A box with an open top is to be made from a square piece of cardboard by cutting equal squares from the corners and turning up the sides. If the piece of cardboard measures 12 cm on the side, find the size of the squares that must be cut out to yield the maximum volume of the box.

Okay, so im not understanding how you went from V'(x)=(12-2x)²+2x(12-2x)(-2) to V'(x)=12(x²-8)... ive done this problem many times and am stuck on this part. i always end up with V'(x)= 12x²-96x+144 => 12(x²-8x+12)

so i end up with x=2 and x=6... which doesn't make sense because x=6 would make the base 0??? Help please!!!

Trisha, your answer is correct; x=2 or 6. However this is where we have to use intuition and realize that x=6 is indeed extraneous (since a base of 0 will not be a box). Therefore x=2 is the answer!

To find the size of the squares that must be cut out to yield the maximum volume of the box, we need to maximize the volume of the box.

Let's denote the side length of the square to be cut out as "x". Then, the dimensions of the resulting box would be:

Length = (12 - 2x) cm
Width = (12 - 2x) cm
Height = x cm

To find the volume of the box, we multiply the length, width, and height:

Volume = Length * Width * Height
= (12 - 2x) * (12 - 2x) * x
= (144 - 24x - 24x + 4x^2) * x
= (144x - 48x^2 + 4x^3)

Now we have an equation for the volume of the box in terms of "x". To find the value of "x" that maximizes the volume, we can take the derivative of the volume equation with respect to "x" and set it equal to 0:

dV/dx = 144 - 96x + 12x^2

Setting dV/dx = 0:

144 - 96x + 12x^2 = 0

Dividing this equation by 12:

12x^2 - 8x + 12 = 0

Simplifying further:

3x^2 - 2x + 3 = 0

This is a quadratic equation that can be solved using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

Plugging in the values a = 3, b = -2, and c = 3:

x = (-(-2) ± sqrt((-2)^2 - 4(3)(3))) / (2(3))
x = (2 ± sqrt(4 - 36)) / 6
x = (2 ± sqrt(-32)) / 6

Since the square root of a negative number is imaginary, there are no real solutions for "x" in this case. Thus, we cannot cut out any squares to maximize the volume.

Hence, there are no squares that must be cut out to yield the maximum volume of the box.

Let the length of the (four) squares cut from the corner be x.

The box will then have dimensions 12-2x, 12-2x and x once the open box is made.
The volume is therefore:
V(x)=x(12-2x)²
Differentiate with respect to x and equate to zero to get the greatest voume:
V'(x)=(12-2x)²+2x(12-2x)(-2)
=12(x²-8)
Equate to zero and solve for x:
12(x²-8)=0
=>
x=sqrt(8)