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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

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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.)


Math - net of box 60 cm by 30 cm




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:





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Jun 11, 2013
Maximum Volume
by: Staff

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Part II


Math - net of box constructed from material which measures 60 cm by 30 cm




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

Maximum volume of box as a function of dimensions



The first derivative of the volume:

V'(x) = 12x² - 360x + 1800

Maximum volume of box as a function of dimensions - first derivative









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Jun 11, 2013
Maximum Volume
by: Staff


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Part III


Set the first derivative equal to zero, and then solve for x

12x² - 360x + 1800 = 0

x² - 30x + 150 = 0


Maximum volume of box as a function of dimensions - set first derivative equal to zero





use the quadratic formula to solve for x

Format of quadratic equation:

Format of Quadratic Equation




Quadratic formula:


Format of Quadratic Formula





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Jun 11, 2013
Maximum Volume
by: Staff


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Part IV



For this problem:

x² - 30x + 150 = 0

x = unknown

a = 1

b = -30

c = 150


Format of Quadratic Formula - coefficients




Substitute the values of the coefficients a, b, and c into the quadratic formula


Quadratic Formula - enter coefficients




Solve for x


Quadratic Formula - solve for x







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Jun 11, 2013
Maximum Volume
by: Staff


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Part V


Quadratic Formula - solution:  values of x




Calculation of maximum volume


Maximum volume of box - calculated using both solutions of x




Since the volume cannot be a negative value, the final answer is:


Maximum volume of box - final answer







Thanks for writing.

Staff
www.solving-math-problems.com


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