therefore, the radius of the drum that minimizes the cost of the material used is approximately 0.398 meters.