Find minimum and maximum value of $\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$

Question

Blockquote

Find min,max of function $$T=\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$$

Max

By squaring both side and AMGM:

$$T^2=7+2\sqrt{5\cos^2x+1}\cdot \sqrt{5\sin^2x+1}$$

$$\le 7+5\left(\cos^2x+\sin^2x\right)+2=14$$

Or $$T\le \sqrt {14}$$

About minimal value:By $\sqrt x +\sqrt y \ge \sqrt{x+y}$

So $T\ge \sqrt{5+1+1}=\sqrt 7$

I think the minimal value has a little wrong. Pls help me.

Does this answer your question? The maximum and minimum values of the expression — Martin R, May 18 '20 at 13:01
The minimum is right. It's the maximum that may be wrong. You should have $1+\sqrt 6.$ — Allawonder, May 18 '20 at 14:07
@Allawonder: Unless I am mistaken, the maximum is right and the minimum is $1+\sqrt 6$. It is the case $a=\sqrt 6$, $b=1$ from the above-referenced (and possible duplicate target) Q&A. — Martin R, May 18 '20 at 14:19
In fact, you are right. My above comment about the maximum is wrong. The maximum is $\sqrt 14.$ I made a silly mistake. — Allawonder, May 18 '20 at 15:17
@MartinR Yes, I was wrong about the maximum. However, the minimum is $\sqrt 7,$ which is attained when $\sin x=-1/5.$ — Allawonder, May 18 '20 at 15:19
@Allawonder: If $\sin x = -1/5$ then $\sin^2 x = 1/25$ and $\cos^2 x = 24/25$. I do not see how you get $T=\sqrt 7$ in that case. – Also in your answer it seems that you substituted $x$ for $\sin^2 x$, and that cannot be negative. — Martin R, May 18 '20 at 15:26
@MartinR Oh, that's true. I was all the while thinking of a substitution $\sin x\mapsto x.$ Thanks. — Allawonder, May 18 '20 at 15:44

score 1 · Accepted Answer · answered May 18 '20 at 14:20

1

For minimum,

$$T=\sqrt{(\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1})^2}\\ =\sqrt{7+2\sqrt{25\cos^2x\sin^2x+6}}\\ \ge \sqrt{7+2\sqrt6}=1+\sqrt6 $$

answered May 18 '20 at 14:20

Quanto

97,352

score 1 · Answer 2 · answered May 18 '20 at 15:39

The quantity $T(x)\geq0$ is a ${\pi\over2}$-periodic function of $x$. It is extremal, i.e., maximal or minimal, iff $$T^2=7+2\sqrt{(5\cos^2 x+1)(5\sin^2 x+1)}$$ is extremal, and this is the case iff the radicand on the RHS is extremal. Letting $\sin^2 x=:t$ this means that $$\phi(t):=\bigl(5(1-t)+1\bigr)(5t+1)=6+25(t-t^2)\qquad(0\leq t\leq1)$$ should be extremal. This $\phi(t)$ takes its minimum $6$ at $t=0$ and $t=1$, and takes its maximum $6+{25\over4}={49\over4}$ at $t={1\over2}$.

From $T^2=7+2\sqrt{6}=\bigl(1+\sqrt{6}\bigr)^2$ it follows that the minimum of $T$ is $1+\sqrt{6}$. Similarly, from $$T^2=7+2\cdot\sqrt{49\over4}=14$$ it follows that the maximum of $T$ is $\sqrt{14}$.

Allawonder · Answer 3 · 2020-05-18T15:45:32.767

0

Hint. That's equivalent to extremising $$\sqrt{6-5x}+\sqrt {1+5x}$$ on the interval $[0,1].$ Can you now continue?

edited May 18 '20 at 15:45

answered May 18 '20 at 13:56

Allawonder

13,327

but if $x=1/2$ we got $T=\sqrt {14}>1+\sqrt 6$ – user774564 May 18 '20 at 14:09
@user774564 That was a mistake. The maximum is $\sqrt 14$ and the minimum $\sqrt 7.$ – Allawonder May 18 '20 at 15:16
@user774564 Note that the previous domain I gave was not optimal. Check out the edit. So it stands that the minimum is indeed $1+\sqrt 6,$ and the maximum still $\sqrt{14}.$ – Allawonder May 18 '20 at 15:46

Find minimum and maximum value of $\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$

3 Answers3