# Break-even point - using Newton method

• Mar 26th 2009, 10:54 PM
billyso23
Break-even point - using Newton method
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.

• Mar 27th 2009, 12:44 AM
CaptainBlack
Quote:

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.

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
• Mar 27th 2009, 01:04 AM
earboth
Quote:

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.

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

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

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

$x_0 = 100$
.
.
$x_4 = 138.0647696$
• Mar 27th 2009, 06:15 PM
billyso23
Thanks the for helps.

earboth , i have 1 questions for you.

• Mar 27th 2009, 11:12 PM
earboth
Quote:

Originally Posted by billyso23
Thanks the for helps.

earboth , i have 1 questions for you.

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.
$\begin{array}{lcl}x_1&=&100 \\ x_1&=&143.3617641 \\ x_2&=& 138.1229449 \\ ...&=&... \\ x_4&=& 138.0647696\end{array}$