Cho $36x^2+16y^2=9$, tìm Min, max của: $A=y-2x+5$ :) 01/11/2021 Bởi Alaia Cho $36x^2+16y^2=9$, tìm Min, max của: $A=y-2x+5$ 🙂
Đáp án + giải thích các bước giải: Áp dụng bất đẳng thức Bunhiacopxki `(1/16+1/9)(16y^2+36x^2)>=[1/4 . 4y+1/3 . (-6x)]^2` `->25/144 . 9>=(y-2x)^2` `->(y-2x)^2-(5/4)^2<=0` `->(y-2x-5/4)(y-2x+5/4)<=0` `->`\(\left[ \begin{array}{l}\left\{\begin{matrix} y-2x-\dfrac{5}{4}\le0\\y-2x+\dfrac{5}{4}\ge0 \end{matrix}\right.\\\left\{\begin{matrix} y-2x-\dfrac{5}{4}\ge0\\y-2x+\dfrac{5}{4}\le0 \end{matrix}\right.\end{array} \right.\) `->`\(\left[ \begin{array}{l}\left\{\begin{matrix} y-2x\le \dfrac{5}{4}\\y-2x\ge \dfrac{-5}{4} \end{matrix}\right.\\\left\{\begin{matrix} y-2x\ge \dfrac{5}{4}\\y-2x\le \dfrac{-5}{4} \end{matrix}\right.\end{array} \right.\) `->-5/4 <= y-2x <= 5/4` `->15/4<=A<=25/4` Vậy `A` đạt giá trị lớn nhất là `25/4` khi $ \left\{\begin{matrix} \dfrac{4y}{\dfrac{1}{4}}=\dfrac{-6x}{\dfrac{1}{3}}\\y-2x=\dfrac{5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} 16y=-18x\\y-2x=\dfrac{5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} y=\dfrac{-9}{8}x\\\dfrac{-9}{8}x-2x=\dfrac{5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} y=\dfrac{9}{20}\\x=\dfrac{-2}{5}\end{matrix}\right. $ `A` đạt giá trị nhỏ nhất là `15/4` khi $ \left\{\begin{matrix} \dfrac{4y}{\dfrac{1}{4}}=\dfrac{-6x}{\dfrac{1}{3}}\\y-2x=\dfrac{-5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} 16y=-18x\\y-2x=\dfrac{-5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} y=\dfrac{-9}{8}x\\\dfrac{-9}{8}x-2x=\dfrac{-5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} y=\dfrac{-9}{20}\\x=\dfrac{2}{5}\end{matrix}\right. $ Bình luận
Ta có:$-2x+y=-6x.\dfrac13+4y.\dfrac14$ $\to (-2x+y)^2=\left(-6x.\dfrac13+4y.\dfrac14\right)^2$ Áp dụng BĐT Bunhiacopxki, ta có:$\left(-6x.\dfrac13+4y.\dfrac14\right)^2\le(36x^2+16y^2)\left(\dfrac19+\dfrac1{16}\right)=9.\dfrac{25}{144}=\dfrac{25}{16}$ $\to (-2x+y)^2\le \dfrac{25}{16}$$\to -\dfrac54\le -2x+y \le \dfrac54$$\to \dfrac{15}{4}\le y-2x+5\le \dfrac{25}{4}$ Vậy $\begin{cases}Min_A=\dfrac{15}{4}\\Max_A=\dfrac{25}{4}\end{cases}$ Bình luận
Đáp án + giải thích các bước giải:
Áp dụng bất đẳng thức Bunhiacopxki
`(1/16+1/9)(16y^2+36x^2)>=[1/4 . 4y+1/3 . (-6x)]^2`
`->25/144 . 9>=(y-2x)^2`
`->(y-2x)^2-(5/4)^2<=0`
`->(y-2x-5/4)(y-2x+5/4)<=0`
`->`\(\left[ \begin{array}{l}\left\{\begin{matrix} y-2x-\dfrac{5}{4}\le0\\y-2x+\dfrac{5}{4}\ge0 \end{matrix}\right.\\\left\{\begin{matrix} y-2x-\dfrac{5}{4}\ge0\\y-2x+\dfrac{5}{4}\le0 \end{matrix}\right.\end{array} \right.\)
`->`\(\left[ \begin{array}{l}\left\{\begin{matrix} y-2x\le \dfrac{5}{4}\\y-2x\ge \dfrac{-5}{4} \end{matrix}\right.\\\left\{\begin{matrix} y-2x\ge \dfrac{5}{4}\\y-2x\le \dfrac{-5}{4} \end{matrix}\right.\end{array} \right.\)
`->-5/4 <= y-2x <= 5/4`
`->15/4<=A<=25/4`
Vậy `A` đạt giá trị lớn nhất là `25/4` khi $ \left\{\begin{matrix} \dfrac{4y}{\dfrac{1}{4}}=\dfrac{-6x}{\dfrac{1}{3}}\\y-2x=\dfrac{5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} 16y=-18x\\y-2x=\dfrac{5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} y=\dfrac{-9}{8}x\\\dfrac{-9}{8}x-2x=\dfrac{5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} y=\dfrac{9}{20}\\x=\dfrac{-2}{5}\end{matrix}\right. $
`A` đạt giá trị nhỏ nhất là `15/4` khi $ \left\{\begin{matrix} \dfrac{4y}{\dfrac{1}{4}}=\dfrac{-6x}{\dfrac{1}{3}}\\y-2x=\dfrac{-5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} 16y=-18x\\y-2x=\dfrac{-5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} y=\dfrac{-9}{8}x\\\dfrac{-9}{8}x-2x=\dfrac{-5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} y=\dfrac{-9}{20}\\x=\dfrac{2}{5}\end{matrix}\right. $
Ta có:
$-2x+y=-6x.\dfrac13+4y.\dfrac14$
$\to (-2x+y)^2=\left(-6x.\dfrac13+4y.\dfrac14\right)^2$
Áp dụng BĐT Bunhiacopxki, ta có:
$\left(-6x.\dfrac13+4y.\dfrac14\right)^2\le(36x^2+16y^2)\left(\dfrac19+\dfrac1{16}\right)=9.\dfrac{25}{144}=\dfrac{25}{16}$
$\to (-2x+y)^2\le \dfrac{25}{16}$
$\to -\dfrac54\le -2x+y \le \dfrac54$
$\to \dfrac{15}{4}\le y-2x+5\le \dfrac{25}{4}$
Vậy $\begin{cases}Min_A=\dfrac{15}{4}\\Max_A=\dfrac{25}{4}\end{cases}$