Find the maximum value of $2x + 2\sqrt{x(1-x)}$ when $0 \leq x \leq 1.$
I'm trying to find the upper bound of the expression, but I'm having trouble on how to start/how to solve the problem. Can someone help? Thanks in advance!
Find the maximum value of $2x + 2\sqrt{x(1-x)}$ if $0 \leq x \leq 1.$
algebra-precalculus
Related Solutions
It's essentially a parameter substitution. Let $c = 2b$. Then the question becomes
Maximise $\tfrac{1}{2}ac$ given that $a+c=3$.
By symmetry you can argue that the answer must be $a=c$ and thus $a=3/2$, $c=3/2$.
The reason for this is that if you had an asymmetrical solution, $a=x$ and $c=3-x$ then there would be another solution, $a=3-x$ and $c=x$. Therefore there is no solution at all (e.g. consider the same problem with "minimize" instead of "maximize") or there is a unique, symmetric solution.
Finally you undo the substitution, getting $b = c/2 = 3/4$.
Graphing the constraints in blue and the iso-$z$ lines in red, we get:
the values of $z$ increasing for larger $x$ and $y$, as expected from the formula. We can find the maximum $z$ at $(5,11)$.
Best Answer
Hint : Try to plug $x = cos^{2}(\theta)$ and use some trigonometry to see what can be the maximum value of $sin(A) + cos(A)$.
I hope this helps. If you are stuck, let me know!
Update : Since the OP seems not to follow the hint and work through it, I'll give the complete solution:
Plug in $x = cos^{2}(\theta)$, then we get:
$ 2x + 2\sqrt{x(1-x)} = 2cos^{2}(\theta) + 2 \sqrt{cos^{2}(\theta)(1-cos^{2}(\theta))} = 2cos^{2}(\theta) + 2 \sqrt{cos^{2}(\theta)sin^{2}(\theta)}$ $\mbox{ } $ , using the identity $cos^{2}(\theta) + sin^{2}(\theta) = 1$
$ = 2cos^{2}(\theta) + 2 sin(\theta) cos(\theta) = [1 + cos(2\theta)] + sin(2\theta)$ $\mbox{ }$ (using the identities : $cos(2\theta) = 1 - 2cos^{2}(\theta)$ and $sin( 2\theta) = 2 sin(\theta) cos(\theta)$ )
Now we find maximum of $cos(2\theta) + sin(2\theta)$ :
$cos(2\theta) + sin(2\theta) = \sqrt{2} [1/\sqrt{2} sin(2\theta) + 1/\sqrt{2} cos(2\theta)] = \sqrt{2} [cos(\pi/4) sin(2\theta) + sin(\pi/4) cos(2\theta)] = \sqrt{2}sin( \pi/4 + 2\theta)$
using the identity $sin( A +B) = sin(A)cos(B) + cos(A)sin(B)$
Since maximum value of $sin( \pi/4 + 2\theta) = 1$, $cos(2\theta) + sin(2\theta) \leq \sqrt{2}$.
Thus, $ 2x + 2\sqrt{x(1-x)} \leq 1 + \sqrt{2}$.