Question Statement

A rectangular field has an area of 2500 square meters, and the length of the field is 20 meters longer than the width. Find the width of the field using the Newton-Raphson method (correct to two decimal places). The initial guess is $x_0=20$.

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Background and Explanation

The Newton-Raphson method is an iterative technique for finding successively better approximations to the roots (or zeroes) of a real-valued function $f(x)=0$. It requires an initial guess and uses the function's derivative to converge rapidly to the solution.

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Solution

Setting Up the Equation

Let $x$ meters be the width of the rectangle. Then the length of the rectangle is $(x+20)$ meters.

Given that the area of the rectangle is 2500 square meters:

$x(x+20)=2500$

Expanding and rearranging into standard form:

$x^2+20x-2500=0$

Thus, we define the function:

$f(x)=x^2+20x-2500$

Deriving the Iteration Formula

The Newton-Raphson formula is:

$x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime }(x_n)}$

where $f^{\prime }(x_n)\ne 0$ and $n=0,1,2,\dots$

First, we find the derivative of $f(x)$:

$f^{\prime }(x)=2x+20$

Therefore:

• $f(x_n)=x_n^2+20x_n-2500$
• $f^{\prime }(x_n)=2x_n+20$

Substituting into the Newton-Raphson formula:

$\begin{array}{rl}{x}_{n+1}& =x_n-\frac{x_n^2+20x_n-2500}{2x_n+20}\\ & =\frac{x_n(2x_n+20)-(x_n^2+20x_n-2500)}{2x_n+20}\\ & =\frac{2x_n^2+20x_n-x_n^2-20x_n+2500}{2x_n+20}\\ x_{n+1}& =\frac{x_n^2+2500}{2x_n+20}\end{array}$

Iteration 1 ($n=0$)

Using the initial guess $x_0=20$ in equation (i):

$\begin{array}{rl}{x}_1& =\frac{x_0^2+2500}{2x_0+20}\\ & =\frac{(20{)}^2+2500}{2(20)+20}\\ & =\frac{400+2500}{40+20}\\ & =\frac{2900}{60}\\ x_1& =48.3333\end{array}$

Iteration 2 ($n=1$)

Using $x_1=48.3333$:

$\begin{array}{rl}{x}_2& =\frac{x_1^2+2500}{2x_1+20}\\ & =\frac{(48.3333{)}^2+2500}{2(48.3333)+20}\\ & =\frac{2336.111+2500}{96.6666+20}\\ & =\frac{4836.111}{116.6666}\\ x_2& =41.4524\end{array}$

Iteration 3 ($n=2$)

Using $x_2=41.4524$:

$\begin{array}{rl}{x}_3& =\frac{x_2^2+2500}{2x_2+20}\\ & =\frac{(41.4524{)}^2+2500}{2(41.4524)+20}\\ & =\frac{1718.301+2500}{82.9048+20}\\ & =\frac{4218.301}{102.9048}\\ x_3& =40.9922\end{array}$

Iteration 4 ($n=3$)

Using $x_3=40.9922$:

$\begin{array}{rl}{x}_4& =\frac{x_3^2+2500}{2x_3+20}\\ & =\frac{(40.9922{)}^2+2500}{2(40.9922)+20}\\ & =\frac{1680.360+2500}{81.9844+20}\\ & =\frac{4180.360}{101.9844}\\ x_4& =40.9901\end{array}$

Conclusion

Comparing the values:

• $x_3=40.9922\approx 40.99$ (to two decimal places)
• $x_4=40.9901\approx 40.99$ (to two decimal places)

Since $x_3$ and $x_4$ agree to two decimal places, the iteration has converged.

Therefore, the width of the rectangular field is 40.99 meters (correct to two decimal places).

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Key Formulas or Methods Used

• Newton-Raphson Method: $x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime }(x_n)}$
• Quadratic Equation Setup: $f(x)=x^2+20x-2500=0$ derived from area constraint $x(x+20)=2500$
• Derivative: $f^{\prime }(x)=2x+20$
• Simplified Iteration Formula: $x_{n+1}=\frac{x_n^2+2500}{2x_n+20}$

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Summary of Steps

1. Formulate the equation: Set up $x(x+20)=2500$ and rearrange to $f(x)=x^2+20x-2500=0$
2. Find the derivative: Calculate $f^{\prime }(x)=2x+20$
3. Derive the iteration formula: Substitute into Newton-Raphson to get $x_{n+1}=\frac{x_n^2+2500}{2x_n+20}$
4. Iterate: Start with $x_0=20$ and calculate successive approximations:
   • $x_1=48.3333$
   • $x_2=41.4524$
   • $x_3=40.9922$
   • $x_4=40.9901$
5. Check convergence: Stop when consecutive values agree to the required precision (two decimal places)
6. State the answer: The width is approximately 40.99 meters