Question

Two triangular pens are built against a barn. Two hundred meters of fencing are to be used for the three sides and the diagonal dividing fence. What dimensions of maximize the area of the pen? What is the area of the pen? What is the length of the dividing fence? b- Newton's Method is a method to approximate the roots of a function. It states: 1. Choose an initial approximation X, as close to the root as possible, 2. For n = 0,1,2,3,..... f(x) 3. The root x, is found when x,, is very very close to X41. Usually within 4 decimal places. Find the root of f(x) = x - 5x + 1 with an initial guess of the root to be Xo = -2 I

Answer 1

According to the HOMEWORKLIB POLICY allowed to only one question solution

So I help with 3 Question

Here x3 - 5x + 1 = 0 Let f(x) = x3 - 5x + 1 + 4(x - 5x + 1) = 3x2.5 :: f (x) = 3x2-5 хо = 2 1st iteration : f(xo) = f(-2) = - 23 - (5 - 2) +1 = 3 f(x) = f(-2) = 3 -22-5 = 7 f(xo) X1 = Xo- Ư (xo) 3 X1 2 IN X1 = - 2.4286 2nd iteration : f(x1) = f(-2.4286) = - 2.42863 - (5 - 2.4286) + 1 = - 1.1808 f (x1) = f(-2.4286) = 3 - 2.42862 - 5 = 12.6939 X2 = X1 - f(x1) f (x2) - 1.1808 X2 = -2.4286 - 12.6939 Yg Sca2.3356 Caitseamer

3rd iteration : f(x2) = f(-2.3356) = - 2.33563 - (5 - 2.3356) + 1 = -0.0622 f (x2) = f(-2.3356) = 3 - 2.33562 - 5 = 11.3644 X3 = X2 - f(x2) f (x2) -0.0622 X3 = - 2.3356 - 11.3644 Х3 = = -2.3301 4th iteration : f(x3) = f(-2.3301) = - 2.33013 - (5 - 2.3301) + 1 = -0.0002 f (x3) = f(-2.3301) = 3 - 2.33012 - 5 = 11.2878 X4 = X3 - f(x3) (x3) -0.0002 X4 = -2.3301 - 11.2878 X4 = -2.3301 Approximate root of the equation x3 - 5x + 1 = 0 using Newton Raphson mehtod is -2.3301 n хо f(xo) f (xo) X1 Update لی -2 3 7 -2.4286 Xo = X1 -2.4286 -1.1808 12.6939 N -2.3356 Xo = X1 3 -2.3356 -0.0622 11.3644 -2.3301 Xo = X1 14 56-2.3301ith -0.0002 11.2878 -2.3301 Xo = X1 CamScanner

For A)

We have two pieces of fencing of length x, one piece of length y
and one piece of length √(x²+y²) (the diagonal)
Total fencing = 200
2x + y + √(x²+y²) = 200
Solve for y:
√(x²+y²) = 200 − 2x − y
Square both sides:
x² + y² = 4x² + 4xy − 800x + y² − 400y + 40000
x² = 4x² + 4xy − 800x − 400y + 40000
400y − 4xy = 3x² − 800x + 40000
(400−4x) y = 3x² − 800x + 40000
y = (3x² − 800x + 40000) / (400−4x)

Area = x * y
We can express area as a function of x only, by using value of y found above:
A = x (3x² − 800x + 40000) / (400−4x)
A = (3x³ − 800x² + 40000x) / (400−4x)
A' = (−3x³ + 850x² − 80000x + 2000000) / (2(100−x)²) = 0
−3x³ + 850x² − 80000x + 2000000 = 0
x ≈ 38.8142
y = (3x² − 800x + 40000) / (400−4x) ≈ 55.0302
√(x²+y²)≈ 67.341406532

Area of pen is maximized when
two ends = 38.8142 m
end opposite barn = 55.0302 m
diagonal = 67.3414 m

Maximum area = x * y = 2135.95 m²