Sử dụng HÌNH HỌC GIẢI TÍCH OXY tìm MIN, MAX.

Đề bài (Câu hỏi của bạn Riemann Hypothesis hỏi trên facebook Trợ Giúp Toán Học)
$\begin{array}{l}
Cho\,\left\{ \begin{array}{l}
a,b,c,d \in R\\
a + 2b = 9;\,c + 2d = 4
\end{array} \right..\,Min\,P = \sqrt {{a^2} + {b^2} - 12a - 8b + 52}  + \sqrt {{a^2} + {b^2} + {c^2} + {d^2} - 2ac - 2bd} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \sqrt {{c^2} + {d^2} - 4c + 8d + 20}  = ?
\end{array}.$
Giải:
$Do\,P = \sqrt {{{\left( {a - 6} \right)}^2} + {{\left( {b - 4} \right)}^2}}  + \sqrt {{{\left( {a - c} \right)}^2} + {{\left( {b - d} \right)}^2}}  + \sqrt {{{\left( {c - 2} \right)}^2} + {{\left( {d + 4} \right)}^2}} .$

$\begin{array}{*{20}{l}}
{Coi{\mkern 1mu} \left\{ {\begin{array}{*{20}{l}}
{M\left( {a;b} \right) \Rightarrow M \in {d_1}:{f_1}\left( {x,y} \right) = x + 2y - 9 = 0}\\
{N\left( {c;d} \right) \Rightarrow N \in {d_2}:{f_2}\left( {x,y} \right) = x + 2y = 4}\\
{A\left( {6;4} \right),{\mkern 1mu} D\left( {2; - 4} \right)}
\end{array}} \right.}\\
{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{P = AM + MN + ND}\\
{{d_1}//{d_2}}
\end{array}} \right.}\\
{Do{\mkern 1mu} \left\{ {\begin{array}{*{20}{l}}
{{f_1}\left( A \right) = 6 + 8 - 9 = 5 > 0}\\
{{f_2}\left( A \right) = 6 + 8 - 4 = 10 > 0}
\end{array}} \right.}
\end{array}.$

Khi đó A nằm phía trên 2 đường thẳng ${d_1}\& {d_2}.$
Mặt khác, $Do\,\left\{ \begin{array}{l}
{f_1}\left( D \right) = 2 - 8 - 9 =  - 15 < 0\\
{f_2}\left( D \right) = 2 - 8 - 4 =  - 10 < 0
\end{array} \right.$
Khi đó D nằm phía dưới 2 đường thẳng ${d_1}\& {d_2}.
$\[\begin{array}{l}
 \Rightarrow P\,\min  \Leftrightarrow \left\{ \begin{array}{l}
AM\,\min  \Leftrightarrow AM = AB = d\left( {A \to {d_1}} \right)\left( {AB \bot {d_1}} \right) \Leftrightarrow M \equiv B\\
MN\,\min  \Leftrightarrow MN = d\left( {{d_1};{d_2}} \right) = BC\\
ND\,\min  \Leftrightarrow ND = CD = d\left( {D \to {d_2}} \right)\left( {CD \bot {d_2}} \right) \Leftrightarrow N \equiv D
\end{array} \right. \Rightarrow AB \equiv BC \equiv CD \equiv AD\\
 \Rightarrow Min\,P = d\left( {A \to {d_1}} \right) + d\left( {{d_1};{d_2}} \right) + d\left( {D \to {d_2}} \right) = \frac{5}{{\sqrt 5 }} + \frac{5}{{\sqrt 5 }} + \frac{{10}}{{\sqrt 5 }} = 4\sqrt 5 \\
 \Rightarrow AD:2x - y = 8 \Rightarrow \left\{ \begin{array}{l}
B = AD \cap {d_1}:\left\{ \begin{array}{l}
2x - y = 8\\
x + 2y = 9
\end{array} \right. \Leftrightarrow B\left( {5;2} \right)\\
C = AD \cap {d_2}:\left\{ \begin{array}{l}
2x - y = 8\\
x + 2y = 4
\end{array} \right. \Leftrightarrow C\left( {4;0} \right)
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = 5\\
b = 2\\
c = 4\\
d = 0
\end{array} \right.
\end{array}\]Một cách giải khác hay hơn, các bạn tham khảo nhé!