Snacks will be provided in a box with a lid (made by removing squares from each corner of a rectangular piece of card and then folding up the sides)

link to image
imageshack com /a/img661/6094/fZUQXg.jpg

You have a piece of cardboard that is 40cm by 40 cm – what dimensions would give the maximum volume?

To find the dimensions that would give the maximum volume, we need to come up with a mathematical expression for the volume and then optimize it.

Let's assume that we remove squares with side length "x" from each corner of the rectangular piece of card. This would leave us with a width of (40 - 2x) cm and a length of (40 - 2x) cm for the bottom of the box, and a height of "x" cm for the sides.

The volume of a rectangular box is calculated by multiplying its length, width, and height. So, the volume of the box in terms of "x" is given by:

V(x) = (40 - 2x) * (40 - 2x) * x

To find the value of "x" that maximizes the volume, we need to take the derivative of V(x) with respect to "x" and set it equal to zero.

Let's simplify the equation and find the derivative:

V(x) = (1600 - 160x + 4x^2) * x
V(x) = 4x^3 - 160x^2 + 1600x

Now, let's take the derivative:

V'(x) = 12x^2 - 320x + 1600

Next, let's solve for "x" by setting V'(x) equal to zero:

12x^2 - 320x + 1600 = 0

We can now use the quadratic formula to solve for "x":

x = (-(-320) ± √((-320)^2 - 4*12*1600)) / (2*12)

After solving the equation, you will get two values for "x." Since the sides cannot have negative lengths, choose the positive value for "x."

Once you have the optimized value for "x," you can substitute it back into the equation (40 - 2x) to find the length and width dimensions of the box.

Note: Since you mentioned a link to an image, it is not possible for me to view or access it as an AI text-based model. Please refer to the visual representation as needed while following the explanation above.

To find the dimensions that give the maximum volume, we need to determine the dimensions of the removed squares from each corner.

Let's assume the length of the square removed from each corner is x cm.

The width of the resulting base (after folding up the sides) will be 40 - 2x cm (because each removed square removes x cm from both sides).

Similarly, the length of the resulting base (after folding up the sides) will also be 40 - 2x cm.

The height of the box will be x cm (the height of the removed square).

The volume (V) of the box can be calculated using the formula:

V = (length of base) * (width of base) * (height)
= (40 - 2x) * (40 - 2x) * x

To find the maximum volume, we need to find the value of x that maximizes the volume V.

Taking the derivative of V with respect to x and setting it equal to zero will give us the critical points. Let's do that:

dV/dx = 0
First, apply the product rule to differentiate V with respect to x:

dV/dx = 2(40 - 2x)(-2) * x + (40 - 2x)(-2)

Simplifying the equation further:

0 = -8x(40 - 2x) + 2(40 - 2x)
0 = -8x(40 - 2x) + 80 - 4x
0 = -320x + 16x^2 + 80 - 4x
0 = 16x^2 - 324x + 80

To solve this quadratic equation, we can either factor it or use the quadratic formula. Since factoring is not straightforward, let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Where a = 16, b = -324, and c = 80.

Calculating the values:

x = (-(-324) ± √((-324)^2 - 4*16*80)) / (2*16)
x = (324 ± √(104976 - 5120)) / 32
x = (324 ± √100856) / 32

Now simplify further:

x = (324 ± 317.619) / 32

We have two possible solutions:

1. x = (324 + 317.619) / 32 = 641.619 / 32 = 20.05 cm (approximately)
2. x = (324 - 317.619) / 32 = 6.381 / 32 = 0.199 cm (approximately)

Since the dimensions cannot be negative and the original side length of the cardboard is 40 cm, we can conclude that the square removed from each corner should be approximately 20.05 cm to maximize the volume of the box.

Note: Please ignore any inaccuracies in the calculations as they are based on approximation.