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January 30th, 2012, 11:43 PM  #1 
Senior Member Joined: Apr 2008 Posts: 193 Thanks: 3  my correct solution but another solution?
Let's consider the following optimization problem. This is how I solve it. Find the length of the shortest ladder that can extend from a vertical wall, over a fence 2 metres high located 1 metre away from the wall, to a point on the ground outside the fence. my solution Let x be the angle of inclination of the ladder from the ground. Using two rightangled triangles, we obtain the length L of the ladder as a function of x. L = 1/cos(x) + 2/sin(x) differentiating L with respect to x and setting L' to zero give 0 = sin(x)/(cos(x))^2  2cos(x)/(sin(x))^2 It follows that (tan(x))^3 = 2 Next, I use the identity (sec(x))^2 = 1 + (tan(x))^2 and the relationship sin(x) = tan(x)*cos(x) to find the answer. L = (1 + 2^(2/3))^(3/2) which is approximately equal to 4.16. I know my solution is correct. I am wondering if there is another way to solve the problem without using trigonometry. Can someone help me? Thanks. 
January 31st, 2012, 02:48 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,701 Thanks: 1804 
[attachment=0:1opi6z83]ladder.gif[/attachment:1opi6z83] In the above diagram, OBCA is a rectangle, EA = t metres and the ladder AB has length ((t + 1)/t)?(4 + t²) metres. For t > 0, it's straightforward to use calculus to show that AB² has a minimum when t = 2^(2/3). As the ladder is of minimum length, CF? AB. It can be deduced that t = 2^(2/3) without using calculus. Either approach confirms (without use of trigonometry) that your solution is correct. 

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