diff --git a/codes/cpp/chapter_greedy/CMakeLists.txt b/codes/cpp/chapter_greedy/CMakeLists.txt
index 96ee4d218..91788668d 100644
--- a/codes/cpp/chapter_greedy/CMakeLists.txt
+++ b/codes/cpp/chapter_greedy/CMakeLists.txt
@@ -1,2 +1,3 @@
 add_executable(coin_change_greedy coin_change_greedy.cpp)
-add_executable(fractional_knapsack fractional_knapsack.cpp)
\ No newline at end of file
+add_executable(fractional_knapsack fractional_knapsack.cpp)
+add_executable(max_capacity max_capacity.cpp)
\ No newline at end of file
diff --git a/codes/cpp/chapter_greedy/max_capacity.cpp b/codes/cpp/chapter_greedy/max_capacity.cpp
new file mode 100644
index 000000000..99926c42a
--- /dev/null
+++ b/codes/cpp/chapter_greedy/max_capacity.cpp
@@ -0,0 +1,39 @@
+/**
+ * File: max_capacity.cpp
+ * Created Time: 2023-07-21
+ * Author: Krahets (krahets@163.com)
+ */
+
+#include "../utils/common.hpp"
+
+/* 最大容量：贪心 */
+int maxCapacity(vector<int> &ht) {
+    // 初始化 i, j 分列数组两端
+    int i = 0, j = ht.size() - 1;
+    // 初始最大容量为 0
+    int res = 0;
+    // 循环贪心选择，直至两板相遇
+    while (i < j) {
+        // 更新最大容量
+        int cap = min(ht[i], ht[j]) * (j - i);
+        res = max(res, cap);
+        // 向内移动短板
+        if (ht[i] < ht[j]) {
+            i++;
+        } else {
+            j--;
+        }
+    }
+    return res;
+}
+
+/* Driver Code */
+int main() {
+    vector<int> ht = {3, 8, 5, 2, 7, 7, 3, 4};
+
+    // 贪心算法
+    int res = maxCapacity(ht);
+    cout << "最大容量为 " << res << endl;
+
+    return 0;
+}
diff --git a/codes/java/chapter_greedy/max_capacity.java b/codes/java/chapter_greedy/max_capacity.java
new file mode 100644
index 000000000..85e743399
--- /dev/null
+++ b/codes/java/chapter_greedy/max_capacity.java
@@ -0,0 +1,38 @@
+/**
+ * File: max_capacity.java
+ * Created Time: 2023-07-21
+ * Author: Krahets (krahets@163.com)
+ */
+
+package chapter_greedy;
+
+public class max_capacity {
+    /* 最大容量：贪心 */
+    static int maxCapacity(int[] ht) {
+        // 初始化 i, j 分列数组两端
+        int i = 0, j = ht.length - 1;
+        // 初始最大容量为 0
+        int res = 0;
+        // 循环贪心选择，直至两板相遇
+        while (i < j) {
+            // 更新最大容量
+            int cap = Math.min(ht[i], ht[j]) * (j - i);
+            res = Math.max(res, cap);
+            // 向内移动短板
+            if (ht[i] < ht[j]) {
+                i++;
+            } else {
+                j--;
+            }
+        }
+        return res;
+    }
+
+    public static void main(String[] args) {
+        int[] ht = { 3, 8, 5, 2, 7, 7, 3, 4 };
+
+        // 贪心算法
+        int res = maxCapacity(ht);
+        System.out.println("最大容量为 " + res);
+    }
+}
diff --git a/codes/python/chapter_greedy/max_capacity.py b/codes/python/chapter_greedy/max_capacity.py
new file mode 100644
index 000000000..12377b408
--- /dev/null
+++ b/codes/python/chapter_greedy/max_capacity.py
@@ -0,0 +1,33 @@
+"""
+File: max_capacity.py
+Created Time: 2023-07-21
+Author: Krahets (krahets@163.com)
+"""
+
+
+def max_capacity(ht: list[int]) -> int:
+    """最大容量：贪心"""
+    # 初始化 i, j 分列数组两端
+    i, j = 0, len(ht) - 1
+    # 初始最大容量为 0
+    res = 0
+    # 循环贪心选择，直至两板相遇
+    while i < j:
+        # 更新最大容量
+        cap = min(ht[i], ht[j]) * (j - i)
+        res = max(res, cap)
+        # 向内移动短板
+        if ht[i] < ht[j]:
+            i += 1
+        else:
+            j -= 1
+    return res
+
+
+"""Driver Code"""
+if __name__ == "__main__":
+    ht = [3, 8, 5, 2, 7, 7, 3, 4]
+
+    # 贪心算法
+    res = max_capacity(ht)
+    print(f"最大容量为 {res}")
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diff --git a/docs/chapter_greedy/max_capacity_problem.md b/docs/chapter_greedy/max_capacity_problem.md
new file mode 100644
index 000000000..5219873f1
--- /dev/null
+++ b/docs/chapter_greedy/max_capacity_problem.md
@@ -0,0 +1,158 @@
+# 最大容量问题
+
+!!! question
+
+    输入一个数组 $ht$ ，数组中的每个元素代表一个垂直隔板的高度。数组中的任意两个隔板，以及它们之间的空间可以组成一个容器。容器的容量等于高度和宽度的乘积（即面积），其中高度由较短的隔板决定，宽度是两个隔板的数组索引之差。
+    
+    请在数组中选择两个隔板，使得组成的容器的容量最大，返回最大容量。
+
+![最大容量问题的示例数据](max_capacity_problem.assets/max_capacity_example.png)
+
+**第一步：问题分析**
+
+容器由任意两个隔板围成，**因此本题的状态为两个隔板的索引，记为 $[i, j]$** 。
+
+根据定义，容量等于高度乘以宽度，其中高度由短板决定，宽度是两隔板的索引之差。设容量为 $cap[i, j]$ ，可得计算公式：
+
+$$
+cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
+$$
+
+设数组长度为 $n$ ，两个隔板的组合数量（即状态总数）为 $C_n^2 = \frac{n(n - 1)}{2}$ 个。最直接地，**我们可以穷举所有状态**，从而求得最大容量，时间复杂度为 $O(n^2)$ 。
+
+**第二步：贪心策略确定**
+
+当然，这道题还有更高效率的解法。如下图所示，现选取一个状态 $[i, j]$ ，其满足索引 $i < j$ 且高度 $ht[i] < ht[j]$ ，即 $i$ 为短板、 $j$ 为长板。
+
+![初始状态](max_capacity_problem.assets/max_capacity_initial_state.png)
+
+我们发现，**如果将长板 $j$ 向短板 $i$ 靠近，则容量一定变小**。这是因为在移动长板 $j$ 后：
+
+- 宽度 $j-i$ 肯定变小；
+- 高度由短板决定，因此高度只可能不变（ $i$ 仍为短板）或变小（移动后的 $j$ 成为短板）；
+
+![向内移动长板后的状态](max_capacity_problem.assets/max_capacity_moving_long_board.png)
+
+反向思考，**我们只有向内收缩短板 $i$ ，才有可能使容量变大**。因为虽然宽度一定变小，**但高度可能会变大**（移动后的短板 $i$ 变长了）。
+
+![向内移动长板后的状态](max_capacity_problem.assets/max_capacity_moving_short_board.png)
+
+由此便可推出本题的贪心策略：
+
+1. 初始状态下，指针 $i$ , $j$ 分列与数组两端。
+2. 计算当前状态的容量 $cap[i, j]$ ，并更新最大容量。
+3. 比较板 $i$ 和 板 $j$ 的高度，并将短板向内移动一格。
+4. 循环执行第 `2.` , `3.` 步，直至 $i$ 和 $j$ 相遇时结束。
+
+=== "<1>"
+    ![最大容量问题的贪心过程](max_capacity_problem.assets/max_capacity_greedy_step1.png)
+
+=== "<2>"
+    ![max_capacity_greedy_step2](max_capacity_problem.assets/max_capacity_greedy_step2.png)
+
+=== "<3>"
+    ![max_capacity_greedy_step3](max_capacity_problem.assets/max_capacity_greedy_step3.png)
+
+=== "<4>"
+    ![max_capacity_greedy_step4](max_capacity_problem.assets/max_capacity_greedy_step4.png)
+
+=== "<5>"
+    ![max_capacity_greedy_step5](max_capacity_problem.assets/max_capacity_greedy_step5.png)
+
+=== "<6>"
+    ![max_capacity_greedy_step6](max_capacity_problem.assets/max_capacity_greedy_step6.png)
+
+=== "<7>"
+    ![max_capacity_greedy_step7](max_capacity_problem.assets/max_capacity_greedy_step7.png)
+
+=== "<8>"
+    ![max_capacity_greedy_step8](max_capacity_problem.assets/max_capacity_greedy_step8.png)
+
+=== "<9>"
+    ![max_capacity_greedy_step9](max_capacity_problem.assets/max_capacity_greedy_step9.png)
+
+代码实现如下所示。最多循环 $n$ 轮，**因此时间复杂度为 $O(n)$** 。变量 $i$ , $j$ , $res$ 使用常数大小额外空间，**因此空间复杂度为 $O(1)$** 。
+
+=== "Java"
+
+    ```java title="max_capacity.java"
+    [class]{max_capacity}-[func]{maxCapacity}
+    ```
+
+=== "C++"
+
+    ```cpp title="max_capacity.cpp"
+    [class]{}-[func]{maxCapacity}
+    ```
+
+=== "Python"
+
+    ```python title="max_capacity.py"
+    [class]{}-[func]{max_capacity}
+    ```
+
+=== "Go"
+
+    ```go title="max_capacity.go"
+    [class]{}-[func]{maxCapacity}
+    ```
+
+=== "JavaScript"
+
+    ```javascript title="max_capacity.js"
+    [class]{}-[func]{maxCapacity}
+    ```
+
+=== "TypeScript"
+
+    ```typescript title="max_capacity.ts"
+    [class]{}-[func]{maxCapacity}
+    ```
+
+=== "C"
+
+    ```c title="max_capacity.c"
+    [class]{}-[func]{maxCapacity}
+    ```
+
+=== "C#"
+
+    ```csharp title="max_capacity.cs"
+    [class]{max_capacity}-[func]{maxCapacity}
+    ```
+
+=== "Swift"
+
+    ```swift title="max_capacity.swift"
+    [class]{}-[func]{maxCapacity}
+    ```
+
+=== "Zig"
+
+    ```zig title="max_capacity.zig"
+    [class]{}-[func]{maxCapacity}
+    ```
+
+=== "Dart"
+
+    ```dart title="max_capacity.dart"
+    [class]{}-[func]{maxCapacity}
+    ```
+
+**第三步：正确性证明**
+
+之所以贪心比穷举更快，是因为每轮的贪心选择都会“跳过”一些状态。
+
+比如在状态 $cap[i, j]$ 下，$i$ 为短板、$j$ 为长板。若贪心地将短板 $i$ 向内移动一格，会导致以下状态被“跳过”，**意味着之后无法验证这些状态的容量大小**。
+
+$$
+cap[i, i+1], cap[i, i+2], \cdots, cap[i, j-2], cap[i, j-1]
+$$
+
+![移动短板导致被跳过的状态](max_capacity_problem.assets/max_capacity_skipped_states.png)
+
+观察发现，**这些被跳过的状态实际上就是将长板 $j$ 向内移动的所有状态**。而在第二步中，我们已经证明内移长板一定会导致容量变小，也就是说这些被跳过的状态的容量一定更小。
+
+也就是说，被跳过的状态都不可能是最优解，**跳过它们不会导致错过最优解**。
+
+以上的分析说明，**移动短板的操作是“安全”的**，贪心策略是有效的。
diff --git a/mkdocs.yml b/mkdocs.yml
index c6224906f..b79ebcd4d 100644
--- a/mkdocs.yml
+++ b/mkdocs.yml
@@ -269,6 +269,8 @@ nav:
     - 15.1. &nbsp; 贪心算法: chapter_greedy/greedy_algorithm.md
     # [status: new]
     - 15.2. &nbsp; 分数背包问题: chapter_greedy/fractional_knapsack_problem.md
+    # [status: new]
+    - 15.3. &nbsp; 最大容量问题: chapter_greedy/max_capacity_problem.md
   - 16. &nbsp; 附录:
     # [icon: material/help-circle-outline]
     - chapter_appendix/index.md