Find min and max for $f(x)=x^2*(35-x)^5$. We may use calculus.

Question

I have got the following function: $f(x)=x^2(35-x)^5$

I need to find the points that
$$f'(x)=0$$ in order to find extrema points, but I cannot find the derivative due to exponents!

Problem: I broke $(35-x)^5$ via binomial expansion ,but I think the approach was of no use..I have got higher order of polynomial expression .I dont know how to solve

So how should I approach in order to find its extrema using its derivative point in the least costly way??

Thank You

It is not difficult to find the roots of the derivative, they should be 35, 0 and 10.... — dmtri, Aug 12 '18 at 16:10
Just use product rule. Expanding the term manually and derivating will make it hard to factorise. — prog_SAHIL, Aug 12 '18 at 16:27
You're probably confusing product rule with chain rule (where order matters). — kingW3, Aug 13 '18 at 06:42

mvw · Accepted Answer · 2018-08-12T16:29:37.983

0

Hint: $$ \begin{align} (f\circ g)'(x) = f(g(x))' &= f'(g(x))\, g'(x) \\ f(x) &= x^5 \\ g(x) &= 35-x \end{align} $$ Hint: $$ (f g)'(x) = f'(x) g(x)+f(x)g'(x) $$ Hint: $$ f'(x) = 7x(10-x)(35-x)^4 $$

edited Aug 12 '18 at 16:29

answered Aug 12 '18 at 16:00

mvw

34,562

ok Fine i derived the $(35-x)^5$ using above..But we got $x^2*(35-x)^5$ then which function should be choosen as first function here..I got LIATE rule to do which prevent me from moving foreward..I am confuse to choose ,as they both are algeberic functions – Shivanshu Raj Aug 12 '18 at 16:16
You might review your rules for taking a derivative. – mvw Aug 12 '18 at 16:17
But what about it...the first function...i mean if i got like $ sin^2\theta*(35-x)^2$then i wont ask such question.. – Shivanshu Raj Aug 12 '18 at 16:19

score 0 · Answer 2 · answered Aug 12 '18 at 16:15

0

$\frac{dy}{dx}$=$(-1)x^{2}(5(35-x)^{4})+((35-x)^{5})(2x)$

=$(35-x)^{4}(-5x^{2}+70x-2x^{2})$

$(35-x)^{4}=0$ or $(-7x^{2}+70x)=0$

$ x=0, x=10, x=35 $

Should be easy solve from here hope this helps.

answered Aug 12 '18 at 16:15

Robbie Meaney

141
9

So you choosed $x^2$ as first function and derivied the equation – Shivanshu Raj Aug 12 '18 at 16:23
I used the product rule. let u=$x^{2}$ and v=$(35-x)^{5}$ – Robbie Meaney Aug 12 '18 at 16:25
Then $f'(x)$=u$\frac{dv}{dx}$+v$\frac{du}{dx}$ where the chain rule was applied to differentiate v – Robbie Meaney Aug 12 '18 at 16:26
how did you assumed the first function and second function to apply product rule??I mean both of them are algeberic function – Shivanshu Raj Aug 12 '18 at 16:30
when you have a function in the form f(x)=u(x).v(x) you can apply product rule and u v can be in any order. Not being rude but you should familarise yourself with basic calculus before attempting to find max/mins of tougher functions. – Robbie Meaney Aug 12 '18 at 16:57

score 0 · Answer 3 · answered Aug 13 '18 at 06:15

0

No need for calculus: $p(x)=x^2(35-x)^5$ has a double root at $x=0$ and a quintuple root at $x=35$; these points are clearly stationary points. There is no stationary point outside the interval $[0,35]$ since $p(x)$ is monotonic for $x>35$ or $x<0$. In the interval $[0,35]$ there is a unique maximum for $p(x)\geq 0$: by the AM-GM inequality

$$\begin{eqnarray*} x^2(35-x)^5 &=& \frac{4}{25}\left[\left(\tfrac{5}{2}x\right)\left(\tfrac{5}{2}x\right)(35-x)(35-x)(35-x)(35-x)(35-x)\right]\\&\color{red}{\leq}&\frac{4}{25}\left[\frac{2\cdot\left(\tfrac{5}{2}x\right)+5\cdot(35-x)}{7}\right]^5=\frac{4}{25}\left[25\right]^5 = 4\cdot 5^8 = 1562500\end{eqnarray*}$$ and $\color{red}{\leq }$ holds as $\color{red}{=}$ iff $\frac{5}{2}x=35-x$, i.e. iff $x=10$.

answered Aug 13 '18 at 06:15

Jack D'Aurizio

353,855

I can't get with AM-GM inequilaty – Shivanshu Raj Aug 14 '18 at 08:16
@ShivanshuRaj: what can't you get? – Jack D'Aurizio Aug 14 '18 at 08:18
and how you can say that $p(x)$ is monotonic without knowing its derivative? – Shivanshu Raj Aug 14 '18 at 08:19
Am-GM is sounds new are you talking about Arthemetic mean or geometric mean?? – Shivanshu Raj Aug 14 '18 at 08:20
@ShivanshuRaj: the product of two positive and increasing functions is positive and increasing, you do not need derivatives to be able to state it. AM-GM is the standard name of the inequality between the arithmetic mean and the geometric mean. – Jack D'Aurizio Aug 14 '18 at 08:20

Find min and max for $f(x)=x^2*(35-x)^5$. We may use calculus.

3 Answers3