Chủ đề này có 2 trả lời

[minhmin09]

Cho $x^{2}+2y^{2}= \frac{8}{3}$$(x,y>0)$. Tìm Max $P=7(x+2y)-4\sqrt{x^{2}+2xy+8y^{2}}$

Bài viết đã được chỉnh sửa nội dung bởi minhmin09: 11-12-2018 - 21:45

[DOTOANNANG]

Từ $\it{:}$ $\it{x}^{\,\it{2}}+ \it{2}\,\it{y}^{\,\it{2}}= \frac{\it{8}}{\it{3}}\,\,\Rightarrow \,\,\it{(}\,\,\it{x}- \it{2}\,\it{y}\,\,\it{)}\it{(}\,\,\it{x}- \frac{\it{4}}{\it{3}}\,\,\it{)}\geqq \it{0}\,\,,\,\,\it{2}\,{\it{y}}'= -\,\frac{\it{x}}{\it{y}}$ $\it{.}$ Đặt $\it{:}$ $\it{f}\,\it{(}\,\,\it{x}\,\,\it{)}= \it{7}\,\it{(}\,\,\it{x}+ \it{2}\,\it{y}\,\,\it{)}- \it{4}\,\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}}$ $\it{.}$ Khi đó $\it{:}$

${\it{f}}'\,\it{(}\,\,\it{x}\,\,\it{)}= \it{7}\,\it{(}\,\,\it{1}+ \it{2}\,{\it{y}}'\,\,\it{)}- \frac{\it{4}\,\it{(}\,\,\it{2}\,\it{x}+ \it{2}\,\it{y}+ \it{2}\,\it{x}{\it{y}}'+ \it{16}\,\it{y}{\it{y}}'\,\,\it{)}}{\it{2}\,\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}}}=$

$= \it{7}\,\it{(}\,\,\it{1}- \frac{\it{x}}{\it{y}}\,\,\it{)}- \frac{\it{2}\,\it{(}\,\,\it{2}\,\it{x}+ \it{2}\,\it{y}- \frac{\it{x}^{\,\it{2}}}{\it{y}}- \it{8}\,\it{x}\,\,\it{)}}{\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}}}=$

$= \it{7}\,\it{(}\,\,\it{2}- \frac{\it{x}}{\it{y}}\,\,\it{)}+ \frac{\it{2}\,\it{(}\,\,\it{6}\,\it{xy}- \it{2}\,\it{y}^{\,\it{2}}+ \it{x}^{\,\it{2}}\,\,\it{)}}{\it{y}\,\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}}}- \it{7}=$

$= \frac{\it{1}}{\it{y}}\,\left \{ -\,\it{7}\,\it{(}\,\,\it{x}- \it{2}\,\it{y}\,\,\it{)}+ \frac{\it{4}\,\it{(}\,\it{6}\,\it{xy}- \it{2}\,\it{y}^{\,\it{2}}+ \it{x}^{\,\it{2}}\,\,\it{)}^{\,\it{2}}- \it{49}\,\it{y}^{\,\it{2}}\,\it{(}\,\,\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}\,\,\it{)}^{\,\it{2}}}{\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}}\left [ \it{2}\,\it{(}\,\it{6}\,\it{xy}- \it{2}\,\it{y}^{\,\it{2}}+ \it{x}^{\,\it{2}}\,\,\it{)}+ \it{7}\,\it{y}\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}} \right ]} \right \}=$

$= -\,\frac{\it{x}- \it{\it{2}\,\it{y}}}{\it{y}}\,\left \{ \,\,\underbrace{\it{7}- \frac{\it{4}\,\it{x}^{\,\it{3}}+ \it{56}\,\it{x}^{\,\it{2}}\it{y}+ \it{191}\,\it{xy}^{\,\it{2}}+ \it{188}\,\it{y}^{\,\it{3}}}{\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}}\left [ \it{2}\,\it{(}\,\it{6}\,\it{xy}- \it{2}\,\it{y}^{\,\it{2}}+ \it{x}^{\,\it{2}}\,\,\it{)}+ \it{7}\,\it{y}\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}} \right ]}}_{> \it{0}}\,\, \right \}= $

Ta sẽ chứng minh bất đẳng thức trong dấu ngoặc nhọn luôn đúng $\it{(}\,\,\it{!}\,\,\it{)}$ Thật vậy $\it{,}$ với $\it{t}= -\,\it{2}\,{\it{y}}'> 0$ thì $\it{:}$

$\frac{\mathrm{d} }{\mathrm{d} \it{t}}\left ( \,\,\underbrace{\it{2}\,\it{(}\,\,\it{6}\,\it{t}- \it{2}+ \it{t}^{\,\it{2}}\,\,\it{)}+ \it{7}\,\sqrt{\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}}\,\,\it{)}}_{> \it{0}\,\,\Leftarrow \,\,\it{(}\,\,\star\,\,\it{)}}\,\, \right )= \frac{\it{14}\,\it{(}\,\,\it{t}+ \it{1}\,\,\it{)}}{\it{2}\,\sqrt{\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}}}+ \it{4}\,\it{(}\,\,\it{t}+ \it{3}\,\,\it{)}> \it{0}\,\,\it{(}\,\,\star\,\,\it{)}$

$\it{(}$ Hay $\it{)}$ tương đương với $\it{:}$

$\it{14}\,\it{(}\,\,\it{6}\,\it{t}- \it{2}+ \it{t}^{\,\it{2}}\,\,\it{)}\,\sqrt{\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}}> \it{4}\,\it{t}^{\,\it{3}}+ \it{56}\,\it{t}^{\,\it{2}}+ \it{191}\,\it{t}+ \it{188}- \it{49}\,\it{(}\,\,\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}\,\,\it{)}$

Xét $\it{0}< \it{t}\leqq \it{2}$ thì$\it{:}$

$\it{14}\,\it{(}\,\,\it{6}\,\it{t}- \it{2}+ \it{t}^{\,\it{2}}\,\,\it{)}\,\sqrt{\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}}\geqq \it{4}\,\it{t}^{\,\it{3}}+ \it{241}\,\it{t}^{\,\it{2}}+ \it{192}\,\it{t}+ \it{188}- \it{49}\,\it{(}\,\,\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}\,\,\it{)}> \it{4}\,\it{t}^{\,\it{3}}+ \it{56}\,\it{t}^{\,\it{2}}+ \it{191}\,\it{t}+ \it{188}- \it{49}\,\it{(}\,\,\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}\,\,\it{)}$

Xét $\it{t}\geqq \it{2}$ thì $\it{:}$

$\it{14}\,\it{(}\,\,\it{6}\,\it{t}- \it{2}+ \it{t}^{\,\it{2}}\,\,\it{)}\,\sqrt{\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}}\geqq \it{4}\,\it{t}^{\,\it{3}}+ \it{56}\,\it{t}^{\,\it{2}}+ \it{562}\,\it{t}+ \it{188}- \it{49}\,\it{(}\,\,\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}\,\,\it{)}> \it{4}\,\it{t}^{\,\it{3}}+ \it{56}\,\it{t}^{\,\it{2}}+ \it{191}\,\it{t}+ \it{188}- \it{49}\,\it{(}\,\,\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}\,\,\it{)}$

Ta được $\it{:}$ $\it{(}\,\,\it{x}- \it{2}\,\it{y}\,\,\it{)}{\it{f}}'\,\it{(}\,\,\it{x}\,\,\it{)}\leqq \it{0}\,\,\Rightarrow\,\, \it{(}\,\,\it{x}- \frac{\it{4}}{\it{3}}\,\,\it{)}{\it{f}}'\,\it{(}\,\,\it{x}\,\,\it{)}\leqq \it{0}\,\,\Rightarrow\,\, \it{f}\,\it{(}\,\,\it{x}\,\,\it{)}\leqq \it{f}\,\it{(}\,\,\frac{\it{4}}{\it{3}}\,\,\it{)}= \it{8}$

$\lceil$ Ý nghĩa hình học của đạo hàm $\it{(}$ $\it{!}$ $\it{)}$   $\rfloor$

By H-a-i-D-a-n-g-e-l $\it{(}$ $\it{D-...-A-N-G}$ $\it{)}$

 

[DOTOANNANG]

$\lceil$ https://math.stackex.../3275678/685500 $\rfloor$