Course Content
第1章：行列式
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1.1 Determinant Expension 行列式的展开
1.2 Law of Determinants 行列式的性质
1.3 Cramer’s Rule 克兰姆法则
第二章：矩阵
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2.1 Addition, Subtraction and Multiplication of Matrices 矩阵的加减法和乘法
2.2 Inverse Matrices逆矩阵
2.3 Solve Linear Simultaneous Equations by Using Matrices 用矩阵解线性方程式
第七章：二项式定理
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7.1 Binomial Theorem with Natural Number Power 指数为自然数的二项式定理
7.2 General Term Formula for Binomial Expansion 二项展开式的通项公式
7.3 Binomial Theorem with Rational Power 指数为有理数的二项式定理
7.4 Application of Binomial Theorem in Approximate Calculation 二项式定理在近似值计算中的应用
第十二章：极限
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12.1 Concept of Limits 极限的概念
12.2 Limit of Functions 函数的极限
12.3 Limit of Infinities 无穷的极限
12.4 Limit of Series 数列的极限
12.5 Continuous Functions 连续函数
第十三章：微分法（一）
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13.1 Derivatives 导数
13.2 Continuous of Functions 函数的连续性
13.3 Differentiation Formulas 微分法则
13.4 Chain Rles链导法
13.5 Higher Degree Derivatives 高阶导数
13.6 Sandwich Theorem 挟挤定理
13.7 Derivatives of Trigonometric Functions 三角函数的导数公式
第十四章：微分法的应用（一）
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14.1 Tangent and Normal 切线与法线
14.2 The Increasing, decreasing & Extreme Values of Functions 函数的增减性&极值
14.3 Graph Sketching for Rational Functions 有理函数的作图
14.4 Optimization 最佳化
第十五章：不定积分（一）
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15.1 Fundamental Indefinite Integral 基本不定积分
15.2 Indefinite Integral of Trigonometric Functions 三角函数的不定积分
15.3 The Substitution Rules of Integration 换元积分法
第十六章：定积分及其应用（一）
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16.1 Fundamental Definite Integral 基本定积分
16.2 The Substitution Rules of Integration 换元积分法
16.3 Calculation of Area 面积的计算
16.4 Volume of Rotating Solid 旋转体的体积
高二数学 | 高级数学
About Lesson

14.4 Optimization 最佳化

 

1.
If 1200 cm^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

若用1200 cm^2 的材料要做一个底为正方形且无盖的盒子，求盒子的最大可能体积。

Answer :
4000cm^3
2.
If x = t – 3 and y = 2t^2 – 2t, find the extreme value of z where z = 3x^2 + y and state the nature.

若x = t – 3 和 y = 2t^2 – 2t, 求 z 的极值其中 z = 3x^2 + y 并说明其性质。

Answer :
7, min

Answer :
(b) 4, 2\sqrt{3}
4.
Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola y = 8 – x^2.

求使长方形的面积最大时的尺寸，若其底在x-轴且另两个顶点在x-轴以上并落在抛物线y = 8 – x^2 上。

Answer :
\frac{4}{5}\sqrt{6} x 5\frac{1}{3}
5.
A sector with central angle \theta radians is cut from a circle of radius 12cm, and the resulting angles are brought together to form a cone with height h and the radius r.
从一半径为12cm的圆剪掉一圆心角\theta 弧度的扇形，剩余的角连接起来形成一圆锥体，其高为 h 和半径为 r.

(a) State the relations between
说明以下的关系

(i) r and \theta,
r 和 \theta,

(ii) r and h.
r and h,

(b) Find the magnitude of r so that the volume of cone is maximum.
求r 的大小能使圆锥面积最大。

(c) Accurate to 2 decimal places, find the value of \theta when the volume of cone is maximum.
准确至2个小数位，求当圆锥体体积最大时\theta的值。

Answer :

6.
A piece of wire of length 8 cm is cut into two pieces, one of length x cm, the other of length (8 –x) cm. The piece of length x cm is bent to form a circle with circumference x cm. The other piece is bent to form a square with perimeter (8 –x) cm. Show that, as x varies, the sum of the areas enclosed by these two pieces of wire is a minimum when the radius of the circle is \frac{4}{4+\pi} cm.

一条8cm 长的铁线被剪成两段, 一段长 x cm, 另一段长 (8 –x)cm. x cm 长的那段被弯曲成一圆周为 x cm的圆. 另外一段则被弯成一周长为 (8 –x)cm的正方形. 试证, 当 x 改变, 被两段铁线所包围的总面积最小时圆的半径是 \frac{4}{4+\pi} cm.

 

7.
A piece of wire, 60 cm long, is bent to form the shape consist of a semicircular arc, radius r cm, and three sides of a rectangle of height x cm. Determine to 3 significant figures, the value of r for which the area of the window is a maximum .

一条铁线60 cm 长, 被弯曲成一形状包括一半圆的弧, 其半径 r cm, 和一高为x cm的长方形的三边. 计算准确至 3 位有效数字, r 的值。

Answer :
8.40cm
8.
Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r.

求在半径为r的圆可内接最大面积的等腰三角形之尺寸。

Answer :
\sqrt{3}r, \sqrt{3}r, \sqrt{3}r

Answer :
(c)  

(i)   4, 16  

(ii)   半径变小，高变大          

(d) 不，比例保持不变

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