题目内容
已知|
|=4,|
|=6,
=x
+y
,且x+2y=1,∠AOB是钝角,若f(t)=|
-t
|的最小值为2
,则|
|的最小值是______.
| OA |
| OB |
| OC |
| OA |
| OB |
| OA |
| OB |
| 3 |
| OC |
f(t)=|
-t
|的最小值为2
,
∴根据图形知,当
-t
⊥
时,f(t)=|
-t
|的最小值为2
,
∵|
|=4,∴∠AOB=120°,
∵
=x
+y
,且x+2y=1,
∴|
|2=x2
2+y2
2+2xy
•
=16x2+36y2-24xy=16(1-2y)2+36y2-24(1-2y)y
=148y2-88y+16≥
.
∴|
|的最小值是
;
故答案为
.

| OA |
| OB |
| 3 |
∴根据图形知,当
| OA |
| OB |
| OB |
| OA |
| OB |
| 3 |
∵|
| OA |
∵
| OC |
| OA |
| OB |
∴|
| OC |
| OA |
| OB |
| OA |
| OB |
=16x2+36y2-24xy=16(1-2y)2+36y2-24(1-2y)y
=148y2-88y+16≥
| 108 |
| 37 |
∴|
| OC |
6
| ||
| 37 |
故答案为
6
| ||
| 37 |
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