Problem Statement: Given the following problem diagram of a 5 meter ladder leaning against a 1 meter block and the side of a building, find the maximum height up the building the ladder can reach to the nearest 0.1 meter.

Using the Pythagorean Theorem (assuming angle DAE = 90 degrees) with the given diagram we can be developed Equation # 1:

and from the two semilar triangles of the problem diagram

we can develop Equation # 2:

By expanding Eq # 1

and reducing, we get a second degree polynomial in two unknowns:

Now when we substitute Eq # 2 in for " y " in the above reduction, we get

Multiplying by the lowest common denominator

results in

which can be rearranged to the 4th Degree Polynomial

Now the question is " HOW TO SOLVE IT ? "

Open Microsoft Excel and choose positive values for x in increments of 0.05 for the
interval [1 , 5]. These values of x were chosen because the ladder is only 5 meters long
and the cube is a 1 meter square, so 1 <= x+1 <= 5.

You can see there is a sign change from - to + between 3.8 and 4.

By changing the step size from 0.2 to 0.005 and looking at the interval [ 3.8 , 4 ]
we get a more accurate estimate of the root of the cubic polynomial.

The solution is somewhere between 3.835 and 3.84.

Newton's Method comes from " the Calculus "

Make an initial guess at the zero of the given polynomial and use Newton's iterative
algorithm to generate better guesses at the solution. When the Function Value is equal
to zero, you know you have the solution.

Newton's Method uses the following algorithm:

where x(0) is your initial guess at the root, f(x(0)) is the value of the polynomial function
at your initial guess, and f ' (x(0)) is the value of the first derivative of the polynomial
function at the initial guess.

Using Newton's Method took 8 iterations to get eight decimal places of accuracy.

Solution: x = 3.83850116

Using Algebra Expressor to graph Equation # 1 and Equation # 2 we get:

Which clearly shows 4 intersections between the two functions. Since the negative intercepts
don't make any sence for the problem, we will concentrate on the two positive intercepts.

And " ZOOM " ing in on the larger value

we get an answer of approximately x = 3.84.

CONCLUSIONS:

Multiple approaches can be used to solve this problem. By far, Algebra Expressor gives
you the best graphic of what is going on.


NOTE: Newton's Method and the Bisection Method using MATLAB will be added in the near future. These will be computer programs that will run on the standard version of MATLAB currently being used by the Mathematics Department.