∴HM=ND=50-40cosθ.AM=50-40sinθ.
∴S=(50-40cosθ)(50-40sinθ)
=100[25-20(sinθ+cosθ)+16sinθcosθ],(0≤θ≤
| π |
| 2 |

令t=sinθ+cosθ=
| 2 |
| π |
| 4 |
则sinθcosθ=
| t2-1 |
| 2 |
| 2 |
∴S=100[25-20t+8(t2-1)]
=800(t-
| 5 |
| 4 |
又∵t∈[1,
| 2 |
∴当t=1时,S取最大值500.
此时,
| 2 |
| π |
| 4 |
∴sin(θ+
| π |
| 4 |
| ||
| 2 |
∵
| π |
| 4 |
| π |
| 4 |
| 3π |
| 4 |
∴θ+
| π |
| 4 |
| π |
| 4 |
| 3π |
| 4 |
即θ=0或θ=
| π |
| 2 |
答:当点H在
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| EF |


