On a rectangular piece of cardboard with perimeter 11 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, use a graphing calculator to find the value of that maximizes the volume enclosed by this box. Then give the maximum volume. Round your responses to two decimal places.
values of x that maximizes volume = in
Maximum volume= in^3
I would start with a drawing as I do for most problems. Draw a reactangle 4x one side and b the other.
The perimeter is then 4x+b+4x+b=11
8x+2b=11
if this is folded to a tube then the volume of the tube is bx^2, i.e. a tube with cross sectional area x^2 and length b.
so V=bx^2
rearrangen and substitute for b into the equation above gives
8x+2V/(x^2) = 11
or
8x^3+2V=11x^2
or
V=5.5x^2-4x^3
which you can plot to find max V
I got 1.54 in^3 as the max volume
but check the maths!
To solve this problem using a graphing calculator, we need to plot a graph of the volume of the box as a function of the distance between the creases, x, and find the maximum point on the graph. Here's what you need to do:
1. Start by understanding the dimensions of the rectangular piece of cardboard. Let's assume the length of the cardboard is L and the width is W.
2. Since the perimeter of the cardboard is given as 11 inches, we can write the equation: 2L + 2W = 11. Rearranging the equation, we have: L = (11 - 2W)/2.
3. Next, consider the dimensions of the folded box. When the cardboard is folded along the creases, the length of the box will be equal to L, and the width will be equal to x (since there are three equally spaced creases).
4. The height of the box is the distance between the creases, which is x. So, the volume of the box can be expressed as V = L * x * x = x^2 * (11 - 2W)/2.
5. Enter the equation for the volume of the box into your graphing calculator. You can also utilize online graphing tools or software like Desmos or GeoGebra.
6. Once you have plotted the graph, look for the point that represents the maximum volume. This would correspond to the maximum point on the graph.
7. Read the x-coordinate of the maximum point on the graph. This value represents the distance between the creases that maximizes the volume.
8. Use the x-coordinate value you obtained to calculate the maximum volume.
Plug the x-coordinate value into the formula for the volume: V = x^2 * (11 - 2W)/2.
Calculate the maximum volume by substituting the x-coordinate value into the equation.
Make sure to round your responses to two decimal places, as requested.
This step-by-step process should enable you to use a graphing calculator to find the value of x that maximizes the volume and determine the maximum volume of the box.