(Half cylinder) The length was 32 ft more than 324 times it’s width. If the original tunnel volume was 62,231,040 ft^3, what’s the original width?

Question

Algebra 2, polynomial word problem. HELP

Answer 1

Okay, this one is not as bad as it seems. The two key things are the formula for the volume of a half cylinder (the tunnel) and the fact that the radius is one half the width. The formula is V = (πr²/2) x L. The length (L) is 324w + 32, but w = 2r, so substitute 2r for the w and you get (324 x 2r) + 32 for the length, or 648r + 32 = L. Now plug that into the formula for volume and you get  V = (πr²/2) x (648r + 32). That gives V = (648πr³ + 32πr²)/2 which simplifies to V = (324πr³ + 16πr²) when you divide by 2. Multiply by pi and the equation becomes V = 1017.9r³ + 50.26r² . The next step is to put this in the standard form of a cubic equation, and then plug the coefficients into cubic equation solving function on your calculator. The equation becomes 62231040 = 1017.9r3 + 50.26r2, and then set it equal to zero giving, 0 = 1017.9r³ + 50.26r² - 62231040. Now just use the magic of the calculator because nobody memorizes the cubic formula for good reason. You will get three answers because cubic equations have three roots. Only the positive one counts because you cannot have a negative length, so the radius of the tunnel is 39.38 ft, making the width 78.76 feet. To get the length, plug that back into the equation L = 324w = 32, and you get 25,550 feet, or a pretty long tunnel.