Đáp án:

`P_{max}={7\sqrt{3}}/3` khi `A=B≈73°13'; C≈33°34'`

Giải thích các bước giải:

`P=2cosA+2cosB+2\sqrt{3}cosC`

`=2. 2 cos\ {A+B}/2.cos \ {A-B}/2 +2\sqrt{3} cosC`

`=4 cos\ {180-C}/2 cos \ {A-B}/2 +2\sqrt{3} cosC`

`=4 cos\ (90- C/2). cos \ {A-B}/2 +2\sqrt{3} cosC`

`=4sin\ C/ 2 cos \ {A-B}/2 +2\sqrt{3} cosC`

`=4sin\ C/ 2 cos \ {A-B}/2 +2\sqrt{3}. (1-2sin^2\  C/2)`

`=4sin\ C/ 2 cos \ {A-B}/2 -4\sqrt{3} sin^2\  C/2 +2\sqrt{3}`

`\le 4sin\ C/2-4\sqrt{3} sin^2\  C/2 +2\sqrt{3}`

(vì `cos\ {A-B}/2\le 1`)

`\le -4\sqrt{3}. (sin^2 \ C/2 - 2. sin\ C/2 1/{2\sqrt{3}} + 1/{12})+1/\sqrt{3}+2\sqrt{3}`

`\le -4\sqrt{3} (sin\ C/2-1/{2\sqrt{3}})^2+7/\sqrt{3}`

`\le 7/{\sqrt{3}}={7\sqrt{3}}/3`

Dấu "=" xảy ra khi:

 `\qquad sin\ C/ 2 -1/{2\sqrt{3}}=0`

`<=>sin \ C/2=1/{2\sqrt{3}}`

`<=>C≈33°34'`

`\qquad A=B={180°-C}/2≈73°13'`

Vậy `P_{max}={7\sqrt{3}}/3` khi `A=B≈73°13'`

`C≈33°34'`

$\begin{array}{*{20}{l}} {P = 2\cos A + 2\cos B + 2\sqrt 3 \cos C}\\ {{\rm{ \;}} \to \dfrac{P}{{2\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }}\left( {\cos A + \cos B} \right) + \cos C}\\ {{\rm{ \;}} \to \dfrac{P}{{2\sqrt 3 }} = \dfrac{2}{{\sqrt 3 }}\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) + 1 - 2{{\sin }^2}\dfrac{C}{2}}\\ {{\rm{ \;}} \to \dfrac{P}{{2\sqrt 3 }} \le \dfrac{2}{{\sqrt 3 }}\sin \dfrac{C}{2} + 1 - 2{{\sin }^2}\dfrac{C}{2} = {\rm{ \;}} - 2\left( {{{\sin }^2}\dfrac{C}{2} - \dfrac{1}{{\sqrt 3 }}\sin \dfrac{C}{2} + \dfrac{1}{{12}}} \right) + \dfrac{7}{6}}\\ { = {\rm{ \;}} - {{\left( {\sin \dfrac{C}{2} - \dfrac{{\sqrt 3 }}{6}} \right)}^2} + \dfrac{7}{6} \le \dfrac{7}{6}}\\ {{\rm{ \;}} \to P \le \dfrac{{7\sqrt 3 }}{3}}\\ {{\rm{ \;}} \to \max P = \dfrac{{7\sqrt 3 }}{3}} \end{array}$

Dấu bằng xảy ra khi và chỉ khi:

$\left\{ {\begin{array}{*{20}{l}} {\sin \dfrac{C}{2} = \dfrac{{\sqrt 3 }}{6}}\\ {A = B} \end{array}} \right.$