cho \(0\le x\le4;0\le y\le3\)Tìm GTLN của:
\(Q=\left(3-y\right)\left(4-x\right)\left(2y+3z\right)\)
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Cho \(0< x< y\le z\le1\) và \(3x+2y+z\le4\). Tìm Max \(S=3x^2+2y^2+z^2\) - Hoc24
Tham khảo
Khai triển Abel ta có:
\(S=\left(z-y\right)z+\left(y-x\right)\left(z+2y\right)+x\left(3x+2y+z\right)\)
\(\le\left(z-y\right).1+\left(y-x\right).3+4x=x+2y+z\)
\(=\left(1-1\right)z+\left(1-\dfrac{1}{3}\right)\left(2y+z\right)+\dfrac{1}{3}\left(3x+2y+z\right)\)
\(\le\dfrac{2}{3}.3+\dfrac{1}{3}.4=\dfrac{10}{3}\)
Dấu = xảy ra khi \(x=\dfrac{1}{3},y=z=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=x\left(3x+2y+z\right)+\left(y-x\right)\left(2y+z\right)+\left(z-y\right).y\)
\(S\le4x+3\left(y-x\right)+z-y=x+2y+z\)
\(S\le\dfrac{1}{3}\left(3x+2y+z\right)+\dfrac{2}{3}\left(2y+z\right)\le\dfrac{1}{3}.4+\dfrac{2}{3}.3=\dfrac{10}{3}\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\dfrac{1}{3};1;1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Do \(-2\leq x,y\leq 1\) nên:
\(\left\{\begin{matrix} (x+2)(x-1)\leq 0\\ (y+2)(y-1)\leq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x^2+x-2\leq 0\\ y^2+y-2\leq 0\end{matrix}\right.\)
\(\Rightarrow x^2+x-2+y^2+y-2\leq 0\)
\(\Leftrightarrow x^2+y^2\leq 4-(x+y)=4\)
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+y^2-2x-4y-4=0\\ \Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2-9=0\\ \Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=9=0^2+3^2=0^2+\left(-3\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=0\\y-2=3\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=3\\y-2=0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=0\\y-2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-3\\y-2=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow-2\le x\le4\left(y\in R\right)\)
Ta có \(S=3x+4y\)
Mà \(x\ge-2;y\ge-1\Leftrightarrow S\ge3\cdot\left(-2\right)+4\cdot\left(-1\right)=-6-4=-10\)
Vậy GTNN của S là \(-10\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)
Lời giải:
ĐKĐB $\Leftrightarrow (x^2-2x+1)+(y^2-4y+4)-9=0$
$\Leftrightarrow (x-1)^2+(y-2)^2-9=0$
$\Rightarrow (x-1)^2=9-(y-2)^2\leq 9$
$\Rightarrow -3\leq x-1\leq 3$
$\Leftrightarrow -2\leq x\leq 4$
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Đặt $x-1=a; y-2=b$ thì bài toán trở thành:
Cho $a,b$ thực thỏa mãn $a^2+b^2=9$
Tìm min $S=3a+4b+11$
Áp dụng BĐT Bunhiacopxky:
$(3a+4b)^2\leq (a^2+b^2)(3^2+4^2)=9.25$
$\Rightarrow -15\leq 3a+4b\leq 15$
$\Rightarrow 3a+4b\geq -15$
$\Rightarrow S=3a+4b+11\geq -4$
Vậy $S_{\min}=-4$ khi $x=\frac{-4}{5}; y=\frac{-1}{5}$
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=(6-2x)(12-3y)(2x+3y)/6
<=(6-2x+12-3y+2x+3y)3/(6.27)
=183/(6.27)=36
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(y=4x^3-x^4=x^3\left(4-x\right)=x.x.x.\left(4-x\right)\).
Vì vậy: \(3y=x.x.x.\left(12-4x\right)\).
Với \(0\le x\le4\) thì \(\left\{{}\begin{matrix}x\ge0\\12-4x\ge0\end{matrix}\right.\).
Áp dụng bất đẳng thức cô si cho bốn số: x,x,x, 12 - 3x ta có:
\(x.x.x.\left(12-3x\right)\le\left(\dfrac{x+x+x+12-3x}{4}\right)^4=81\).
Dấu bằng xảy ra khi: \(x=12-3x\)\(\Leftrightarrow4x=12\)\(\Leftrightarrow x=3\).
Như vậy: \(3y\le81\) \(\Leftrightarrow y\le27\) nên max của y bằng 27 khi x = 3.