This problem is a simpler version of the Knapsack problem . It can be solved recursively.Let $Cap(x)$ be the maximum amount of water that can be carried for a weight limit of $x$ . And let $v$ be the set of the values of the smaller bottles.We can define recursively as $Cap(x)=max( \left\{ Cap(x-{v}_{0})+{v}_{0} , Cap(x-{v}_{1})+{v}_{1},Cap(x-{v}_{2})+{v}_{2} ...,Cap(x-{v}_{i})+{v}_{i} \right\} )$

We are basically subtracting each value of the set of bottle( $v)$ values from the limit( $15kg$ ) and seeing which difference yields the maximum value..

This can be implemented easily in python..

Save = {} #Always save values when using recursive functions
          #it saves a lot of time and avoids recomputation
def Cap(Capacity,values):
    if Capacity in Save:
        return Save[Capacity]
    vs = []
    if Capacity in values:
        return Capacity
    if Capacity <= 0:
        return 0
    for value in values:
        if value > Capacity:
            vs.append(0)
        else:
            vs.append(Cap(Capacity-value,values)+value)
    Save[Capacity] = max(vs)
    return max(vs)


print Cap(1500,[77,110,340,700])/100.0

Note:I multiplied each bottle value by $100$ ,since it is easier to deal with integers.

x=0 y=0 z=0 w=0 R=0 for x in range (0,19) : for y in range (0,13) : for z in range (0,4) : for w in range(0,2): M = 0.77 x + 1.1 y + 3.4 z + 7 w if (R<M and M<=15) : R=M print "Max weight that can be carried is %s" % (R)

Here is a solution i made in python. Have a look.

capacity = 0

capacities = [ ]

for a in xrange(0,15):

for b in xrange(0,15):

  for c in xrange(0,15):

    for d in xrange(0,15):

      capacity = (a*0.77)+(b*1.10)+(c*3.4)+(d*7.0)

      if capacity <=15.0:

        capacities.append(capacity)

print max(prices)

It returns the value as 14.96.

/ C++ code /

include<stdio.h>

void main() { float i,j,l,m; float val=0,max=0; for(i=0;i<20;i++) { for(j=0;j<14;j++) { for(l=0;l<5;l++) { for(m=0;m<3;m++) { val=i 0.77+j 1.10+l 3.4+m 7; if((val<=15)&&(val>max)) { max=val; } } } } } printf("max value=>%f",max); }

up as a code block

Solution in C++ (or C if you replace cout with stdio)

int main()
{
    float lim = 15,maxcost = 0;
    float costA = 7, costB = 3.4,costC = 1.1,costD = 0.77;
    for(int a = 0; costA*a <= lim ; ++a)
    {
        for(int b = 0 ;costB*b <= lim-costA*a ; ++b)
        {
            for(int c = 0 ;costC*c <= lim-costA*a-costB*b ; ++c)
            {
                for(int d = 0 ; costD*d <= lim-costA*a-costB*b-costC*c ; ++d)
                    maxcost = std::max(maxcost,costA*a+costB*b+costC*c+costD*d);
            }
        }
    }
    cout << maxcost << endl;
}

Python Code:

    list=[]

    for i in range(15):

             for j in range(15):

                     for k in range(5):

                               for l in range(3):

                                        if 0.77*i+1.10*j+3.4*k+l*7<15.00:

                                                  list.append(0.77*i+1.10*j+3.4*k+l*7)

      print max(list)

Wrote in Powershell :

First found the maximum possible number of each bottle (taken only one size of bottle at a time)

0.77 * 19 = 14.63 ----- <15 1.10 13 = 14.3 ----- <15 3.4 4 = 13.6 ----- <15 7*2 = 14 ----- <15

The code goes like this:

PS C:\Windows\system32> [float]$a1 = 0; [float]$a2 = 0; [float]$a3 = 0; [float]$a4 = 0; [float]$z = 0; [float]$largest = 0; for($a1 = 0;$a1 -le 19;$a1++) { for($a2 = 0;$a2 -le 13;$a2++) { for($a3 = 0;$a3 -le 4;$a3++) { for($a4 = 0;$a4 -le 2;$a4++) { $z = $a1 0.77 + $a2 1.10 + $a3 3.4 + $a4 7; if( ($z -gt $largest) -and ($z -le 15)) { $largest = $z; } } } } }

Write-host $largest

14.96

code in python

def d(n): i=x=y=0 while i<=19: j=0 while j<=13: k=0 while k<=4: m=0 while m<=2: x = 0.77 i+1.10 j+3.40 k+7.00 m if x>y and x<=n: y=x m=m+1 k=k+1 j=j+1 i=i+1 print y d(15.00)

c++ code for mingw.not efficient but I love recursive brute force.

btw divide solution by 100 ;-)

include <iostream>

int mx=1500,t=77,s=110,m=340,l=700;

int findmax(int a){

int profit=a;

if(a<=mx){

    if((a+t)<=mx){

        profit = max(profit,findmax(a+t));

    }

    if((a+s)<=mx){

        profit = max(profit,findmax(a+s));

    }

    if((a+m)<=mx){

        profit = max(profit,findmax(a+m));

    }

    if((a+l)<=mx){

        profit = max(profit,findmax(a+l));

    }

    return profit;

} else

    return 0;

}

int main(){

cout<<findmax(0)<<endl;

}

Python Solution:

maxWater = 0
for a in range(int(15/0.77) + 2):
    for b in range(int(15/1.1) + 2):
        for c in range(int(15/3.4) + 2):
            for d in range(int(15/7) + 2):
                totalWater = 0.77*a + 1.1*b + 3.4*c + 7*d
                if totalWater <= 15:
                    if totalWater > maxWater:
                        maxWater = totalWater
print maxWater

THIS IS A C# SOLUTION :

static void Main(string[] args) { double a = 7, b=3.4, c=1.1, d=.77; double n = 14.94;

        for ( int i = 0 ; i < 20 ; i++)
        {
            for(int j =0 ; j <20 ; j++)
            {
                for (int k=0 ; k<20 ; k++)
                {
                    for (int l=0 ; l<20 ; l++)
                    {
                        double x = a * i + b * j + c * k + d * l;
                        if (x > n  &&  x<=15)
                        {
                            n = x;
                        }

                    }
                }
            }
        }

        Console.WriteLine(n);
        Console.ReadLine();

The output is : 14.96

for i in range (0,10): for j in range(0,10): for k in range(0,10): for l in range(0,10): if tiny * i + small * j + medium *k + large *l <=15 and tiny * i + small * j + medium *k + large *l >=14: print i,j,k,l ,tiny * i + small * j + medium *k + large *l

then i appended the output in a list and sorted out and checked which was the highest value. The highest value was 14.96 for i=8, j=8, k= 0, l=0

Enter all the 4 options or Start entering from 15.00 and go backwards (this is small version for MCQ - type)

 #include<iostream.h>
 int main()
 {
 int a,b,c,d;
 float e,f;
 cin>>f;

 for(a=0;a<20;a++)
 {
  for(b=0;b<15;b++)
  {
   for(c=0;c<15;c++)
   {
    for(d=0;d<15;d++)
    {
     e=(a*0.77)+(b*1.1)+(c*3.4)+(d*7);
     if(e==f)
     cout<<a<<"\n"<<b<<"\n"<<c<<"\n"<<d<<"\n\n";
     }
   }
  }
 }
 return(0);
 }

15 - no combination.

14.98 - no combination.

14.96 - two combinations.

14.94 - one combination.

Ask in comments if you have any doubts.