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Find the maximum value of $ \ P(t) \ $ given by

$P(t)=0.0000000 219 t^4-0.0000 167 t^3+0.00155t^2+0.002t+0.23 \ $

For what value of $ \ t\ $ , $ \ P(t) \ $ is maximum ?

Answer:

For maximum value of $ P(t) \ $,

$ P(t)'=0 \\ \Rightarrow 4 \times 0.0000000219 \ t^3-3 \times 0.0000167 \ t^2+0.00155 \ t+0.002 =0 \ $

But from this how to get the solution ?

Is Newton-raphson method applicable here ?

I need help .

MAS
  • 10,638
  • your derivative is with error, check it again then factorize using technique explained here:https://math.stackexchange.com/questions/2118291/how-to-factorize-this-cubic-equation; you will get two or more intervals, in each intervals find the sign of the slope/derivative and then find max – Vikram Oct 11 '17 at 06:01

1 Answers1

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$$P(t)'=0 \Rightarrow 4 \times 0.0000000219 \ t^3-3 \times 0.0000167 \ t^2+2\times 0.00155 \ t+0.002 =0 $$

$t_1= -0.638564; \; t_2 = 71.3329;\; t_3 = 501.223$

To understand which is maximum second derivative test is the best method

$P''(t)=0.0031 - 0.0001002 t + 2.628\times 10^{-7} t^2$

$P''(t_1)=0.00316409>0$ then $t_1$ is a minimum

$P''(t_2)=-0.00271033<0$ then $t_2$ is a maximum

$P''(t_3)=0.0188993>0$ then $t_3$ is another minimum

Hope this can be useful

Edit

How did I find the roots?

Well, I'll show $t_2$ which is the maximum

First I made a table of values to look for signs changing. Like this

$ \begin{array}{l|l} x & f'(x) \\ \hline 10 & 0.0280776 \\ 20 & 0.0446608 \\ 30 & 0.0522752 \\ 40 & 0.0514464 \\ 50 & 0.0427 \\ 60 & 0.0265616 \\ 70 & 0.0035568 \\ 80 & -0.0257888 \\ 90 & -0.0609496 \\ 100 & -0.1014 \\ \end{array} $

Therefore a root is in the interval $[70,80]$

I set $t_0=70$ and then, recursively

$$t_{n+1}=t_n-\frac{f'(t_n)}{f''(t_n)}$$

we get

$ \begin{array}{l|l} n & t_n\\ \hline 0 & 70 \\ 1 & 71.3543 \\ 2 & 71.3329 \\ 3 & 71.3329 \\ \end{array} $

so $t\approx 71.3329$

Raffaele
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