Wanted to cut planks of wood. The size of the possible cuts was fixed. Wanted to utilize as much of the wooden planks as possible. Like any other engineer I took this to the editor :P .

optimalWoodCut.js

// CONSTANTS
// =====================

const a = 210; // length of a possible cut 
const b = 156; // length of another possible cut 
const c = 138; // length of another possible cut
const K = 590; // length of the wood plank to be cut in pieces of the above lengths



// LOGIC
// =====================

let permutations = findAllWoodCuttingPermutations(a, b, c, K);
let optimalPermutation = findPermWithMinRemainder(permutations);

let utilization = parseFloat(((K - optimalPermutation.remainder) / K) * 100).toFixed(2);

console.log(`Optimal solution for cutting a wooden plank of length ${K} is to cut it into: \n`
              + `${optimalPermutation.x} pieces of length ${a} \n`
              + `${optimalPermutation.y} pieces of length ${b} \n`
              + `${optimalPermutation.z} pieces of length ${c} \n\n`
              + `This results in leftover length of ${optimalPermutation.remainder} length units`
              + `which is a utilization of ${utilization}%.`
            );



// FUNCTION DECLARATIONS
// =====================

function findAllWoodCuttingPermutations(a, b, c, K) {
  let permutations = [];
  for (let x=0; x*a<=K; x++) {
    for (let y=0; ((x*a)+(y*b))<=K; y++) {
      for (let z=0; ((x*a)+(y*b)+(z*c))<=K; z++) {
        let remainder = K - ((x*a) + (y*b) + (z*c));
        if ((remainder <= K) && (remainder >= 0)) {
          permutations.push({
            'x': x,
            'y': y,
            'z': z,
            'remainder': remainder,
          });
        }
      }
    } 
  }
  return permutations;
}

function findPermWithMinRemainder(permutations) {
  let minRemainder = Number.MAX_SAFE_INTEGER;
  let permWithMinRemainder;

  for (perm of permutations) {
    if (perm.remainder <= minRemainder) {
      minRemainder = perm.remainder;
      permWithMinRemainder = perm;
    }
  }
  return permWithMinRemainder;
}