IB Mathematics: Optimisation Problems with Quadratic Functions | IB数学：二次函数的最优化问题

📚 IB Mathematics: Optimisation Problems with Quadratic Functions | IB数学：二次函数的最优化问题

Quadratic functions are among the most powerful tools in mathematics for modelling real-world situations where a maximum or minimum value is required. From maximising the area enclosed by a fence to minimising production costs, the parabolic shape of a quadratic function provides a clear, visual representation of optimisation.

二次函数是数学中解决现实最优化问题的强大工具之一，无论是围栏围成的最大面积，还是生产成本的最小化，抛物线的形状都为最大或最小值提供了一种清晰直观的表示。

1. The Vertex as the Optimum Point | 顶点作为最优点

Every quadratic function $y = ax^2 + bx + c$ (with $a \neq 0$) has a single turning point called the vertex. If $a > 0$, the vertex represents the minimum value of the function; if $a < 0$, it represents the maximum value.

每个二次函数 $y = ax^2 + bx + c$（其中 $a \neq 0$）都有一个唯一的转折点，称为顶点。若 $a > 0$，顶点对应函数的最小值；若 $a < 0$，顶点对应函数的最大值。

The x-coordinate of the vertex is found using:

顶点的 x 坐标由下式求出：

x_vertex = -b / (2a)

Substituting this value back into the quadratic equation gives the corresponding y-coordinate, which is the optimal value sought.

将该值代回二次方程，即可得到对应的 y 坐标，也就是所寻求的最优值。

2. Completing the Square: Vertex Form | 配方法：顶点式

Writing a quadratic in the form $y = a(x – h)^2 + k$ reveals the vertex directly at $(h, k)$. This is called the vertex form, and it is especially useful in optimisation because it gives the maximum or minimum value without further calculation.

将二次函数写成 $y = a(x – h)^2 + k$ 的形式，可直接看出顶点坐标为 $(h, k)$。这种形式称为顶点式，在优化问题中尤其有用，因为无需额外计算即可获得最大值或最小值。

For example, $y = 2x^2 – 8x + 3$ can be rewritten as $y = 2(x – 2)^2 – 5$. Therefore the minimum value is -5, occurring at $x = 2$.

例如，$y = 2x^2 – 8x + 3$ 可改写为 $y = 2(x – 2)^2 – 5$。因此最小值为 -5，在 $x = 2$ 处取得。

3. Using the Derivative to Find Extrema | 用导数求极值

For a quadratic function $f(x)$, the derivative $f'(x) = 2ax + b$ is a linear function. Setting $f'(x) = 0$ yields the same vertex x-coordinate, $-b / (2a)$. This calculus approach is essential when the optimisation problem involves more complex functions in IB.

对二次函数 $f(x)$，其导数 $f'(x) = 2ax + b$ 是一次函数。令 $f'(x) = 0$，得到同样的顶点 x 坐标 $-b / (2a)$。当最优化问题涉及更复杂的函数时，这种微积分方法在 IB 中尤为重要。

The second derivative $f”(x) = 2a$ determines the nature of the critical point: if $a > 0$, it is a minimum; if $a < 0$, it is a maximum.

二阶导数 $f”(x) = 2a$ 用于判断临界点的性质：若 $a > 0$，则为极小值；若 $a < 0$，则为极大值。

4. The Discriminant and Feasible Ranges | 判别式与可行域

In optimisation problems, the discriminant $\Delta = b^2 – 4ac$ helps determine the number of intersections with the x-axis. However, its more subtle role is to ensure that the mathematical optimum lies within a physically or economically meaningful domain.

在优化问题中，判别式 $\Delta = b^2 – 4ac$ 用于判断二次函数与 x 轴的交点个数。然而，其更细致的作用在于确保数学上的最优点落在具有实际物理或经济意义的范围内。

For instance, when modelling profit $P(x) = -2x^2 + 40x – 100$, the vertex occurs at $x = 10$. The discriminant tells us whether the profit is positive for that production level, which is a necessary feasibility check.

例如，在建模利润 $P(x) = -2x^2 + 40x – 100$ 时，顶点位于 $x = 10$。判别式可以判断在该产量水平下利润是否为正值，这是必要的可行性检验。

5. Optimising Area with a Fixed Perimeter | 固定周长下的面积最大化

A classic optimisation problem: a rectangular enclosure is to be built using 100 metres of fencing. If one side uses no fencing because it lies along a river, what dimensions maximise the area?

一个经典的优化问题：用 100 米篱笆围成一个矩形场地。如果靠河的一边不需要围栏，那么怎样的尺寸能使面积最大？

Let the two equal sides be $x$ and the side parallel to the river be $100 – 2x$. The area is $A(x) = x(100 – 2x) = -2x^2 + 100x$. Completing the square or using the vertex formula gives $x = 25$, so the maximum area is 1250 m².

设两条相等的边为 $x$，平行于河边的边为 $100 – 2x$。面积为 $A(x) = x(100 – 2x) = -2x^2 + 100x$。配方法或顶点公式可得 $x = 25$，故最大面积为 1250 平方米。

6. Maximising Revenue from a Price Increase | 涨价策略中的收益最大化

Business applications often involve a trade-off between price and quantity sold. Suppose a shop sells 200 items at $10 each. For every $0.50 increase in price, 5 fewer items are sold. The revenue is $R(x) = (10 + 0.5x)(200 – 5x)$, where $x$ is the number of $0.50 price increases.

商业应用中经常涉及价格与销量之间的权衡。假设一家商店以每个 10 美元卖出 200 件商品。价格每提高 0.50 美元，销量就减少 5 件。收入为 $R(x) = (10 + 0.5x)(200 – 5x)$，其中 $x$ 表示 0.50 美元涨价的次数。

Expanding gives $R(x) = -2.5x^2 + 50x + 2000$. The vertex occurs at $x = 10$, meaning 10 price increases (to $15) maximise revenue, giving $R = 2250$ dollars.

展开得 $R(x) = -2.5x^2 + 50x + 2000$。顶点位于 $x = 10$，即涨价 10 次（售价 15 美元）使收入最大，最大收入为 2250 美元。

7. Optimising Distance and Projectile Motion | 距离与抛体运动的优化

In physics, the height of a projectile is modelled by $h(t) = -5t^2 + 20t + 1$. The maximum height is found at $t = 2$ seconds, giving $h(2) = 21$ metres. The time to hit the ground is found by solving $h(t) = 0$, with the positive root being the feasible answer.

在物理中，抛体的高度可建模为 $h(t) = -5t^2 + 20t + 1$。最大高度出现在 $t = 2$ 秒，即 $h(2) = 21$ 米。落地时间通过求解 $h(t) = 0$ 得出，取正根作为可行答案。

Optimisation also applies to finding the minimum distance from a point to a curve. The squared distance is usually a quadratic function, and minimising it avoids square roots.

最优化也适用于求点到曲线的最小距离。距离的平方通常是二次函数，通过最小化它可避免平方根运算。

8. Minimising Surface Area for a Given Volume | 给定体积下的表面积最小化

A cylindrical can with fixed volume $V$ can be designed to minimise its surface area. The surface area as a function of radius $r$ is $S(r) = 2\pi r^2 + 2V/r$, which is not quadratic. However, for a rectangular prism with a square base and fixed volume, the surface area simplifies to a quadratic in the side length.

给定体积 $V$ 的圆柱形罐子可以设计为使表面积最小。表面积随半径变化的关系为 $S(r) = 2\pi r^2 + 2V/r$，它不是二次的。然而，对于底面为正方形的长方体，若体积固定，其表面积可以简化为关于边长的二次函数。

Let the base side be $x$ and height be $h$, with $x^2 h = V$. The surface area $A = 2x^2 + 4xh = 2x^2 + 4V/x$. Differentiating and setting to zero gives the optimal dimension. Though the final step involves non-linear equations, the theory of quadratic extrema underpins the initial reasoning.

设底面边长为 $x$，高为 $h$，且 $x^2 h = V$。表面积 $A = 2x^2 + 4xh = 2x^2 + 4V/x$。求导并令其为零可得到最优尺寸。虽然最后一步涉及非线性方程，但二次极值的理论是初始推理的基础。

9. Inequalities and Restricted Domains | 不等式与受限定义域

Real-world optima are not always attained at the vertex. If the domain is restricted, such as $x \in [0, 10]$, the maximum might occur at an endpoint. Therefore always check both the vertex and the boundaries.

现实中的最优点不一定在顶点处取得。如果定义域受限，例如 $x \in [0, 10]$，最大值可能在端点处取得。因此，必须同时检查顶点和边界。

For example, $f(x) = -x^2 + 8x + 5$ has its vertex at $x = 4$, which lies inside the domain, so the maximum is $f(4) = 21$. But if the domain were $[0, 3]$, the maximum would be $f(3) = 20$, not the vertex value.

例如，$f(x) = -x^2 + 8x + 5$ 的顶点位于 $x = 4$，在定义域内，所以最大值为 $f(4) = 21$。但如果定义域为 $[0, 3]$，最大值则为 $f(3) = 20$，而不是顶点值。

10. Worked IB-Style Problem | IB 风格例题

A farmer wants to create a rectangular paddock along a straight riverbank. No fence is needed along the river. The farmer has 240 metres of fencing. Find the maximum possible area and the corresponding dimensions.

一位农场主想沿着笔直的河岸建一个矩形牧场。沿河的一边不需要围栏。现有 240 米篱笆。求最大可能面积及相应的尺寸。

Let the two sides perpendicular to the river be $x$, and the side parallel to the river be $240 – 2x$. The area is $A = x(240 – 2x) = -2x^2 + 240x$. The vertex is at $x = -240 / (2 \times -2) = 60$. Thus the parallel side is $240 – 120 = 120$. The maximum area is $60 \times 120 = 7200$ m².

设垂直于河岸的两条边为 $x$，平行于河岸的边为 $240 – 2x$。面积为 $A = x(240 – 2x) = -2x^2 + 240x$。顶点在 $x = -240 / (2 \times -2) = 60$ 处。因此平行边为 $240 – 120 = 120$。最大面积为 $60 \times 120 = 7200$ 平方米。

This example illustrates the complete process: define the variable, write the quadratic model, identify the vertex, and interpret the result within the physical context.

此例展示了完整的解题过程：设定变量、建立二次模型、确定顶点，并在实际情境中解释结果。

11. Common Mistakes and Exam Tips | 常见错误与考试建议

Forgetting to check the sign of $a$: if $a < 0$, the vertex is a maximum; if $a > 0$, it is a minimum. Switching these leads to incorrect answers.

忘记检查 $a$ 的符号：若 $a < 0$，顶点为最大值；若 $a > 0$，则为最小值。混淆这一点会导致答案错误。
Ignoring the physical domain: sometimes the vertex x-coordinate is outside the feasible region. Always state the domain from the problem statement.

忽略实际定义域：有时顶点的 x 坐标不在可行范围内。务必根据题目条件写出定义域。
Misreading the variable in price-demand problems: clearly define what $x$ represents, such as the number of price increases, not the final price itself.

在价格-需求问题中误读变量：清晰定义 $x$ 的含义，例如是涨价的次数，而不是最终价格。
Not verifying the nature of the optimal point using either the second derivative or the sign of $a$ in vertex form.

未通过二阶导数或顶点式中的 $a$ 符号验证最优点性质。

12. Summary and Further Practice | 总结与巩固练习

Quadratic optimisation in IB Mathematics requires a solid grasp of vertex form, derivative tests, and the ability to translate real-world constraints into a mathematical model. The vertex formula $x = -b/(2a)$ is the key computational tool, but domain awareness ensures that the answer makes practical sense.

IB 数学中的二次函数最优化问题要求牢固掌握顶点式、导数检验法，以及将实际约束转化为数学模型的能力。顶点公式 $x = -b/(2a)$ 是关键的计算工具，而定义域意识则确保答案具有实际意义。

To master this topic, practise a variety of problems: geometry, business, physics, and pure graph analysis. For each problem, always answer the “real-world” question — not just the mathematical extremum.

要精通这一主题，需要练习多种类型的问题：几何、商业、物理以及纯图形分析。在每道题中，始终要回答“现实世界”的问题，而不仅仅是数学上的极值。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

IB Mathematics: Optimisation Problems with Quadratic Functions | IB数学：二次函数的最优化问题