you can make an open box from a piece of flat cardboard. First cut congruent squares from the four corners of the cardboard. Then fold and tape the sides. let x equal the side of each congruent squares as x increases so does the depth of the box the useable area of the cardboard decreases as x increases, and so do the length and width of the box. what happens to the volume of the box? does it increase or deacreas as x increases? would the answer both suprise you? what size square should you cut from the corners to maximize the volume of your box?What are the dimensions of the box in centimeters?

Question

you can make an open box from a piece of flat cardboard. First cut congruent squares from the four corners of the cardboard. Then fold and tape the sides. let x equal the side of each congruent squares as x increases so does the depth of the box the useable area of the cardboard decreases as x increases, and so do the length and width of the box. what happens to the volume of the box? does it increase or deacreas as x increases? would the answer both suprise you? what size square should you cut from the corners to maximize the volume of your box?What are the dimensions of the box in centimeters?

Answer 1

length of bottom = L-2x

width of bottom = W -2x

v = x (L-2x)(W-2x)
= x(LW-2Wx-2Lx +4x^2)
= LW x -2Wx^2 -2Lx^2 +4x^3

This is a cubic polynomial and will have maxima and minima. Just looking at it the volume would be huge as x got huge because of the x^3. However x can not be bigger than W/2 in practice so it is not that simple.
Using calculus to find max or min:
dv/dx = LW -4Wx-4Lx + 12 x^2
that is 0 at a max or min
12 x^2 -4(L+W)x + LW = 0
for a given L and W, solve that quadratic for x of min or max
for example for a square sheet of 10 cm on a side

12 x^2 -4(20) + 100 = 0
3 x^2 -20 + 25 = 0
x = [20 +/- sqrt (400 -300) ]/6
x = [20 +/-10]/6
x = 30/6 or 10/6
30/6 is 5 which is half the width so zero volume
so 10/6 is what we have, 1 2/3 deep

Answer 2

To determine what happens to the volume of the box as the side length of the squares, x, increases, we can follow these steps:

1. Visualize the scenario: Imagine a piece of flat cardboard with dimensions L (length) and W (width).
2. Cut congruent squares from each corner. The side length of each square is x.
3. Fold and tape the sides of the cardboard to form an open box.

Let's analyze the impact of x on the volume of the box:

The height of the box is determined by the side length of the squares, so it will be x.

The length of the box will be L - 2x, as we have removed two squares (x by x) from both ends of the cardboard.

The width of the box will be W - 2x, for the same reason as the length.

Therefore, the volume of the box will be: V = (L - 2x)(W - 2x)x.

As x increases, the volume of the box will initially increase because the height increases. However, as x continues to increase, there will be a point where the usable area of the cardboard decreases significantly, causing the length and width to decrease. This decrease in length and width outpaces the increase in height, resulting in a decrease in volume.

So, in summary, the volume of the box increases initially but eventually decreases as x continues to increase.

To maximize the volume of the box, we need to determine the value of x that gives the largest possible volume. This can be achieved by finding the maximum of the volume function and solving for x.

As for the specific dimensions of the box in centimeters, we would need to know the initial length and width of the cardboard (L and W) and the value of x provided to provide the exact dimensions.