Tiggsy

The question says that x and y are real numbers, it does not say that they are integers (necessarily).

The minimum is 3 + 6sqrt(2), approximately 11.585, at x = 1 + 3sqrt(2), y = 1 + 3sqrt(2)/2, approximately (5.243, 3.121).

For the chord of length a, take the isosceles triangle formed by its end points and the centre of the circle, and drop a vertical from the centre of the circle to its mid-point. Then, from one of the smaller triangles created,

\(\displaystyle \frac{(a/2)}{r}=\sin(15)\quad \text{so}\quad a=2r\sin(15)\)

where r is the  radius of the circle and the angle (and all following angles) is in degrees.

Similarly,

\(\displaystyle b=2r\sin(45)\quad\text{and}\quad c=2r\sin(\theta),\)

where theta is half of the angle subtended by the chord of length c at the centre of the circle.

Since c = a + b,

\(\displaystyle 2r\sin(\theta)=2r\sin(15)+2r\sin(45) \\\sin(\theta)=\sin(15)+\sin(45)=2\sin(30)\cos(15)\\ \qquad \text{(using the identity sin(A) + sin(B) = 2sin((A+B)/2)cos((A-B)/2) ) }\\ \hspace{25pt} = 2.(1/2).\sin(75) \\ \hspace{25pt}=\sin(75).\)

So the angle subtended by the chord length c at the centre of the circle is 2 times 75 = 150 deg.

Square the equation and replace the cos squared term using the identity

\(\displaystyle \cos^{2}(A/2)=(1/2)(1+\cos(A)).\)

Use the cosine rule to replace the cosine term with (b^2 + c^2 - a^2)/(2bc), and simplify.

That should get you to 1 - a^2/((b + c)^2)), so finally cos^2 = 1 - sin^2.

Sorry, but your assumption is not correct.

There are two mistakes in your application of the Newton-Raphson formula.

The formula is

\(\displaystyle x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}.\)

The sign in front of the second term on the rhs is a negative.

The other mistake is in the differentiation of f(x), the term xsin(x) should be differentiated as a product, 

there should be three terms on the bottom line of that fraction.

The second part of the question is asking you if you understand the basis of the method, and that sometimes it doesn't work.

Draw the tangent to curve at x = 5, the second approximation will be the value of x where this crosses the x-axis, somewhere just above x = 4.

Repeat the process, the next approximation will be where the tangent to the curve for this value of x crosses the x-axis, somewhere close to x = 5.

Repeating the process just gets you an oscillation about the minimum point of the curve, you never get down to the root close to x = 1.

Here's a different method, Newton Interpolation, sort of.

Let

 \(\displaystyle f(x) = a_{0}+a_{1}(x-1)+a_{2}(x-1)(x-2)+a_{3}(x-1)(x-2)(x-3)+(x-1)(x-2)(x-3)(x-4)\)

(The coefficient for the final term will be 1 since the polynomial is monic).

Sustituting x = 1, 2, 3 ,4, gets us

\(\displaystyle f(1)=1=a_{0}, \qquad\text{ so }\quad a_{0}=1,\\ f(2)=2=a_{0}+a_{1},\qquad \text{ so }\quad a_{1}=1,\\ f(3)=3=a_{0}+2a_{1}+2a_{2}\qquad \text{ so }\quad a_{2}=0,\\ f(4)=4=a_{0}+3a_{1}+6a_{2}+6a_{3} \qquad\text{ so }\quad a_{3}=0.\)

So,

\(\displaystyle f(x)=1+(x-1)+(x-1)(x-2)(x-3)(x-4)\\ \text{and therefore}\\ f(6)=1 + 5+5*4*3*2=126.\)

From #1, the difference between the roots 

\(\displaystyle = 2\sqrt{(b/2)^{2}-c}=\sqrt{b^{2}-4c}= |b-2c|.\)

Squaring,

\(\displaystyle b^{2}-4c=b^{2}-4bc+4c^{2},\)

from which

\(\displaystyle c(c-b+1)=0,\\ \text{so that} \\c=0, \text{ or }c=b-1.\)

\(\displaystyle c=0 \text{ is excluded, so }c=b-1.\)

The equation can be written as 

\(\displaystyle x^{2}+bx+b-1=(x+1)(x+(b-1))=0.\)

The roots will be

\(\displaystyle -1 \text{ and }1-b,\\\text{ so the modulus of the difference will be }\\|2-b|=|b-2(b-1)|=|b-2c|.\)

My guess, Guests, and Melody, is that it's the area of the triangle.

I know pretty much nothing about critical path analysis so this is simply a common sense view of the situation.

The problem seems to be with the join of C and D and specifically F from that point.

F depends upon D but not upon C, so activity F can be started as soon as D is finished and before C is finished.

That means you have to separate that join of C and D in some way. 

I don't know how, conventionally,  this is represented.

Your first step should be the substitution z = x + y.

That gets you the equation

\(\displaystyle \frac{dz}{dx}+ z=z^{2}e^{2x}. \)

Now use the usual Bernoulli substitution.

Thanks Melody,

  -  love the pic, what did you use to draw it ?