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Let $x\in \mathbb{R}$, find the function maximum of the value $$f(x)=\dfrac{\sqrt{3}}{4}x^2+\dfrac{\sqrt{(9-x^2)(x^2-1)}}{4}$$

my attemp $$x^2=5+4\sin{t},t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$ then $$f=\dfrac{5\sqrt{3}}{4}+2\sin{\left(t+\frac{\pi}{6}\right)}\le 2+\dfrac{5}{4}\sqrt{3}$$

My Question:this function have other methods to find this maximum? such as AM-GM,Cauchy-Schwarz inequality and so on?

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  • $\begingroup$ I don't understand your working. Are you trying to find a local maximum of $f(x)$? $\endgroup$ – actinidia Nov 27 '17 at 4:30
  • $\begingroup$ @TiwaAina,yes ,it is $\endgroup$ – communnites Nov 27 '17 at 4:31
  • $\begingroup$ Did you use derivative? $\endgroup$ – Nosrati Nov 27 '17 at 4:34
  • $\begingroup$ I think your approach might work, but be careful. Domain of your function is $[-3,-1]\cup[1,3]$ and therefore if corresponding parameter $t$ provides an $x$ in this domain then you are done. $\endgroup$ – Bumblebee Nov 27 '17 at 4:37
  • $\begingroup$ The domain of function is $D_f=[-3,-1]\cup[1,3]$ then $1\leq x^2\leq9$ and with $ab\leq\left(\dfrac{a+b}{2}\right)^2$ we have $$f(x)=\dfrac{\sqrt{3}}{4}x^2+\dfrac{\sqrt{(9-x^2)(x^2-1)}}{4}\leq\dfrac{\sqrt{3}}{4}9+\dfrac{8}{4}$$ so the maximum is less than this value! $\endgroup$ – Nosrati Nov 27 '17 at 4:38
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There's always the classic, but perhaps computationally icky derivative approach.$$f(x)=\dfrac{\sqrt{3}}{4}x^2+\dfrac{\sqrt{(9-x^2)(x^2-1)}}{4}$$ $$f'(x)= \frac{\sqrt3}2 x + \frac{20 x - 4 x^3}{ 8 \sqrt{(9 - x^2) (-1 + x^2)}} = \frac 12 x \, \Big(\sqrt{3} + \frac{5 - x^2}{\sqrt{- x^4 - 10 x^2 - 9 }}\Big)$$

Let $f'(x) = 0$ and solve to find the local extrema:

Clearly $x=0$ is a solution, but doesn't exist in the domain of $f$, which is $[−3,−1]∪[1,3]$.

So we find solutions to $$\sqrt{3} + \frac{5 - x^2}{\sqrt{- x^4 - 10 x^2 - 9 }} = 0$$ or equivalently $$ \sqrt3 \sqrt{-x^4+10 x^2-9} = 5 - x^2 $$ which comes to $$3 (-x^4+10 x^2-9)=(5-x^2)^2$$ $$-3 x^4+30 x^2-27=x^4-10 x^2+25$$ $$-4 x^4+40 x^2-52=0$$ Complete the square to get $$(x^2-5)^2=12$$ so $$x = \pm \sqrt{5 \pm 2 \sqrt 3}$$

Checking for extraneous solutions reveals that the only values of $x$ satisfying $f'(x) = 0$ are $$x = \pm \sqrt{5 +2 \sqrt 3}$$

Now plug in these values to get

$$f_{\text{max}} = f(\pm \sqrt{5 +2 \sqrt 3}) = \frac{8 + 5 \sqrt3}4$$ which is in line with MyGlasses's observation that $f_{\text{max}}$ must be no more than $\dfrac{\sqrt{3}}{4}9+\dfrac{8}{4}$.

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Consider the function and its first derivative$$f(x)=\frac{\sqrt{3}}{4}x^2+\frac{\sqrt{(9-x^2)(x^2-1)}}{4}$$ $$f'(x)=\frac{1}{2} x \left(\frac{5-x^2}{\sqrt{-x^4+10 x^2-9}}+\sqrt{3}\right)$$ The first derivative cancels for $x=0$ which has to be excluded.

Now, let $x^2=y$ and solve $$\frac{5-y}{\sqrt{-y^2+10 y-9}}+\sqrt{3}=0\implies 5-y=-\sqrt{3}\sqrt{-y^2+10 y-9}$$ and square to get $$4 y^2-40 y+52=0 \implies y=5\pm2 \sqrt{3}$$ Only $y=5+2 \sqrt{3}$ must be kept because of the real domain.

$$f(\sqrt{5+2 \sqrt{3}})=2+\frac{5 \sqrt{3}}{4}$$

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Hint:

$$4f(x)-5\sqrt3=\sqrt{16-(x^2-5)^2}+\sqrt3(x^2-5)$$

Now set $x^2-5=4\cos t,0\le t\le\dfrac\pi2$

We can prove $$a\cos t+b\sin t\le\sqrt{a^2+ b^2}$$

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  • $\begingroup$ How did you arrive at that substitution? $\endgroup$ – actinidia Nov 27 '17 at 5:08
  • $\begingroup$ @Tiwa, After completing the square , I found $$(x^2-5)^2\le34$$ $\endgroup$ – lab bhattacharjee Nov 27 '17 at 5:09
  • $\begingroup$ That's wrong: $(9-x^2)(x^2-1)=-x^4+10x^2\color{red}{-9}=16-(x^2-5)^2$, and you'll end up with the same solution as the OP. $\endgroup$ – Professor Vector Nov 27 '17 at 7:38
  • $\begingroup$ @ProfessorVector, Thanks for your rectification. But at what point I'm mistaken resulting in wrong answer? $\endgroup$ – lab bhattacharjee Nov 27 '17 at 7:47
  • $\begingroup$ I've marked it in red. Your wrong completing the square would mean a $+9$ in that place. $\endgroup$ – Professor Vector Nov 27 '17 at 7:53
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We can simplify the problem by finding the minimun of

$$ g(t) = \frac{\sqrt{3}}{4}t + \frac{\sqrt{(9-t)(t-1)}}{4} $$

with $t = x^2, \ t \in [1,9]$

A straightforward approach is just to take the first derivative

$$ g'(t) = \frac{\sqrt{3}}{4} + \frac{5-t}{4\sqrt{-9+10t-t^2}} $$

Solving $g'(t)=0$ leads to

$$ \sqrt{16 - (t-5)^2} = \frac{t-5}{\sqrt{3}} \quad (t > 5) $$

which gives $t = 5 + 2\sqrt{3}$

Thus, the maximum is $$g(5+2\sqrt{3}) = \frac{\sqrt{3}}{4}(5+2\sqrt{3}) + \frac{1}{2} = 2 + \frac{5\sqrt{3}}{4}$$

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