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

Algebra ->  Surface-area -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
lets start with
(i) v+=+%28pi%29r%5E2%2Ah
we know v = 50, so we get
(ii) 50=+%28pi%29r%5E2%2Ah
If we solve for h, we get
(iii) h+=+50%2F%28pi%2Ar%5E2%29
now we look at surface area as
(iv) sa+=+2%28pi%29%2Ar%2Ah%2B2%28pi%29%2Ar%5E2
substitute (iii) into (iv) to get
(v) sa+=+2%2Api%2Ar%2A%2850%2F%28pi%2Ar%5E2%29%29+%2B+2%2Api%2Ar%5E2
now everything is in terms of r.
getting a common denominator as pir^2, we get
(vi) sa+=+%28100%2Api%2Ar+%2B+2%2Api%5E2%2Ar%5E4%29%2F%28pi%2Ar%5E2%29
now, factoring, we get
(vii) sa+=+%282%2Api%2Ar%2850%2Bpi%2Ar%5E3%29%29%2F%28pi%2Ar%5E2%29
reducing we get
(viii) sa+=+2%2850%2Bpi%2Ar%5E3%29%2Fr
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) About Me  (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.
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Eliminate either r or h, then minimize Area wrt the remaining variable.
V = pi*r^2h = 50
h = 50/(pi*r^2)
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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
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It's a tough problem.