<p>This was on part of a test I had today. I'm not sure if many people got a solution, though. This wasn't the exact wording of the problem, but it's all the same information:</p>
<p>A 6 ft. pole and an 8 ft. pole are on level ground -- their bases are 10 ft. apart. A wire attached to the top of the 8 ft. pole connects to the ground at some point between the two poles and then runs back up and attaches to the top of the 6 ft. pole.</p>
<p>a) What is the minimum length of the wire?</p>
<p>b) Where on the ground should the wire be attached so that the wire is as short as possible?</p>
<p>And keep in mind that this is an Applications of Derivatives Exam (if there are other ways of doing this using integrals or something). The only way I was able to solve was by using L'Hopital's Rule and my calculator, but no way algebraically. Maybe that's the only way that was possible... but I doubt that our teacher would give us something that we couldn't solve algebraically.</p>
<p>First, there's way too much algebra for me to even attempt showing it here. However, here's how you can do it.</p>
<p>First, create a function for the length of the wire according to the distance of the ground point from one of the poles. I imagined the 8 ft. pole on the left and the 6 ft. pole on the right and the function was of how far the ground point was from the 8 ft pole. f(x)=sqr(64+x^2)+sqr(36+(10-x)^2) Since this respresents the length of the wire, and you want to find the smallest possible length, all you need is to find the minimum. Just find the first derivative and solve for zero, right? Right, but the algebra is cumbersome. You'll end up getting the answer to your (b) question first, then you plug that into the original function to find the shortest wire length. Game set and match. Since you'll be finding the real zeroes of a second order polynomial, you'll get two x-values. One of them can't be the right one because it'll be more than 10 ft. I got 5.71 ft. from the base of the 8 ft. pole. and the min. wire length was 17.2 ft.</p>
<p>And in attempting to solve for L' = 0 or DNE, I couldn't get an answer algebraically. Were you actually able to solve that for X? I wasn't able to factor it at all, but perhaps I did something wrong while I was trying to solve it. The only way I was able to solve it was using Newton's Method (I just realized that I accidentally said that I used L'Hopital's Rule before -- sorry about that) and my calculator.</p>
<p>okay, i tossed my scrap paper that i solved it on, but i think i can walk you thru it. First, find the LCD for the derivative and make it one fraction. Once that's done, then solve for zero. Since it's a fraction, we only need to concentrate on the numerator and solve THAT for zero. Make an equation where the numerator is equal to zero and redistribute so that the two quanitites with like roots are on one side and the other root is by itself. Square both sides of that to get much simpler terms. The third and fourth roots cancel on both sides, leaving you with a second order polynomial. Redistribute that so that everything equals zero and use the quadratic formula or your calculator, depending on how your teacher wants the answer. If you read it all at once, it sounds quite confusing, but go slowly step by step and you should see what I mean in each step and be able to do it. The two hard parts of calculus are figuring out how to approach a problem, then doing the algebra. In this case, both are quite hard, but the calculus itself is quite easy.</p>