To the Max

Calculus Level 2

What is the maximum value of $\sqrt{x(4-x)}$ ?

-2 None of these choices 2 The maximum value does not exist 3

2 solutions

Rishabh Jain
Jun 15, 2016

$x(4-x)\ge 0\implies x\in[0,4]$ $\sqrt{x(4-x)}\le \dfrac{\not x+(4-\not x)}{2}=2$ $\mathcal{ \color{#D61F06}{By~GM\le AM}}$

$\therefore \left[\sqrt{x(4-x)}~\right]_{\text{max}}=\large{2}~@~x=2$

$\mathcal{ALT:-}$ Since $x\in[0,4]$ , substitute $x=4\sin^2\alpha$ s.t $\alpha\in\left[0,\dfrac{\pi}2\right]$ .

$f(\alpha)=4\sin~\alpha\cos~\alpha=2\sin 2\alpha$

whose maximum is obviously $2$ @ $x=\dfrac{\pi}4$ .

Rishabh Jain - 5 years ago

Rishabh Tiwari
Jun 15, 2016

Given ,

$f(x)= \sqrt{4x - x^{2}}$ ,

Now, inside the square root we can see a quadratic expression whose $' a ' < 0$ ,

Therefore a maxima occurs at $x= \dfrac{-b}{2a}$ $=$ $\dfrac{-4}{-2}$ $= 2$ ; also $x=2$ is within the domain of the function , hence we proceed.

Plugging in $x=2$ in $f(x)$ gives $f(2)= \color{#3D99F6}{\boxed{2}}$ ; Hence the answer.