Sources: Visual - various maths sites (credits to original creator) Questions - Dong Zong's Textbook suitable for SUEC (Maths), SPM (Maths and Add Maths) too

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