  # Math 3c Derivation - Maximum Volume of Box

English is not my primary language so bear with me as I struggle to translate this.

Jonas is going to make a box without a top which measures 60 cm x 30 cm.

He is going to cut away equally big squares in each corner, and then fold up the sides.

The squares have the sides of x.

Help Jonas to calculate the side x so that the boxes volume gets as big as it can get.

How I tried to solve it:

x(60 - 2x)(30 - 2x)

= (60x - 2x^2)(30 - 2x)

= 1800x - 120x^2 - 60x^2 + 4x^3

y' = 1800 - 240x - 120x - 12x^2

12x^2 - 360x + 1800 = 0

According to PQ :

x1 = 354,92856

x2 = 5,07144

according to the definition x1 is to big

x = 5,07144

### Comments for Math 3c Derivation - Maximum Volume of Box

 Jun 11, 2013 Maximum Volume by: Staff Answer Part I The problem statement The problem statement is very unusual. It gives the length and width of the of the final (assembled) box, rather than the dimensions of the material used to make the box: "Jonas is going to make a box without a top which measures 60 cm x 30 cm. He is going to cut away equally big squares in each corner, and then fold up the sides. " Generally, calculus problems of this type give the dimensions of the material used construct the box, rather than the length and width of the final (assembled) box. However, if the problem statement is correct, a net for the problem can be drawn as follows: (A net is a two-dimensional figure which can be folded into a three-dimensional object.) Calculating the maximum volume of the box is straight forward. It does not require calculus or differential equations: Volume = length * width * height Volume = 60 * 30 * x x can be any value, since it is not restricted by the dimensions of the material used to make the box. Therefore, the maximum volume of the box is infinity (∞). Restatement of the problem I think this is what you intended to say: "Jonas is going to make a box without a top out of a piece of construction material which measures 60 cm x 30 cm. He is going to cut away equally big squares in each corner of the construction material, and then fold up the sides. " A net for this revised problem statement can be drawn as follows: ---------------------------------------------------------

 Jun 11, 2013 Maximum Volume by: Staff --------------------------------------------------------- Part II Volume = length * width * height Volume = L * W * x L = 60 - 2x W = 30 - 2x Volume = (60 - 2x) * (30 - 2x) * x Volume = 4x³ - 180x² + 1800x The volume as a function of x: V(x) = 4x³ - 180x² + 1800x The first derivative of the volume: V'(x) = 12x² - 360x + 1800 ---------------------------------------------------------

 Jun 11, 2013 Maximum Volume by: Staff --------------------------------------------------------- Part III Set the first derivative equal to zero, and then solve for x 12x² - 360x + 1800 = 0 x² - 30x + 150 = 0 use the quadratic formula to solve for x Format of quadratic equation: Quadratic formula: ---------------------------------------------------------

 Jun 11, 2013 Maximum Volume by: Staff --------------------------------------------------------- Part IV For this problem: x² - 30x + 150 = 0 x = unknown a = 1 b = -30 c = 150 Substitute the values of the coefficients a, b, and c into the quadratic formula Solve for x ---------------------------------------------------------

 Jun 11, 2013 Maximum Volume by: Staff --------------------------------------------------------- Thanks for writing. Staff www.solving-math-problems.com