Tìm giá trị nhỏ nhát của A= 3x^2 +2y^2 biết 4x+y=1 04/10/2021 Bởi Samantha Tìm giá trị nhỏ nhát của A= 3x^2 +2y^2 biết 4x+y=1
Đáp án: Giải thích các bước giải: \(\begin{array}{l}A = 3{x^2} + 2{y^2}\,\,\,\,\,\,\,\,va\,\,\,\,\,\,\,4x + y = 1\\Ta\,co:\\4x + y = 1 \Rightarrow y = 1 – 4x\\ \Rightarrow A = 3{x^2} + 2{y^2}\\ = 3{x^2} + 2.{\left( {1 – 4x} \right)^2}\\ = 3{x^2} + 2.\left( {1 – 8x + 16{x^2}} \right)\\ = 3{x^2} + 2 – 16x + 32{x^2}\\ = 35{x^2} – 16x + 2\\ = 35.\left( {{x^2} – \frac{{16}}{{35}}x} \right) + 2\\ = 35.\left( {{x^2} – 2.\frac{8}{{35}}.x + {{\left( {\frac{8}{{35}}} \right)}^2}} \right) + 2 – 35.{\left( {\frac{8}{{35}}} \right)^2}\\ = 35.{\left( {x – \frac{8}{{35}}} \right)^2} + \frac{6}{{35}}\\ \Rightarrow A \ge \frac{6}{{35}}\\ \Rightarrow {A_{\min }} = \frac{6}{{35}} \Leftrightarrow x = \frac{8}{{35}} \Rightarrow y = 1 – 4.\frac{8}{{35}} = \frac{3}{{35}}\end{array}\) Bình luận
$$\eqalign{ & A = 3{x^2} + 2{y^2} \cr & 4x + y = 1 \Leftrightarrow y = 1 – 4x \cr & \Rightarrow A = 3{x^2} + 2{\left( {1 – 4x} \right)^2} \cr & A = 3{x^2} + 2\left( {1 – 8x + 16{x^2}} \right) \cr & A = 35{x^2} – 16x + 2 \cr & A = 35\left( {{x^2} – {{16} \over {35}}x} \right) + 2 \cr & A = 35\left( {{x^2} – 2.x.{8 \over {35}} + {{\left( {{8 \over {35}}} \right)}^2}} \right) – 35{\left( {{8 \over {35}}} \right)^2} + 2 \cr & A = 35{\left( {x – {8 \over {35}}} \right)^2} + {6 \over {35}} \cr & {\left( {x – {8 \over {35}}} \right)^2} \ge 0 \Rightarrow 35{\left( {x – {8 \over {35}}} \right)^2} \ge 0 \Leftrightarrow 35{\left( {x – {8 \over {35}}} \right)^2} + {6 \over {35}} \ge {6 \over {35}} \cr & \Rightarrow A \ge {6 \over {35}} \cr & \Rightarrow {A_{\min }} = {6 \over {35}} \Leftrightarrow x = {8 \over {35}} \cr} $$ Bình luận
Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}A = 3{x^2} + 2{y^2}\,\,\,\,\,\,\,\,va\,\,\,\,\,\,\,4x + y = 1\\Ta\,co:\\4x + y = 1 \Rightarrow y = 1 – 4x\\ \Rightarrow A = 3{x^2} + 2{y^2}\\ = 3{x^2} + 2.{\left( {1 – 4x} \right)^2}\\ = 3{x^2} + 2.\left( {1 – 8x + 16{x^2}} \right)\\ = 3{x^2} + 2 – 16x + 32{x^2}\\ = 35{x^2} – 16x + 2\\ = 35.\left( {{x^2} – \frac{{16}}{{35}}x} \right) + 2\\ = 35.\left( {{x^2} – 2.\frac{8}{{35}}.x + {{\left( {\frac{8}{{35}}} \right)}^2}} \right) + 2 – 35.{\left( {\frac{8}{{35}}} \right)^2}\\ = 35.{\left( {x – \frac{8}{{35}}} \right)^2} + \frac{6}{{35}}\\ \Rightarrow A \ge \frac{6}{{35}}\\ \Rightarrow {A_{\min }} = \frac{6}{{35}} \Leftrightarrow x = \frac{8}{{35}} \Rightarrow y = 1 – 4.\frac{8}{{35}} = \frac{3}{{35}}\end{array}\)
$$\eqalign{
& A = 3{x^2} + 2{y^2} \cr
& 4x + y = 1 \Leftrightarrow y = 1 – 4x \cr
& \Rightarrow A = 3{x^2} + 2{\left( {1 – 4x} \right)^2} \cr
& A = 3{x^2} + 2\left( {1 – 8x + 16{x^2}} \right) \cr
& A = 35{x^2} – 16x + 2 \cr
& A = 35\left( {{x^2} – {{16} \over {35}}x} \right) + 2 \cr
& A = 35\left( {{x^2} – 2.x.{8 \over {35}} + {{\left( {{8 \over {35}}} \right)}^2}} \right) – 35{\left( {{8 \over {35}}} \right)^2} + 2 \cr
& A = 35{\left( {x – {8 \over {35}}} \right)^2} + {6 \over {35}} \cr
& {\left( {x – {8 \over {35}}} \right)^2} \ge 0 \Rightarrow 35{\left( {x – {8 \over {35}}} \right)^2} \ge 0 \Leftrightarrow 35{\left( {x – {8 \over {35}}} \right)^2} + {6 \over {35}} \ge {6 \over {35}} \cr
& \Rightarrow A \ge {6 \over {35}} \cr
& \Rightarrow {A_{\min }} = {6 \over {35}} \Leftrightarrow x = {8 \over {35}} \cr} $$