Let the dimensions be 4x and y
then the perimeter is 8x+2y = p
(where p is the value you have left out...)
Let the folds be parallel to the side of length y, so there are four strips of width x. Thus, the volume of the square tube formed is
v = x^2 * y = x^2 * (p/2 - 4x) = px^2/2 - 4x^3
dv/dx = px - 12x^2
dv/dx = 0 at x = p/12
and it is easy to show that this is a maximum, not a minimum.
now just finish up with your answers.
On a rectangular piece of cardboard with perimeter
inches, three parallel and equally spaced creases are made. The cardboard is then folded along the creases to make a rectangular box with open ends. Letting x
represent the distance (in inches) between the creases, Using a graphing calculator find the value of x
that maximizes the volume enclosed by this box. Then give the maximum volume. Round your responses to two decimal places.
1 answer
1 answer