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January 30th, 2012, 11:43 PM   #1
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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 right-angled 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.
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January 31st, 2012, 02:48 PM   #2
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[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 + t) metres.
For t > 0, it's straightforward to use calculus to show that AB 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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