Cho x, y. z là các số thực không âm thỏa mãn \(12x+10y+15z\le60\). Tìm GTLN của \(P=x^2+y^2+z^2-4x-4y-z\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\Leftrightarrow\frac{x}{5}+\frac{y}{6}+\frac{z}{4}\le1\)
Đặt \(\left(\frac{x}{5};\frac{y}{6};\frac{z}{4}\right)=\left(a;b;c\right)\Rightarrow0\le a;b;c\le1\) và \(a+b+c\le1\)
\(T=25a^2+36b^2+16c^2-20a-24b-4c\)
\(25a\left(a-\frac{32}{25}\right)\le0\Rightarrow25a^2\le32a\)
\(36b\left(b-1\right)\le0\Rightarrow36b^2\le36b\)
\(16c\left(c-1\right)\le0\Rightarrow16c^2\le16c\)
\(\Rightarrow T\le32a+36b+16c-20a-24b-4c=12\left(a+b+c\right)\le12\)
\(T_{max}=12\) khi \(\left\{{}\begin{matrix}a=0\\b=0\\c=1\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}a=0\\b=1\\c=0\end{matrix}\right.\)
\(4x^2+4y^2\ge8xy\)
\(16x^2+z^2\ge8zx\)
\(16y^2+z^2\ge8yz\)
Cộng vế với vế:
\(20x^2+20y^2+2z^2\ge8\left(xy+yz+zx\right)\)
\(\Leftrightarrow10x^2+10y^2+z^2\ge4\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\dfrac{1}{3};\dfrac{1}{3};\dfrac{4}{3}\right)\)
\(\left\{{}\begin{matrix}x;y;z\ge0\\x+y+z=1\end{matrix}\right.\) \(\Rightarrow0\le x;y;z\le1\)
\(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\\z^2\le z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x^2+x+1\le x^2+2x+1\\2y^2+y+1\le y^2+2y+1\\2z^2+z+1\le z^2+2z+1\end{matrix}\right.\)
\(\Rightarrow P\le\sqrt{\left(x+1\right)^2}+\sqrt{\left(y+1\right)^2}+\sqrt{\left(z+1\right)^2}=x+y+z+3=4\)
\(P_{max}=4\) khi \(\left(x;y;z\right)=\left(0;0;1\right)\) và các hoán vị
\(\sqrt{4x+2\sqrt{x}+1}\le\sqrt{4x+\dfrac{1}{2}\left(2^2+x\right)+1}=\sqrt{\dfrac{9x}{2}+3}\)
\(=\dfrac{1}{\sqrt{21}}.\sqrt{21}.\sqrt{\dfrac{9x}{2}+3}\le\dfrac{1}{2\sqrt{21}}\left(21+\dfrac{9x}{2}+3\right)=\dfrac{1}{2\sqrt{21}}\left(\dfrac{9x}{2}+24\right)\)
Tương tự và cộng lại:
\(A\le\dfrac{1}{2\sqrt{21}}\left(\dfrac{9}{2}\left(x+y+z\right)+72\right)=3\sqrt{21}\)
\(A_{max}=3\sqrt{21}\) khi \(x=y=z=4\)
\(A=1\sqrt{4x+2\sqrt{x}+1}+1.\sqrt{4y+2\sqrt{y}+1}+1\sqrt{4z+2\sqrt{z}+1}\)
\(\le\sqrt{\left(1+1+1\right)\left(4\left(x+y+z\right)+2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+3\right)}\)
\(=\sqrt{3.\left[51+\dfrac{4\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}{2}\right]}\)
\(\le\sqrt{3.\left[51+\dfrac{x+y+z+12}{2}\right]}\)
\(=\sqrt{189}\)
Dấu "=" xảy ra <=> x = y = z = 4
\(P=\sqrt{y}\left(\sqrt{x}+2\sqrt{z}\right)+3\sqrt{zx}=\left(6-\sqrt{x}-\sqrt{z}\right)\left(\sqrt{x}+2\sqrt{z}\right)+3\sqrt{zx}\)
\(P=-x+6\sqrt{x}-2z+12z=-\left(\sqrt{x}-3\right)^2-2\left(\sqrt{z}-3\right)^2+27\le27\)
\(P_{max}=27\) khi \(\left(x;y;z\right)=\left(9;0;9\right)\)
Xét \(5P-\left(12x+10y+15z\right)=5x^2-32x+5y^2-30y+5z^2-20z.\)
\(=5x\left(x-6,4\right)+5y\left(y-6\right)+5z\left(z-4\right).\)(1)
Mà \(x,y,z\ge0\)nên từ \(12x+10y+15z\le60\)suy ra \(\hept{\begin{cases}12x\le60\\10y\le60\\15z\le60\end{cases}\Leftrightarrow\hept{\begin{cases}x\le5\\y\le6\\z\le4\end{cases}\Rightarrow}}\hept{\begin{cases}x-6,4< 0\\y-6\le0\\z-4\le0\end{cases}\Rightarrow\hept{\begin{cases}x\left(x-6,4\right)\le0\\y\left(y-6\right)\le0\\z\left(z-4\right)\le0\end{cases}.}}\)(2)
Từ (1) và (2) suy ra \(5P-\left(12x+10y+15z\right)\le0\)
\(\Rightarrow P\le\frac{12x+10y+15z}{5}\le\frac{60}{5}=12.\)
Vậy GTLN của P=12, Dấu '=' xảy ra khi \(\hept{\begin{cases}x\left(x-6,4\right)=y\left(y-6\right)=z\left(z-4\right)=0\\12x+10y+15z=60\end{cases}\Leftrightarrow\orbr{\begin{cases}x=y=0;z=4\\x=z=0;y=6\end{cases}.}}\)