Question

From a thin piece of cardboard 12in. by 12in., square corners are cut out so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? What is the maximum volume? Round to the nearest tenth, if necessary. @MIT 18.01 Single …

Answer 1

Let the side length, in inches, of the square corners that are cut out be x. Thus, the height of the box is x when folded. Additionally, we can deduce that the width and the length will both be 12-2x. The area is given as follows:\[A=wlh=(12-2x)(12-2x)x=4x^3-48x^2+144x\]Let us find the derivative of this function with respect to x and set to zero to find any extrema.\[\frac{dA}{dx}=0=12x^2-96x+144\]\[0=x^2-8x+12=(x-6)(x-2)\]Thus, we have stationary points at x=6 and x=2. Now, we must figure out whether these are maxima, minima, or neither. To do this, we will use the curvature of the function which is given by the second derivative.\[\frac{d^2A}{dx^2}=24x-96\]It is clear that the second derivative is positive for x=6 and negative for x=2. For a point to be a maximum, it must be concave down, and thus have a negative second derivative. Thus, the maximum box size is when the side length of the removed squares is 2. I trust you can figure out the rest.

Answer 2

Max. Dimensions: 8x8x2
Max. Volume: 128in.^3