Question

An open box is to be made by using a piece of sheet metal with the dimensions of 4m by 2m. Determine the dimensions the box should be to maximize volume by cutting out the four corners of the material and folding up the edges to make the box. The material that is cut out is waste and cannot be used.

Answer 1

Dimensions: 0.4 by 3.2 by 1.2

Explanation:

I'm assuming here that we are cutting out squares out of each of the metal's corners:

Let x = the length of each cut-out square,

Each base (of the desired net square folded) is 4-2x, and 2-2x respectively,

Volume = x(4-2x)(2-2x)

= 4x^3 - 12x + 8x

Now we take the derivative:

`(d)/(dx)\left[4x^3-12x^2+8x\right]\\\\= (d)/(dx)\left(4x^3\right)-(d)/(dx)\left(12x^2\right)+(d)/(dx)\left(8x\right)\\\\= 12x^2-24x+8`

We equate to 0 (0 for max volume), and solve using the quadratic formula:

`12x^2-24x+8=0,\\\\x_(1,\:2)=(-\left(-24\right)\pm √(\left(-24\right)^2-4\cdot \:12\cdot \:8))/(2\cdot \:12)\\\\= (-\left(-24\right)\pm \:8√(3))/(2\cdot \:12)\\\\\mathrm{Separate\:the\:solutions}:\\\\x_1=(-\left(-24\right)+8√(3))/(2\cdot \:12),\:x_2=(-\left(-24\right)-8√(3))/(2\cdot \:12)\\\\x =(3+√(3))/(3),\:x=(3-√(3))/(3)\\\\x=1.57735\dots ,\:x=0.42264\dots`

So we approximate the side lengths to be 1.6 and 0.4 respectively. But when we plug in 1.6 for x, we get the volume as negative. Therefore x has to be 0.4.

Side lengths: 0.4, 4-2(0.4) => 3.2, 2-2(0.4) => 1.2