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Câu hỏi: Cho lăng trụ $ABC.A'B'C'$ có thể tích là $V$. $M,N,P$ là các điểm lần lượt nằm trên các cạnh $AA',BB',CC'$ sao cho $\dfrac{AM}{AA'}=\dfrac{1}{3}$, $\dfrac{BN}{BB'}=x$, $\dfrac{CP}{CC'}=y$. Biết thể tích khối đa diện $ABC.MNP$ bằng $\dfrac{2V}{3}$. Giá trị lớn nhất của $xy$ bằng:
A. $\dfrac{17}{21}$.
B. $\dfrac{25}{36}$.
C. $\dfrac{5}{24}$.
D. $\dfrac{9}{16}$.
Ta có ${{V}_{ABC.MNP}}={{V}_{M.ABC}}+{{V}_{M.NBCP}}=\dfrac{1}{3}d\left( M,\left( ABC \right) \right){{S}_{ABC}}+\dfrac{1}{3}d\left( M,\left( NBCP \right) \right){{S}_{BNPC}}$
$=\dfrac{1}{3}\dfrac{AM}{AA'}d\left( A',\left( ABC \right) \right){{S}_{ABC}}+\dfrac{1}{3}\left( \dfrac{BN}{BB'}+\dfrac{CP}{CC'} \right)d\left( A,\left( NBCP \right) \right){{S}_{BB'C'C}}$
$=\dfrac{1}{3}\dfrac{AM}{AA'}{{V}_{ABC.A'B'C'}}+\dfrac{1}{3}\left( \dfrac{BN}{BB'}+\dfrac{CP}{CC'} \right){{V}_{ABC.A'B'C'}}=\dfrac{\dfrac{AM}{AA'}+\dfrac{BN}{BB'}+\dfrac{CP}{CC'}}{3}{{V}_{ABC.A'B'C'}}$
Ta có $\dfrac{{{V}_{ABC.MNP}}}{{{V}_{ABC.A'B'C'}}}=\dfrac{\dfrac{AM}{AA'}+\dfrac{BN}{BB'}+\dfrac{CP}{CC'}}{3}=\dfrac{\dfrac{1}{3}+x+y}{3}=\dfrac{2}{3}\Leftrightarrow x+y=\dfrac{5}{3}\Rightarrow xy\le \dfrac{{{\left( x+y \right)}^{2}}}{4}=\dfrac{25}{36}$.
Đẳng thức xảy ra khi $x=y=\dfrac{5}{6}$.
A. $\dfrac{17}{21}$.
B. $\dfrac{25}{36}$.
C. $\dfrac{5}{24}$.
D. $\dfrac{9}{16}$.
$=\dfrac{1}{3}\dfrac{AM}{AA'}d\left( A',\left( ABC \right) \right){{S}_{ABC}}+\dfrac{1}{3}\left( \dfrac{BN}{BB'}+\dfrac{CP}{CC'} \right)d\left( A,\left( NBCP \right) \right){{S}_{BB'C'C}}$
$=\dfrac{1}{3}\dfrac{AM}{AA'}{{V}_{ABC.A'B'C'}}+\dfrac{1}{3}\left( \dfrac{BN}{BB'}+\dfrac{CP}{CC'} \right){{V}_{ABC.A'B'C'}}=\dfrac{\dfrac{AM}{AA'}+\dfrac{BN}{BB'}+\dfrac{CP}{CC'}}{3}{{V}_{ABC.A'B'C'}}$
Ta có $\dfrac{{{V}_{ABC.MNP}}}{{{V}_{ABC.A'B'C'}}}=\dfrac{\dfrac{AM}{AA'}+\dfrac{BN}{BB'}+\dfrac{CP}{CC'}}{3}=\dfrac{\dfrac{1}{3}+x+y}{3}=\dfrac{2}{3}\Leftrightarrow x+y=\dfrac{5}{3}\Rightarrow xy\le \dfrac{{{\left( x+y \right)}^{2}}}{4}=\dfrac{25}{36}$.
Đẳng thức xảy ra khi $x=y=\dfrac{5}{6}$.
Đáp án B.