# 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 ->  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

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 Geometry: Area and Surface Area Solvers Lessons Answers archive Quiz In Depth

 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. ThanksFound 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(30993)   (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.