# The maximum volume of a box Code Answer

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Trying to write a simple web app to solve the following common calculus problem in JavaScript.

Suppose you wanted to make an open-topped box out of a flat piece of cardboard that is L long by W wide by cutting the same size square (h × h) out of each corner and then folding the flaps to form the box, as illustrated below:

You want to find out how big to make the cut-out squares in order to maximize the volume of the box.

Ideally I want to avoid using any calculus library to solve this.

My initial naive solution:

```// V = l * w * h
function getBoxVolume(l, w, h) {
return (l - 2*h)*(w - 2*h)*h;
}

function findMaxVol(l, w) {
const STEP_SIZE = 0.0001;

let ideal_h = 0;
let max_vol = 0;

for (h = 0; h <= Math.min(l, w) / 2; h = h + STEP_SIZE) {
const newVol = getBoxVolume(l, w, h);
if (max_vol <= newVol) {
ideal_h = h;
max_vol = newVol;
} else {
break;
}
}

return {
ideal_h,
max_vol
}
}

const WIDTH_1 = 20;
const WIDTH_2 = 30;

console.log(findMaxVol(WIDTH_1, WIDTH_2))

// {
//   ideal_h: 3.9237000000038558,
//   max_vol: 1056.3058953402121
// }
```

The problem with this naive solution is that it only gives an estimate because you have to provide STEP_SIZE and it heavily limits the size of the problem this can solve.

After realising that the derivative of the volume function is a second degree polynomial you can apply a quadratic formula to solve for x.

Using calculus, the vertex point, being a maximum or minimum of the function, can be obtained by finding the roots of the derivative

```// V = l * w * h
function getBoxVolume(l, w, h) {
return (l - 2*h)*(w - 2*h)*h;
}

// ax^2 + bx + c = 0
var x1 = (-1 * b + Math.sqrt(Math.pow(b, 2) - (4 * a * c))) / (2 * a);
var x2 = (-1 * b - Math.sqrt(Math.pow(b, 2) - (4 * a * c))) / (2 * a);
return { x1, x2 };
}

function findMaxVol(l, w) {
// V'(h) = 12h^2-4(l+w)h+l*w - second degree polynomial
// solve to get the critical numbers
const result = solveQuad(12, -4*(l + w), l*w)

const vol1 = getBoxVolume(l, w, result.x1);
const vol2 = getBoxVolume(l, w, result.x2);

let ideal_h = 0;
let max_vol = 0;

// check for max
if (vol1 > vol2) {
ideal_h = result.x1;
max_vol = vol1;
} else {
ideal_h = result.x2;
max_vol = vol2;
}

return {
ideal_h,
max_vol
}
}

const WIDTH_1 = 20;
const WIDTH_2 = 30;

console.log(findMaxVol(WIDTH_1, WIDTH_2))

// {
//   ideal_h: 3.9237478148923493,
//   max_vol: 1056.30589546119
// }
```