A 33 by 33 square piece of cardboard is to be made into a box by cutting out equal square corners from each side of the square. What size corners should be cut out so that the volume of the box is maximized?

let x be cut size. sides are thus 33-2x

v = x(33-2x)^2
v = 4x^3 - 132x^2 + 1089x

dv/dx = 12x^2 - 264x + 1089

max/min volume when dv/dx = 0
(2x-11)(2x-33) = 0

I'll let you figure out which root makes sense.

To find the size of the corners that should be cut out to maximize the volume of the box, we can approach it using calculus.

Let's assume that x represents the length of the side of the square being cut out from each corner.

So, when we cut out x by x squares from each corner, the dimensions of the resulting box will be:
Length = 33 - 2x
Width = 33 - 2x
Height = x

The volume of the box is given by multiplying these dimensions:
Volume = Length * Width * Height
Volume = (33 - 2x) * (33 - 2x) * x

Next, we need to maximize this volume by finding the value of x that yields the maximum result. To do this, we can use calculus.

To find the maximum value, we differentiate the volume equation with respect to x, and set it equal to zero:
d(Volume)/dx = 0

Let's calculate these derivatives step by step:

d(Volume)/dx = d/dx [(33 - 2x) * (33 - 2x) * x]
d(Volume)/dx = [(33 - 2x) * (33 - 2x) * 1] + [(33 - 2x) * (33 - 2x) * (-2)] + [(33 - 2x) * 1]
d(Volume)/dx = (33 - 2x)(33 - 2x + (-4x) + 1)
d(Volume)/dx = (33 - 2x)(34 - 6x)
d(Volume)/dx = -12x^2 + 132x - 1122

Now, setting the above equation equal to zero:

-12x^2 + 132x - 1122 = 0

We can solve this quadratic equation to find the values of x. Using the quadratic formula, we get:

x = (-b ± √(b^2 - 4ac)) / (2a)

where a = -12, b = 132, and c = -1122.

Calculating x using the quadratic formula:

x = (-132 ± √(132^2 - 4(-12)(-1122))) / (2(-12))
x = (-132 ± √(17424 - 53712)) / -24
x = (-132 ± √(-36288)) / -24

As we cannot take the square root of a negative value, it means that the equation does not have real solutions. However, it does have complex solutions. Taking the complex roots, we get:

x = (-132 + i√(36288)) / -24 and x = (-132 - i√(36288)) / -24

Since x cannot be complex in this case, we discard these results. It implies that there is no value of x that maximizes the volume within the given constraints.

Hence, the conclusion is that there is no way to cut out equal square corners from the 33 by 33 square piece of cardboard to maximize the volume of the resulting box.