# A rectangular box with a square base and open top is to be made from 1200square meters of material2.1 Find the dimensions of the largest box that can be made. 2.2 use the second derivative test to...

A rectangular box with a square base and open top is to be made from 1200square meters of material

2.1 Find the dimensions of the largest box that can be made.

2.2 use the second derivative test to show that the dimensions in 2.1 give a maximum volume.

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2.1 You should use the following notation for the sides of square base and for the height of the box such that:

x - length of side of square

y - height of the box

You should write the formula of volume of the box such that:

`V = x^2*y`

You need to evaluate the surface area of the opened box such that:

`S = x^2 + 4xy`

The problem provides the amount of material, hence, you should substitute 1200 m^2 for S such that:

`1200 = x^2 + 4xy => 1200 - x^2 = 4xy => y = (1200 - x^2)/(4x)`

You need to substitute `(1200 - x^2)/(4x)` for y in formula of volume such that:

`V = x^2*(1200 - x^2)/(4x) => V = (1200x - x^3)/4`

You need to optimize the volume, hence, you need to find extrema solving the equation `V'(x) = 0` such that:

`V'(x) = (1/4)(1200 - 3x^2)`

`V'(x) = 0 <=> 1200 - 3x^2 = 0 => 3x^2 = 1200 => x^2 = 400 => x_(1,2) = +-20`

You only need to keep the positive value, hence x = 20 m.

You may evaluate the height such that;

`y = (1200 - 20^2)/(4*20) => y = (1200 - 400)/80 = 10 m`

**Hence, evaluating the dimensions of the largest box yields x = 20 m and y = 10 m.**

2.2 Using the second derivative test yields:

`V''(x) = -6x`

**Since the second derivative is negative for all `x>0` , then, the volume is maximum at the length of side of square of 20 m and the length of height of 10 m.**