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

Answer 1

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: $$ f'(x) = 7x(10-x)(35-x)^4 $$

Answer 2

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

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

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

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

Should be easy solve from here hope this helps.

Answer 3

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 $x=10$.