Break-even point - using Newton method

**billyso23** · March 26th 2009, 10:54 PM

Revenue: $R(x) = 10x ^{2/3}$
Cost: $C(x) = 2x-9$

Finding the break-even point, i have the following:

R(x) = C (x)
R(x) - C (x) = 0

Therefore, the new function is $10x^{2/3} - 2x + 9$
The derivative is $(20/3)*x^{-1/3} -2$

It is obvious to me, the answer is negative. However, you can have a negative answer for such question.

Please helps

**CaptainBlack** · March 27th 2009, 12:44 AM

Originally Posted by billyso23

Revenue: $R(x) = 10x ^{2/3}$
Cost: $C(x) = 2x-9$

Finding the break-even point, i have the following:

R(x) = C (x)
R(x) - C (x) = 0

Therefore, the new function is $10x^{2/3} - 2x + 9$
The derivative is $(20/3)*x^{-1/3} -2$

It is obvious to me, the answer is negative. However, you can have a negative answer for such question.

Please helps

That is not at all obvious, it has a root close to 138. Opps.. wrong sign on the 9, should be near 1.225, Opps... original sign was right, just goes to show that you should never pay attention to the other replies (which have been deleted so you won't know what I am talking about).

CB

**earboth** · March 27th 2009, 01:04 AM

Originally Posted by billyso23

Revenue: $R(x) = 10x ^{\frac23}$
Cost: $C(x) = 2x-9$

Finding the break-even point, i have the following:

R(x) = C (x)
R(x) - C (x) = 0

Therefore, the new function is $10x^{\frac23} - 2x + 9$
The derivative is $(20/3)*x^{-1/3} -2$

It is obvious to me, the answer is negative where do you know this from? . However, you can have a negative answer for such question.

Please helps

1. If you use the substitution $x = \sqrt[3]{y}$ you'll get the equation:

$-2y^3+10y^2+9=0~\implies~y^3-5y^2-\frac92 = 0$

You now can use the Cardanic formula to solve this equation: To get the reduced equation you have to use the substitution $y = t+\frac53$ . The discriminant has the value $D= \frac{1243}{48}$ . Thus you know that there is only one real solution (and two conjugate complex solutions)

2. If you aren't familiar with the Cardanic formulae you can use an iteration like Newton method:

The equation f(x) = 0 has the solution

$x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$

With your equation you get:

$x_{n+1} = x_n - \dfrac{10x^{\frac23} - 2x +9}{\frac{20}3 x^{-\frac13} - 2}$

Start with
$x_0 = 100$
.
.
$x_4 = 138.0647696$

**billyso23** · March 27th 2009, 06:15 PM

Thanks the for helps.

earboth , i have 1 questions for you.

a) What's your reason to pick 100 as your initial guess?

**earboth** · March 27th 2009, 11:12 PM

Originally Posted by billyso23

Thanks the for helps.

earboth , i have 1 questions for you.

a) What's your reason to pick 100 as your initial guess?

to a): I've made a rough sketch of the graph of $f(x)=10x^{\frac23}-2x+9$ to estimate the value of the zero.

To answer the other question:
$\begin{array}{lcl}x_1&=&100 \\ x_1&=&143.3617641 \\ x_2&=& 138.1229449 \\ ...&=&... \\ x_4&=& 138.0647696\end{array}$

**CaptainBlack** · March 28th 2009, 02:00 AM

I must say that I disagree with deleting posts (I'm looking at you MrF) because they are wrong, we can all learn a lot from our own and other's mistakes, and there is always the option of editing to correct the mistakes (you are allowed to edit now!)

CB

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