SOLUTION: Suppose the volume must be 50in^3, What are the values for r and h that will minimize the amount of sheet metal required to obtain this volume. I know volume is v= (pi)r^2h. I als

Question 266652: Suppose the volume must be 50in^3, What are the values for r and h that will minimize the amount of sheet metal required to obtain this volume. I know volume is v= (pi)r^2h. I also know the surface area is sa= 2(pi)rh+2(pi)r^2.
I have been working on this all day. Any help would be greatly appreciated. Thanks
Found 2 solutions by drk, Alan3354:
Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
lets start with
(i)
we know v = 50, so we get
(ii)
If we solve for h, we get
(iii)
now we look at surface area as
(iv)
substitute (iii) into (iv) to get
(v)
now everything is in terms of r.
getting a common denominator as pir^2, we get
(vi)
now, factoring, we get
(vii)
reducing we get
(viii)
It turns out that r is minimum at 2.
This means that h = 50/4pi or h ~ 3.97887
This gives us a minimum sa at
sa = 75.1326
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Suppose the volume must be 50in^3, What are the values for r and h that will minimize the amount of sheet metal required to obtain this volume. I know volume is v= (pi)r^2h. I also know the surface area is sa= 2(pi)rh+2(pi)r^2.
----------------
Eliminate either r or h, then minimize Area wrt the remaining variable.
V = pi*r^2h = 50
h = 50/(pi*r^2)
----------------
A = 2pi*rh + 2pi*r^2 = 2pi(rh + r^2)
A = 2pi(r^2 + 50/pi*r)
A = 2pi*r^2 + 100/r
dA/dr = 4pir - 100/r^2 = 0
pi*r - 25/r^2 = 0
r^3 = 25/pi
r =~ 1.99647 inches
h =~ 3.99295 inches
---------------------
It's a tough problem.

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