Tray Volume Maximization

The problem asks us to find the size of the square cutouts from the corners of a larger square sheet so that when the sides are folded up, the resulting open box (or tray) has the maximum possible volume.

We start with a square sheet of side 12 cm. Let the side of the small squares cut from each corner be $x$ cm.

When we cut out squares of side $x$ from each corner, the length of each side of the base of the tray will be the original side length minus two times the side of the cut square (one $x$ from each end). So, the length and width of the base of the tray will be $12 - 2x$ cm.

When we fold up the edges, the height of the tray will be equal to the side length of the square cutouts, which is $x$ cm.

Volume Formula for the Tray

The tray is a rectangular prism (a box). The volume ($V$) of a rectangular prism is given by the formula:

$$V = \text{Length} \times \text{Width} \times \text{Height}$$

Substituting the dimensions of our tray in terms of $x$:

$$V(x) = (12 - 2x) \times (12 - 2x) \times x$$

$$V(x) = x(12 - 2x)^2$$

Finding the Domain of x

For a valid tray to be formed, the dimensions must be positive:

• The side of the cut square must be positive: $x > 0$.
• The side of the base must be positive: $12 - 2x > 0$.

From $12 - 2x > 0$, we get $12 > 2x$, which simplifies to $6 > x$.

So, the value of $x$ must be in the interval $0 < x < 6$.

Maximizing the Volume using Calculus

To find the value of $x$ that maximizes the volume $V(x)$, we need to find the critical points by taking the derivative of $V(x)$ with respect to $x$ and setting it equal to zero.

First, expand $V(x)$: $$V(x) = x(144 - 48x + 4x^2) = 4x^3 - 48x^2 + 144x$$

Now, find the derivative $V'(x)$:

$$V'(x) = \frac{d}{dx}(4x^3 - 48x^2 + 144x)$$

$$V'(x) = 12x^2 - 96x + 144$$

Set the derivative to zero to find critical points:

$$12x^2 - 96x + 144 = 0$$

We can divide the entire equation by 12 to simplify:

$$x^2 - 8x + 12 = 0$$

This is a quadratic equation. We can factor it:

$$(x - 2)(x - 6) = 0$$

This gives us two possible values for $x$: $x = 2$ and $x = 6$.

Recall that the valid domain for $x$ is $0 < x < 6$. The value $x = 6$ is at the boundary of the domain (and would result in a base of size $12 - 2(6) = 0$, giving a volume of 0), so the relevant critical point within the domain is $x = 2$.

Confirming Maximum Volume

To confirm that $x = 2$ corresponds to a maximum volume, we can use the second derivative test or check the sign of the first derivative around $x = 2$.

Let's check the sign of $V'(x) = 12(x - 2)(x - 6)$:

• For $x < 2$ (e.g., $x=1$), $V'(1) = 12(1-2)(1-6) = 12(-1)(-5) = 60 > 0$. The volume is increasing.
• For $x > 2$ but $x < 6$ (e.g., $x=3$), $V'(3) = 12(3-2)(3-6) = 12(1)(-3) = -36 < 0$. The volume is decreasing.

Since the derivative changes from positive to negative at $x = 2$, the volume has a local maximum at $x = 2$. This is the maximum volume within the valid domain.

Alternatively, using the second derivative:

$$V''(x) = \frac{d}{dx}(12x^2 - 96x + 144)$$

$$V''(x) = 24x - 96$$

Evaluate $V''(x)$ at $x = 2$:

$$V''(2) = 24(2) - 96 = 48 - 96 = -48$$

Since $V''(2) < 0$, the volume is maximized at $x = 2$.

The value of $x$ that maximizes the volume of the tray is 2 cm.

Let's check the options provided. The option with the value 2 cm is the correct one.