Quadratic Function Optimisation Problems in IB Maths | IB数学：二次函数的优化问题

📚 Quadratic Function Optimisation Problems in IB Maths | IB数学：二次函数的优化问题

Quadratic functions are one of the most frequently tested topics in IB Mathematics. Optimisation, which involves finding the maximum or minimum value of a quadratic function, is a core skill that appears in both Analysis and Approaches (AA) and Applications and Interpretation (AI). This article explains the key concepts, step-by-step methods, and common exam traps so that you can solve these problems with confidence.

二次函数是IB数学中最常考的内容之一。优化问题，即求二次函数的最大值或最小值，是IB数学分析与方法（AA）以及应用与解释（AI）中的核心技能。本文将讲解关键概念、分步解法以及常见考试陷阱，帮助你自信地解决这类问题。

1. Standard Form of a Quadratic Function | 二次函数的标准形式

A quadratic function can be written in the standard form f(x) = ax² + bx + c, where a ≠ 0. The sign of a determines the shape of the parabola: if a > 0, the graph opens upwards and has a minimum value; if a < 0, it opens downwards and has a maximum value.

二次函数可以写成标准形式 f(x) = ax² + bx + c，其中 a ≠ 0。a 的符号决定了抛物线的形状：当 a > 0 时，图像开口向上，存在最小值；当 a < 0 时，图像开口向下，存在最大值。

In optimisation problems, we are usually asked to find the value of x that gives the extreme value, and then calculate that extreme value itself.

在优化问题中，通常要求我们找到使函数取得极值的 x 值，然后计算该极值本身。

2. The Vertex and the Extremum | 顶点与极值

The vertex of a parabola is the point where the function reaches its maximum or minimum. The x-coordinate of the vertex is given by the formula:

抛物线的顶点是函数达到最大值或最小值的点。顶点的 x 坐标公式为：

x = −b / (2a)

Once you have this x-value, substitute it back into f(x) to find the corresponding y-value, which is the maximum or minimum value of the function.

得到 x 值后，将其代回 f(x) 求出对应的 y 值，该 y 值就是函数的最大值或最小值。

For example, if f(x) = 2x² − 8x + 5, then a = 2 and b = −8, so x = −(−8)/(2·2) = 2. Substituting gives f(2) = 2·4 − 8·2 + 5 = −3. Since a > 0, the minimum value is −3.

例如，若 f(x) = 2x² − 8x + 5，则 a = 2，b = −8，所以 x = −(−8)/(2·2) = 2。代入得 f(2) = 2·4 − 8·2 + 5 = −3。因为 a > 0，最小值为 −3。

3. Completing the Square | 配方法

Completing the square rewrites f(x) = ax² + bx + c into the vertex form f(x) = a(x − h)² + k, where (h, k) is the vertex. This form is especially useful because the maximum or minimum value is simply k.

配方法将 f(x) = ax² + bx + c 改写为顶点式 f(x) = a(x − h)² + k，其中 (h, k) 是顶点。这种形式特别有用，因为最大值或最小值就是 k。

For example, f(x) = x² − 6x + 10 can be written as (x − 3)² + 1. Therefore the vertex is (3, 1), and the minimum value is 1.

例如，f(x) = x² − 6x + 10 可以写成 (x − 3)² + 1。因此顶点为 (3, 1)，最小值为 1。

In IB exams, you may be required to complete the square without a calculator, especially in the non-calculator paper. Practise with fractions and negative coefficients carefully.

在IB考试中，特别是在非计算器试卷中，可能会要求你使用配方法。练习时要特别注意分数和负系数的情况。

4. The Axis of Symmetry | 对称轴

The axis of symmetry is the vertical line x = −b/(2a). It passes through the vertex and divides the parabola into two mirror-image halves.

对称轴是垂直线 x = −b/(2a)。它经过顶点，将抛物线分成两个镜像对称的部分。

When solving optimisation problems, the axis of symmetry tells you the location of the optimal input value. If the problem gives a restricted domain, you must check whether this x-value lies inside that domain.

在解决优化问题时，对称轴告诉你最佳输入值的位置。如果题目给定了定义域限制，你必须检查这个 x 值是否在该定义域内。

For instance, if the domain is 0 ≤ x ≤ 5 and the vertex is at x = 7, then the maximum or minimum on that interval will occur at an endpoint, not at the vertex.

例如，如果定义域为 0 ≤ x ≤ 5，而顶点在 x = 7，那么该区间上的最大值或最小值将出现在端点处，而不是顶点处。

5. Determining Whether It Is a Maximum or Minimum | 判断最大值还是最小值

Simply look at the coefficient a:

只需看系数 a：

If a > 0, the parabola opens upwards, so the vertex gives a minimum value.
如果 a > 0，抛物线开口向上，因此顶点给出最小值。
If a < 0, the parabola opens downwards, so the vertex gives a maximum value.
如果 a < 0，抛物线开口向下，因此顶点给出最大值。

This is often the first step in an optimisation problem. In some word problems, you also need to interpret whether a maximum or minimum makes sense in the given context.

这通常是优化问题的第一步。在一些应用题中，你还需要判断最大值或最小值在给定情境中是否有实际意义。

6. Optimisation on a Restricted Domain | 有限定义域上的优化

Many real-world problems restrict x to a certain interval, such as 0 ≤ x ≤ 10 or x ∈ [1, 8]. In that case, the overall maximum or minimum may not be at the vertex.

许多实际问题会限制 x 在某个区间内，例如 0 ≤ x ≤ 10 或 x ∈ [1, 8]。在这种情况下，整体最大值或最小值可能不在顶点处。

This is an application of the Extreme Value Theorem: if a quadratic function is continuous on a closed interval, the absolute maximum and minimum occur either at the vertex (if it lies inside the interval) or at the endpoints.

这是极值定理的应用：如果二次函数在闭区间上连续，那么绝对最大值和最小值要么出现在顶点（如果顶点在区间内），要么出现在端点处。

Therefore, to solve a restricted optimisation problem:

因此，解决带限制的优化问题步骤如下：

1. Find the vertex x-coordinate. 2. Check if it lies in the domain. 3. Evaluate f at the vertex and at all endpoints. 4. Compare the values.

1. 求出顶点的 x 坐标。2. 检查它是否在定义域内。3. 计算顶点和所有端点处的函数值。4. 比较这些值。

For example, f(x) = −x² + 4x + 1 on [0, 5]. The vertex is at x = 2, which is inside the domain. f(2) = 5, f(0) = 1, f(5) = −4. The maximum is 5 and the minimum is −4.

例如，f(x) = −x² + 4x + 1 在 [0, 5] 上。顶点在 x = 2，位于定义域内。f(2) = 5，f(0) = 1，f(5) = −4。最大值为 5，最小值为 −4。

7. Application: Maximising Area with a Fixed Perimeter | 应用：固定周长下的最大面积

A classic optimisation problem: a farmer has 100 metres of fencing and wants to enclose a rectangular field along a river, so only three sides need fencing. Let the side perpendicular to the river be x. Then the side parallel to the river is 100 − 2x. The area is A = x(100 − 2x) = −2x² + 100x.

一个经典的优化问题：农民有100米围栏，想沿河圈出一块矩形田地，因此只需要围三边。设垂直于河的边为 x，则平行于河的边为 100 − 2x。面积为 A = x(100 − 2x) = −2x² + 100x。

Since a = −2 < 0, the area has a maximum. The vertex occurs at x = −100/(2·(−2)) = 25. Then the parallel side is 100 − 2·25 = 50. The maximum area is 25 × 50 = 1250 m².

因为 a = −2 < 0，面积有最大值。顶点在 x = −100/(2·(−2)) = 25。此时平行边为 100 − 2·25 = 50。最大面积为 25 × 50 = 1250 平方米。

Note the domain: x must be positive and 100 − 2x must also be positive, so 0 < x < 50. The vertex at x = 25 lies inside this interval, so it is valid.

注意定义域：x 必须为正，且 100 − 2x 也必须为正，所以 0 < x < 50。顶点 x = 25 在此区间内，因此有效。

8. Application: Projectile Motion | 应用：抛体运动

Projectile motion problems often involve a quadratic function for height h(t) = −4.9t² + v₀t + h₀, where t is time in seconds, v₀ is initial vertical velocity, and h₀ is initial height.

抛体运动问题通常涉及高度函数 h(t) = −4.9t² + v₀t + h₀，其中 t 是以秒为单位的时间，v₀ 是初始竖直速度，h₀ 是初始高度。

To find the maximum height, find the time at the vertex: t = −v₀ / (2·(−4.9)) = v₀ / 9.8. Substitute this t back into h(t) to get the maximum height.

要求最大高度，先求顶点时间：t = −v₀ / (2·(−4.9)) = v₀ / 9.8。将这个 t 代回 h(t) 即可得到最大高度。

For example, h(t) = −4.9t² + 19.6t + 2. The time of maximum height is t = 19.6/9.8 = 2 s. Then h(2) = −4.9·4 + 19.6·2 + 2 = −19.6 + 39.2 + 2 = 21.6 m.

例如，h(t) = −4.9t² + 19.6t + 2。达到最大高度的时间是 t = 19.6/9.8 = 2 秒。那么 h(2) = −4.9·4 + 19.6·2 + 2 = −19.6 + 39.2 + 2 = 21.6 米。

9. Application: Maximising Profit | 应用：利润最大化

In business contexts, profit P(x) is often modelled as a quadratic function of the number of units sold x. For example, P(x) = −5x² + 200x − 1000. Since a < 0, the profit has a maximum.

在商业情境中，利润 P(x) 通常被建模为销售量 x 的二次函数。例如，P(x) = −5x² + 200x − 1000。因为 a < 0，利润存在最大值。

The optimal number of units is x = −200/(2·(−5)) = 20. The maximum profit is P(20) = −5·400 + 200·20 − 1000 = −2000 + 4000 − 1000 = 1000.

最佳销售数量为 x = −200/(2·(−5)) = 20。最大利润为 P(20) = −5·400 + 200·20 − 1000 = −2000 + 4000 − 1000 = 1000。

Remember to check whether the optimal x is within the feasible range given by the problem. Sometimes x must be an integer, or it cannot exceed a production capacity.

记住要检查最优 x 是否在题目给定的可行范围内。有时 x 必须是整数，或者不能超过生产能力。

10. Common Pitfalls and How to Avoid Them | 常见陷阱与应对方法

Here are the most common mistakes students make in quadratic optimisation problems:

以下是学生在二次函数优化问题中最常见的错误：

Using the wrong sign for a: forgetting that a negative a means a maximum, not a minimum.
把 a 的符号弄错：忘记 a 为负时意味着最大值，而不是最小值。
Forgetting to check the domain: the vertex might lie outside the allowed interval.
忘记检查定义域：顶点可能不在允许的区间内。
Mixing up the x and y coordinates of the vertex: the x-coordinate is the input, the y-coordinate is the extreme value.
混淆顶点的 x 和 y 坐标：x 坐标是输入值，y 坐标是极值。
Substituting incorrectly when completing the square, especially with fractions.
配方法代入错误，特别是处理分数时。
Not writing the units in word problems.
在应用题中忘记写单位。

To avoid these, always write down a, b, and c first. Then use the formula x = −b/(2a) and verify your result by substitution.

要避免这些错误，务必先写出 a、b、c 的值，然后使用公式 x = −b/(2a)，并通过代入验证你的结果。

11. Step-by-Step Problem-Solving Strategy | 分步解题策略

Follow this reliable strategy for any quadratic optimisation problem:

对于任何二次函数优化问题，请遵循以下可靠的策略：

Step 1: Define the variable and the function. Identify what quantity needs to be maximised or minimised.

第一步：定义变量和函数。确定需要最大化或最小化的量。

Step 2: Write the quadratic function in the form ax² + bx + c. Pay attention to the domain from the problem context.

第二步：将二次函数写成 ax² + bx + c 的形式。注意题目背景给出的定义域。

Step 3: Determine the sign of a. This tells you whether the vertex is a maximum or a minimum.

第三步：判断 a 的符号。这告诉你顶点是最大值还是最小值。

Step 4: Calculate x = −b/(2a). Check if this x lies in the domain.

第四步：计算 x = −b/(2a)。检查这个 x 是否在定义域内。

Step 5: Substitute to find the optimum value. If the vertex is outside the domain, evaluate the endpoints instead.

第五步：代入求出最优值。如果顶点在定义域外，则改为计算端点值。

Step 6: Write a conclusion with units and context.

第六步：写出包含单位和情境的结论。

12. Practice and Exam Tips | 练习与考试提示

In IB exams, quadratic optimisation questions can appear in both paper 1 (non-calculator) and paper 2 (calculator). For paper 1, you must complete the square or use the vertex formula by hand. For paper 2, you can use the graphing calculator to check your answer.

在IB考试中，二次函数优化题可能出现在试卷1（非计算器）和试卷2（计算器）中。对于试卷1，你必须手动配方或使用顶点公式。对于试卷2，你可以使用图形计算器检查答案。

Practise problems involving:

练习以下类型的题目：

Finding maximum area or minimum perimeter.
求最大面积或最小周长。
Projectile motion and maximum height.
抛体运动与最大高度。
Profit, revenue, and cost applications.
利润、收入和成本的应用。
Functions with restricted domains from real-world constraints.
由实际约束导致定义域受限的函数。

Always show your working clearly: state the quadratic function, the vertex, and the conclusion. Many exam marks are awarded for method, so even if your arithmetic is wrong, the correct formula can earn you partial credit.

始终清晰地展示你的步骤：写出二次函数、顶点和结论。许多分数是按步骤给的，所以即使计算有误，正确的公式也能帮你获得部分分数。

Finally, remember that optimisation is not just about finding a number; it is about interpreting what that number means in context. Always relate your answer back to the original question.

最后，请记住优化不仅仅是找到一个数值；它还需要解释这个数值在情境中的意义。始终将你的答案与原始问题联系起来。

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Quadratic Function Optimisation Problems in IB Maths | IB数学：二次函数的优化问题