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SUEC 高中 Adv Maths (Extreme Value)
- 1. 𝒑𝒈 𝟕𝟏, 𝟕𝟐 𝒎𝒊𝒏 𝒗𝒂𝒍𝒖𝒆, 𝒚𝒎𝒊𝒏 𝒎𝒂𝒙 𝒗𝒂𝒍𝒖𝒆, 𝒚𝒎𝒂𝒙 极值
- 2. 𝒑𝒈 𝟕𝟐, 𝟕𝟑 = 𝟑 𝒙𝟐 + 𝟐 𝟑 𝒙 + 𝟏 𝒚𝒎𝒊𝒏 = 𝟐 𝟑 𝒂 > 𝟎, 开囗方上, 𝒚𝒎𝒊𝒏 𝒚 = 𝟑𝒙𝟐 + 𝟐𝒙 + 𝟏 = 𝟑 𝒙 + 𝟏 𝟑 𝟐 − 𝟏 𝟑 + 𝟏 = 𝟑 𝒙𝟐 + 𝟐 𝟑 𝒙 + 𝟏 𝟑 𝟐 − 𝟏 𝟑 𝟐 + 𝟏 𝒚 = 𝟑 𝒙 + 𝟏 𝟑 𝟐 + 𝟐 𝟑 𝟐𝟒. 𝒂 𝒚 = 𝟑𝒙𝟐 + 𝟐𝒙 + 𝟏 极值 = −𝟐 𝒙𝟐 + 𝟓 𝟐 𝒙 + 𝟔 𝒂 < 𝟎, 开囗方下, 𝒚𝒎𝒂𝒙 𝒚 = −𝟐𝒙𝟐 − 𝟓𝒙 + 𝟔 = −𝟐 𝒙 + 𝟓 𝟒 𝟐 + 𝟐𝟓 𝟖 + 𝟔 = −𝟐 𝒙𝟐 + 𝟓 𝟐 𝒙 + 𝟓 𝟒 𝟐 − 𝟓 𝟒 𝟐 + 𝟔 𝒚 = −𝟐 𝒙 + 𝟓 𝟒 𝟐 + 𝟕𝟑 𝟖 𝒃 𝒚 = −𝟐𝒙𝟐 − 𝟓𝒙 + 𝟔 ∴ 当𝒙 = − 𝟏 𝟑 时 , 𝒚𝒎𝒂𝒙 = 𝟗 𝟏 𝟖 ∴ 当𝒙 = − 𝟓 𝟒 时 ,
- 3. 𝒑𝒈 𝟕𝟒 极值 𝟏. 𝒂 𝒚 = 𝟐 𝒙 + 𝟏 𝟐 + 𝟐 当 𝒙 = − 1 时 , 𝒚𝒎𝒊𝒏 = 𝟐 𝒃 𝒚 = −𝟑 𝒙 − 𝟏 𝟐 + 𝟏 𝟐 𝟐. 𝒂 𝒚 = 𝟑𝒙𝟐 + 𝟔𝒙 + 𝟕 𝒄 𝒚 = − 𝒙 + 𝟏 𝟐 𝒅 𝒚 = 𝟐 𝒙 − 𝟓 𝟐 − 𝟑 𝒃 𝒚 = −𝒙𝟐 − 𝟒𝒙 + 𝟓 𝒄 𝒚 = 𝟐𝒙𝟐 + 𝟏𝟐𝒙 + 𝟏𝟖 𝒅 𝒚 = −𝟓𝒙𝟐 − 𝟔𝒙 − 𝟒 当 𝒙 = 5 时 , 𝒚𝒎𝒊𝒏 = − 𝟑 当 𝒙 = − 1 时 , 𝒚𝒎𝒊𝒏 = 𝟒 当 𝒙 = − 3 时 , 𝒚𝒎𝒊𝒏 = 𝟎 当 𝒙 = 1 时 , 𝒚𝒎𝒂𝒙 = 𝟏 𝟐 当 𝒙 = − 1 时 , 𝒚𝒎𝒂𝒙 = 𝟎 当 𝒙 = − 2 时 , 𝒚𝒎𝒂𝒙 = 𝟗 当 𝒙 = − 𝟑 𝟓 时 , 𝒚𝒎𝒂𝒙 = − 𝟏𝟏 𝟓
- 4. 𝒑𝒈 𝟕𝟒 极值 = 𝒂 𝒙𝟐 + 𝒃 𝒂 𝒙 + 𝟏𝟎 = 𝒂 𝒙 + 𝒃 𝟐𝒂 𝟐 − 𝒃𝟐 𝟒𝒂 + 𝟏𝟎 = 𝒂 𝒙𝟐 + 𝒃 𝒂 𝒙 + 𝒃 𝟐𝒂 𝟐 − 𝒃 𝟐𝒂 𝟐 + 𝟏𝟎 𝒚 = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝟏𝟎 𝒂, 𝐛 = ? 𝒚𝒎𝒊𝒏 𝒐𝒓 𝒚𝒎𝒂𝒙 当 𝒙 = − 𝟏 𝟐 时 , 𝒚? = 𝟎 𝒃 𝟐𝒂 = 𝟏 𝟐 − 𝒃𝟐 𝟒𝒂 + 𝟏𝟎 = 𝟎 𝒂 = 𝒃 − 𝒃𝟐 𝟒𝒃 + 𝟏𝟎 = 𝟎 𝒃 = 𝟒𝟎 𝒚𝒎𝒊𝒏 = 𝟎 ∴ 𝒂 = 𝒃 = 𝟒𝟎 𝒂 > 𝟎, 开囗方上, 𝒚𝒎𝒊𝒏 𝒂 = 𝒃 = 𝟒𝟎 𝒚𝒎𝒊𝒏 = 𝟎
- 5. 𝒑𝒈 𝟕𝟒 极值 当 𝒙 = 3 时 , 𝒚𝒎𝒂𝒙 = 𝟗 𝟐 对称轴 𝒙 = 𝟑 = 𝒂 𝒙𝟐 + 𝟏 𝒂 𝒙 + 𝟑 = 𝒂 𝒙 + 𝟏 𝟐𝒂 𝟐 − 𝟏 𝟒𝒂 + 𝟑 = 𝒂 𝒙𝟐 + 𝟏 𝒂 𝒙 + 𝟏 𝟐𝒂 𝟐 − 𝟏 𝟐𝒂 𝟐 + 𝟑 𝒚 = 𝒂𝒙𝟐 + 𝒙 + 𝟑 𝒚𝒎𝒊𝒏 𝒐𝒓 𝒚𝒎𝒂𝒙 = ? 𝟏 𝟐𝒂 = − 𝟑 − 𝟏 𝟒𝒂 + 𝟑 𝒂 = − 𝟏 𝟔 𝒂 < 𝟎, 开囗方下, 𝒚𝒎𝒂𝒙 = 𝟔 𝟒 + 𝟑 = 𝟗 𝟐 ∴ 当 𝒙 = 3 时 , 𝒚𝒎𝒂𝒙 = 𝟗 𝟐
- 6. 𝒑𝒈 𝟕𝟒 极值 = 𝒂 𝒙𝟐 + 𝟒 𝒂 𝒙 + 𝟏 = 𝒂 𝒙 + 𝟐 𝒂 𝟐 − 𝟒 𝒂 + 𝟏 = 𝒂 𝒙𝟐 + 𝟒 𝒂 𝒙 + 𝟐 𝒂 𝟐 − 𝟐 𝒂 𝟐 + 𝟏 𝒚 = 𝒂𝒙𝟐 + 𝟒𝒂𝒙 + 𝟏 𝒂 = ? − 𝟒 𝒂 + 𝟏 = 𝟓 − 𝟒 = 𝟒𝒂 𝒚𝒎𝒊𝒏 𝒐𝒓 𝒚𝒎𝒂𝒙 = 𝟓 𝒂 = − 𝟏 𝒂 = − 𝟏
- 7. 𝒑𝒈 𝟕𝟑 = 𝒎 𝒙𝟐 + 𝟒𝒎𝒙 + 𝟓𝒏 当 𝒎 > 𝟎 时 , 𝒚𝒎𝒊𝒏 = − 𝟒𝒎𝟑 + 𝟓𝒏 𝒂 = ? 𝒚𝒎𝒊𝒏 , 𝒚𝒎𝒂𝒙 𝒚 = 𝒎𝒙𝟐 + 𝟒𝒎𝟐𝒙 + 𝟓𝒏 = 𝒎 𝒙 + 𝒎 𝟐 − 𝟒𝒎𝟑 + 𝟓𝒏 = 𝒎 𝒙𝟐 + 𝟒𝒎𝒙 + 𝟐𝒎 𝟐 − 𝟐𝒎 𝟐 + 𝟓𝒏 𝟐𝟎. 𝒄 𝒚 = 𝒎𝒙𝟐 + 𝟒𝒎𝟐 𝒙 + 𝟓𝒏 极值 当 𝒎 < 𝟎 时 , 𝒚𝒎𝒂𝒙 = − 𝟒𝒎𝟑 + 𝟓𝒏 ∴ 𝑺𝒎𝒂𝒙 = 𝟑𝟔 𝒎𝟐 = −𝒙𝟐 + 𝟏𝟐𝒙 = − 𝒙 − 𝟔 𝟐 + 𝟑𝟔 = − 𝒙𝟐 − 𝟏𝟐𝒙 + 𝟔 𝟐 − 𝟔 𝟐 𝟐𝟒 = 𝟐(𝒙 + 𝒛) 𝒙 𝒛 𝒙 + 𝒛 = 𝟏𝟐 𝒛 = 𝟏𝟐 − 𝒙 𝑺 = 𝒙(𝟏𝟐 − 𝒙)
- 8. 𝒑𝒈 𝟕𝟒 极值 A certain farmer wanted to form a rectangular vegetable garden with 80 m long bar. How to enclose the area to maximize the area? = −𝒙𝟐 + 𝟒𝟎𝒙 = − 𝒙 − 𝟐𝟎 𝟐 + 𝟒𝟎𝟎 = − 𝒙𝟐 − 𝟒𝟎𝒙 + 𝟐𝟎 𝟐 − 𝟐𝟎 𝟐 𝟖𝟎 = 𝟐(𝒙 + 𝒛) 𝒙 𝒛 𝒙 + 𝒛 = 𝟒𝟎 𝒛 = 𝟒𝟎 − 𝒙 𝑺 = 𝒙(𝟒𝟎 − 𝒙)
- 9. 𝒑𝒈 𝟕𝟒 极值 当 𝒙 = ? 时 , 𝒚𝒎𝒂𝒙 ∴ 当 𝒙 = 10 𝒎 时 , 𝒚𝒎𝒂𝒙 = 𝟐𝟎𝟎 𝒎𝟐 one side is a wall, and the other three sides are surrounded by a rectangular field with a 40 m long fence = −𝟐𝒙𝟐 + 𝟒𝟎𝒙 = −𝟐 𝒙𝟐 − 𝟐𝟎𝒙 + 𝟏𝟎 𝟐 − 𝟏𝟎 𝟐 𝟒𝟎 = 𝟐𝒙 + 𝒘 𝑺 = 𝒙(𝟒𝟎 − 𝟐𝒙) 𝒘 = 𝟒𝟎 − 𝟐𝒙 = −𝟐 𝒙 − 𝟏𝟎 𝟐 + 𝟐𝟎𝟎 当 𝒙 = 10 𝒎 时 , 𝒚𝒎𝒂𝒙 = 𝟐𝟎𝟎 𝒎𝟐
- 10. 𝒑𝒈 𝟕𝟔 极值 𝒙 = ? 时 , 𝑺𝒎𝒂𝒙 = ? 𝒙 = 𝑹𝟐 𝟐 = 𝟐 𝟐 𝑹 时, 𝑺𝒎𝒂𝒙 = 𝟐𝑹𝟐 𝑺𝟐 = 𝟏𝟔𝒖 (−𝐮 + 𝑹𝟐) 𝑹𝟐 = 𝒙𝟐 + 𝒚𝟐 = 𝟒𝒙 𝑹𝟐 − 𝒙𝟐 𝒚 = 𝑹𝟐 − 𝒙𝟐 − −① 𝑺 = 𝟒𝒙𝒚 𝑳𝒆𝒕 𝒖 = 𝒙𝟐 𝑺𝟐 = 𝟏𝟔𝒙𝟐 (−𝒙𝟐 + 𝑹𝟐 ) = −𝟏𝟔𝒖𝟐 + 𝟏𝟔𝑹𝟐 = −𝟏𝟔 𝒖𝟐 − 𝑹𝟐 + 𝑹 𝟐 𝟐 − 𝑹 𝟐 𝟐 𝑺𝟐 = −𝟏𝟔 𝒖 − 𝑹𝟐 𝟐 𝟐 + 𝟒𝑹𝟐 𝒖 = 𝒙𝟐 = 𝑹𝟐 𝟐 − −② 𝒙 = 𝑹𝟐 𝟐 ② 𝒊𝒏𝒕𝒐 ① 𝒚 = 𝑹𝟐 − 𝑹𝟐 𝟐 = 𝑹𝟐 𝟐
- 11. 𝒑𝒈 𝟕𝟔 极值 𝒙 = ? 时 , 𝑺𝒎𝒂𝒙 = ? 𝑹𝟐 = 𝒙𝟐 + 𝒚𝟐 𝒚 = 𝑹𝟐 − 𝒙𝟐 − −① 𝒖 = 𝒙𝟐 = 𝑹𝟐 𝟐 𝒙 = 𝑹𝟐 𝟐 − −② ② 𝒊𝒏𝒕𝒐 ① 𝒚 = 𝑹𝟐 − 𝑹𝟐 𝟐 = 𝑹𝟐 𝟐 𝒙 = 𝑹𝟐 𝟐 = 𝟐 𝟐 𝑹 时, 𝑺𝒎𝒂𝒙 = 𝟐𝑹𝟐 𝑺 = 𝟒𝒙𝒚 = 𝟒 𝑹𝟐 𝟐 𝑺𝒎𝒂𝒙 = 𝟐𝑹𝟐 𝑺𝟐 = −𝟏𝟔 𝒖 − 𝑹𝟐 𝟐 𝟐 + 𝟒𝑹𝟒 𝑺𝟐 = 𝟒𝑹𝟒 𝑺𝒎𝒂𝒙 = 𝟐𝑹𝟐
- 12. 𝒑𝒈 𝟕𝟒 极值 = −𝒙𝟐 + 𝟑𝟔𝒙 = − 𝒙𝟐 − 𝟑𝟔𝒙 + 𝟏𝟖 𝟐 − 𝟏𝟖 𝟐 𝟑𝟔 = 𝒙 + 𝒛 𝑺 = 𝒙(𝟑𝟔 − 𝒙) 𝒛 = 𝟑𝟔 − 𝒙 = − 𝒙 − 𝟏𝟖 𝟐 + 𝟑𝟐𝟒 the score of 36 is divided into two rounds make 𝒚𝒎𝒂𝒙 for products of two rounds
- 13. 𝒑𝒈 𝟕𝟔 极值 = 𝒙𝟐 + 𝒙𝟐 + 𝟑𝟐𝟒 − 𝟑𝟔𝒙 = 𝟐 𝒙𝟐 − 𝟏𝟖𝒙 + 𝟗𝟐 − 𝟗𝟐 + 𝟑𝟐𝟒 𝟏𝟖 = 𝒙 + 𝒛 𝑺 = 𝒙𝟐 + 𝟏𝟖 − 𝒙 𝟐 𝒛 = 𝟏𝟖 − 𝒙 = 𝟐 𝒙 − 𝟗 𝟐 + 𝟏𝟔𝟐 divide 18 into two numbers so that the sum of the squares of these two numbers, 𝑠𝒎in = 𝟐𝒙𝟐 − 𝟑𝟔𝒙 + 𝟑𝟐𝟒
- 14. 𝒑𝒈 𝟕𝟔 极值 当 𝒙 = 𝟏 𝟐 时 , 𝒚𝒎𝒊𝒏 𝑳𝒆𝒕 𝒙 = 𝑪𝟏 = 𝟐𝝅𝒓𝟏 𝑨𝟏 = 𝝅 𝒓𝟏 𝟐 𝒚 = 𝟏 𝟒𝝅 𝒙𝟐 + 𝒍 − 𝒙 𝟐 = 𝟏 𝟒𝝅 𝒙𝟐 + 𝒙𝟐 − 𝟐𝒍𝒙 + 𝒍𝟐 = 𝟏 𝟐𝝅 𝒙𝟐 − 𝒍𝒙 + 𝒍𝟐 𝟐 𝑳𝒆𝒕 𝒍 − 𝒙 = 𝑪𝟐 = 𝟐𝝅𝒓𝟐 𝒓𝟏 = 𝒙 𝟐𝝅 = 𝝅 𝒙 𝟐𝝅 𝟐 = 𝒙𝟐 𝟒𝝅 𝒓𝟐 = 𝒍 − 𝒙 𝟐𝝅 𝑨𝟐 = 𝝅 𝒓𝟐 𝟐 = 𝝅 𝒍 − 𝒙 𝟐𝝅 𝟐 = 𝒍 − 𝒙 𝟐 𝟒𝝅 = 𝟏 𝟐𝝅 𝒙𝟐 − 𝒍𝒙 + 𝒍 𝟐 𝟐 − 𝒍 𝟐 𝟐 + 𝒍𝟐 𝟐 = 𝟏 𝟐𝝅 𝒙 − 𝒍 𝟐 𝟐 + 𝒍𝟐 𝟖𝝅 ∴ 当 𝒙 = 𝟏 𝟐 时 , 𝒚𝒎𝒊𝒏
- 15. 𝒑𝒈 𝟕𝟒 极值 𝒔 𝒕 ∴ 当 𝒕 = 𝟑 𝟒 秒 , 𝒔𝒎𝒂𝒙 = 𝟗 𝒎 = −𝟏𝟔 𝒕𝟐 − 𝟐𝟒 𝟏𝟔 𝒕 = −𝟏𝟔 𝒕𝟐 − 𝟑 𝟐 𝒕 + 𝟑 𝟒 𝟐 − 𝟑 𝟒 𝟐 𝒔 = 𝟐𝟒𝒕 − 𝟏𝟔𝒕𝟐 = −𝟏𝟔 𝒕 − 𝟑 𝟒 𝟐 + 𝟗 当 𝒕 = ? 秒 , 𝒔𝒎𝒂𝒙 = ? 𝒎 ∴ 当 𝒕 = 𝟑 𝟒 秒 , 𝒔𝒎𝒂𝒙 = 𝟗 𝒎
Editor's Notes
- the diameter of the circle is D, find its inscribed rectangle's largest area the book ans is x = one side of the rectangle?
- the diameter of the circle is D, find its inscribed rectangle's largest area the book ans is x = one side of the rectangle?