solution in java:

public int maximaze() { int a=(int) Math.floor(25000/1579); int b=(int) Math.floor(25000/5899); List<Integer> list= new ArrayList<Integer>();

    int max=0;
    int c=0;

    for(int i=0;i<a;i++)
    {
        for (int j=0;j<b;j++)
        {
            c=1579*i+5899*j;
            if (c>max && c<25000)
            {
                max=c;
            }
        }
    }

    return max;

}

One way is to model the problem as an integer programming problem:

Maximize $1579 x + 5899 y$

subjected to

$1579 x + 5899 y \leq 25000$

$x, y$ integers

And use the branch and bound framework to solve it.

Python:

list=[]

for i in range(16):

    for j in range(4):

        if 1579*i+5899*j<25000:

            list.append(1579*i+5899*j)

print(list)

Here is my solution in Python:

amount = 0     #variable that calculates all amounts under 25000
values = []    #this array stores all amounts under 25000

for a in xrange(0, 20):
    for b in xrange (0, 20):
        amount = (a*1579) + (b*5899)    #calculate the value for amount
        if amount <= 25000:
            values.append(amount)    #add amount to values array if less than 25000

print max(values)

The programme returns 24847 as the answer.

i = 0

X = 0

Y = 0

while i <= 15:

j = 0

while j <= 4:

    X = i*1579 + j*5899

    if X < 25000:

        print X

    j = j + 1 

i = i + 1

print Y

include<stdio.h>

void main() { long i,j,k=0; long price[256],val=0,max; for(i=0;i<16;i++) { for(j=0;j<16;j++) { val=i 1579+j 5899; if(val<=25000) { price[k++]=val; } } } max=price[0]; for(i=0;i<k;i++) { if(price[i]>max) { max=price[i]; } } printf("max value=>%ld",max); }

A quick solution in C++ which will slightly reduce the complexity of the algorithm:

#include <iostream>
using namespace std;

int main() {
    int a, b, big, biggest = -1;
    cin >> a >> b >> big;
    for (int i = 0; i <= big/a; i++) {
        for (int j = 0; j <= (big-a*i)/b; j++) {
            if (biggest < a*i+b*j && a*i+b*j < big) { 
                biggest = a*i+b*j;
                cout << biggest << '\n';
            }
        }
    }

    return 0;
}

*Solution in C++ : *

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    int cost = 0, totalCost[1000], c = 0;
    int cost_of_workstation = 1579;
    int cost_of_server = 5899;

    for(int I = 0; I <= 100; I++)
    {
        for(int J = 0; J < 100; J++)
        {
            cost = I * cost_of_server + J * cost_of_workstation;
            if(cost <= 25000)
                totalCost[c++] = cost;
        }
    }

    int maxCost = totalCost[0];
    for(int I = 0; I < c; I++)
    {
        if(totalCost[I] > maxCost)
            maxCost = totalCost[I];
    }

    cout << maxCost << endl;

    return 0;
}

My Node JS solution, its generic and solves any of these kind of problems, I wish i can still reduce the complexity

input : node max_exp.js 25000 5899 1579; output : result : 1*5899 , 12*1579 = 24847

    var final_total = process.argv[2] || 0,
        items = process.argv.splice(3) || [],
        num_items = items.length;

    var max_count = 0,
        result;

    var cal = function(total, item_price, children, count_arr){

        var max_mul = Math.floor(total/item_price);

        var is_first_item = (children.length + 1)  == num_items,
            is_last_item = children.length == 0;

        for(var i = 0; i <=max_mul; i++){
            if(is_first_item) {
                count_arr = [];
            }
            count_arr[num_items - (children.length + 1)] = i;
            var remaining_price = total - i*item_price;
            if(is_last_item){
                var cal_max_count = final_total - remaining_price;
                if(max_count < cal_max_count){
                    max_count = cal_max_count;
                    result = get_str(count_arr) + " = " + max_count;
                }
                continue;
            }
            cal(remaining_price, children[0], children.slice(1), count_arr);
        }
    };

    // this gets the string to display result
    var get_str = function(count_arr){
        var str = "";
        count_arr.forEach(function(c, i){
            str += c + "*" + items[i] + " , ";
        });
        return str.substr(0, str.length - 3);
    };

    cal(final_total, items[0], JSON.parse(JSON.stringify(items)).splice(1));

    console.log("result : " + result);

int i,j; int m=o,a=0; for(i=0;i<5;i++) { for(j=0;j<14;j++) { a=5899 i+1579 j; if(m>a&&a<25000) a=m; } } cout<<"maximum value is: "<<m;