Find the maximum value of expression $13\sqrt{(x^2-x^4)}+9\sqrt{(x^2+x^4)}$

Question

Find the maximum value of expression $$13\sqrt{(x^2-x^4)}+9\sqrt{(x^2+x^4)}$$ for $0\le x \le 1$

My Attempt
$(x^2)(1 - x^2) \le \frac{x^2+(1 - x^2)}{2}=\frac{1}{2} \Rightarrow 13\sqrt{(x^2-x^4)} \le \frac{13}{\sqrt 2}$
Same way we obtain $9\sqrt{(x^2+x^4)} \le \frac{9}{\sqrt 2}\sqrt{1+2x^2} $
As $0\le x \le 1$ so $\sqrt{1+2x^2} \le \sqrt{3}$ .
Hence, maximum is $\frac{1}{\sqrt{2}}(13+9\sqrt{3})$ occurs at $x=\frac{1}{\sqrt{2}}$.
I then checked the value in wolfram alfa and found that maximum is $16$ and occurs at $x=\frac{2}{\sqrt{5}}$.
Can anyone point my mistake. Thanks in advance.

Answer 1

$f(x)=13\sqrt{x^2-x^4}+9\sqrt{x^2+x^4},\;x\in[0,1]$

$$f'(x)=\frac{9 \left(4 x^3+2 x\right)}{2 \sqrt{x^4+x^2}}+\frac{13 \left(2 x-4 x^3\right)}{2 \sqrt{x^2-x^4}}=0$$ as $x>0$ we can simplify $$\frac{18 x \left(2 x^2+1\right)}{2 x \sqrt{x^2+1}}+\frac{26 x \left(1-2 x^2\right)}{2 x \sqrt{1-x^2}}=0 \to \frac{9 \left(2 x^2+1\right)}{\sqrt{x^2+1}}=\frac{13 \left(2 x^2-1\right)}{\sqrt{1-x^2}}$$ Substitute $x^2= u-1$ as $0\le x^2\le 1$ then $1\le u\le 2$ $$\frac{9 (2 (u-1)+1)}{\sqrt{u}}=\frac{13 (2 (u-1)-1)}{\sqrt{2-u}}$$ square both sides, reorder and simplify

$500 u^3-1500 u^2+1125 u-81=0$

Factor $(5 u-9) \left(100 u^2-120 u+9\right)=0$

gives $u_1=\frac{9}{5},u_2=\frac{3}{10} \left(2-\sqrt{3}\right),u_3=\frac{3}{10} \left(2+\sqrt{3}\right)$

$u_2$ is out of the interval $(1,2)$

so we have $x^2=u_1-1=\dfrac{4}{5} \to x_1=\dfrac{2}{\sqrt 5}$ and

$x^2=u_3-1 \to x_2=\dfrac{1}{10} \sqrt{30 \sqrt{3}-40}$

$f(x_1)=16;\;f(x_3)=\dfrac{1}{10} \sqrt{7974 \sqrt{3}-8168}\approx 7.51$

Maximum value of $f(x)$ in $[0,1]$ is $f\left(\dfrac{2}{\sqrt 5}\right)=16$

hope it is useful

Answer 2

Yes, for $x=\frac{2}{\sqrt5}$ we get a value $16$.

We'll prove that it's a maximal value.

We need to prove that $$13x\sqrt{1-x^2}+9x\sqrt{1+x^2}\leq16$$ or $$44x^4-125x^2+128\geq117x^2\sqrt{1-x^4}$$ or $$(5x^2-4)^2(625x^4+560x^2+1024)\geq0.$$

Also, we can use C-S and AM-GM:

Let $x=\frac{2}{\sqrt5}t$.

Thus, $$13x\sqrt{1-x^2}+9x\sqrt{1+x^2}=\frac{2}{5}t\left(13\sqrt{5-4t^2}+9\sqrt{5+4t^2}\right)=$$ $$=\frac{2}{5}t\left(13\sqrt{5-4t^2}+27\sqrt{\frac{5}{9}+\frac{4}{9}t^2}\right)\leq$$ $$\leq\frac{2}{5}t\sqrt{(13+27)\left(13(5-4t^2)+27\left(\frac{5}{9}+\frac{4}{9}t^2\right)\right)}=$$ $$=16t\sqrt{2-t^2}\leq16\cdot\frac{t^2+2-t^2}{2}=16.$$