**
Math optimization exercise with detailed solution : find the dimensions of the biggest seel container (maximizing volume) taking into account the increase in steel price between 2007 and 2008 and using the minimum steel area (area minimization).
**

1.2 Optimization: The biggest steel container

There is a surge in steel price. Indeed steel prices have almost doubled in the past year : the price of steel was 400 euros per ton in 2007 and rose to reach 700 euros per tonne in 2008 (1 metric tone = 1000 kg). It is therefore critical to save as much steel as possible in order to ensure business profitability. Hence you have to design the largest container with the smallest surface area possible and minimise material waste during cutting steel sheets.

You are asked to find the dimensions of the largest open top container that you could possibly manufacture using a rectangular steel sheet of 24 m^{2} (sheet area = 24 square meters). To do so you must cut equals squares from the 4 corners of the steel sheet and then fold up the sides. One of the sides of the sheet has to measure 6 m.

The following schema shows a steel sheet of 24 m^{2} and how to cut the *x* meters by *x* meters corners.

The steel sheet area measures initially: *Width . Length = W . L = 24 m ^{2}*.

We also know that one of the side of the steel sheet (W or L) has to measure 6 m. Thus let *L = 6*.

When the container is built, its volume is *Container Volume = Width . Lenght . Height = (W - 2x) . (L - 2x) . x*

What must be the value of the first derivative and second derivative of the function *Container Volume* ?

Reminder:

How to find the x and y coordinates of an extrema ? Take the first derivative of the function set equal to zero and solve for If the |

We want to maximize the volume of the container, thus we have to figure out
the x-value that makes the first derivative zero: *(Container Volume)' = 0*. We will get several x-values and we will have to pick among them
the value that makes the second derivative negative, *(Container Volume)' ' < 0.* This is important in order to consider only the x-values in the concave down part of the graph and thus ensure that we have found a maximum volume and not a minimum volume.

The equations we have found until now are:

(1) *24 = W . L*

(2) *L = 6*

(3) *Container Volume = (W - 2x) . (L - 2x) . x*

(4) *(Container Volume)' = 0*

(5) *(Container Volume)' ' < 0*

(1') *W = 24/L = 24/6 = 4*

We are going to replace *W* by 4 and *L* by 6 in equations (3), (4) and (5)

*Container Volume = (W - 2x) . (L - 2x) . x*

(3') *Container Volume = (4 - 2x) . (6 - 2x) . x*

*(Container Volume)' = ((4 - 2x) . (6 - 2x) . x)' = 0*

*(Container Volume)' = -40x + 12x ^{2} + 24 = 0*

By dividing the left and right member of the equation by 4 we get:

(4') *(Container Volume)' = -10x + 3x ^{2} + 6 = 0*

Reminder: the formula giving the roots of a quadratic equation

Given the second degree equation x, can be calculated as follow: _{2} |

We solve equation (4') using the quadratic equation formula (to find the zeros of a second degree equation).

Let us check which of those two x-values satisfies equation (5):

(5) *(Container Volume)' ' < 0*

*(-10x + 3x ^{2} + 6)' < 0*

*-10 + 6x < 0*

*x < 10/6*

*x < 1,667*

Thus we must eliminate the value *x = 2,549* because that x-value is greater than 1,667.
Moreover, note that if we accept *x = 2,549 meters* we would have a container with a side measuring
*W - 2x = 4 - 2 . 2,549 = - 1,097 m*, which is physically impossible. Indeed a container cannot have
a negative width. On the other hand, the value *x = 0,785 m* is correct because it makes simultaneously the first derivative
zero and the second derivative negative (concave down part of the graph). *x = 0,785 m* is then the
x-value that maximize the volume of the container.

The dimensions of the container are:

*Container Height = x = 0,785 m*

*Container Length = L - 2x = 6 - 2 . 0,785 = 4,431 m*

*Container Width = W - 2x = 4 - 2 . 0,785 = 2,431 m*