Hello coders, today we are going to solve **Maximum Production CodeChef Solution** whose Problem Code is **EITA**.

**Task**

Chefland has **7** days in a week. Chef is very conscious about his work done during the week.

There are two ways he can spend his energy during the week. The first way is to do* x* units of work every day and the second way is to do

*(>*

**y***) units of work for the first*

**x****(<**

*d***) days and to do**

*7***(<**

*z**) units of work thereafter since he will get tired of working more in the initial few days.*

**x**Find the maximum amount of work he can do during the week if he is free to choose either of the two strategies.

**Input **

- The first line contains an integer
, the number of test cases. Then the test cases follow.**T** - Each test case contains a single line of input, four integers
,**d**,**x**,**y**.**z**

**Output**

For each testcase, output in a single line the answer to the problem.

**Constraints**

**1 ≤***T*≤ 5⋅10^{3}**1 ≤***d*< 7**1 ≤***z*<*x*<*y*≤ 18

**Subtasks**

**Subtask #1 (100 points): **Original constraints

**Sample Input**

```
3
1 2 3 1
6 2 3 1
1 2 8 1
```

**Sample Output**

```
14
19
14
```

**Explanation**

**Test Case 1:** Using the first strategy, Chef does **2⋅7 = 14** units of work and using the second strategy Chef does **3⋅1 + 1⋅6 = 9** units of work. So the maximum amount of work that Chef can do is **max(14,9) = 14** units by using the first strategy.

**Test Case 2:** Using the first strategy, Chef does **2⋅7 = 14** units of work and using the second strategy Chef does **3⋅6 + 1⋅1 = 19 **units of work. So the maximum amount of work that Chef can do is **max(14,19) = 19 **units by using the second strategy.

**Solution – Maximum Production**

**C++**

#include <iostream> using namespace std; int main() { // your code goes here int t; cin >> t; while(t--){ int d,x,y,z; cin >> d >> x >> y >> z; int way1 = 7*x; int way2 = ((d*y)+((7-d)*z)); cout << max(way1,way2) << "\n"; } return 0; }

**Java**

/* package codechef; // don't place package name! */ import java.util.*; import java.lang.*; import java.io.*; /* Name of the class has to be "Main" only if the class is public. */ class Codechef { public static void main (String[] args) throws java.lang.Exception { // your code goes here Scanner sc = new Scanner(System.in); int t = sc.nextInt(); while(t-->0){ int d,x,y,z; d = sc.nextInt(); x = sc.nextInt(); y = sc.nextInt(); z = sc.nextInt(); int ans = 0; ans = Math.max(7*x,((d*y)+((7-d)*z))); System.out.println(ans); } } }

**Python**

# cook your dish here T = int(input()) for i in range(T): d, x, y, z = map(int, input().split()) first_result = x * 7 second_result = (y*d)+((7-d)*z) result = [first_result, second_result] largest_number = max(result) print(largest_number)

**Disclaimer:** The above Problem **(Maximum Production) **is generated by **CodeChef** but the Solution is Provided by **CodingBroz**. This tutorial is only for **Educational** and **Learning** Purpose.