Maximizing Tray Volume by Optimizing Cutout Size $x$

We are given a square sheet of side 12 cm. Four small squares of side $x$ are cut from the corners to create a tray by folding the edges.

Tray Dimensions and Volume Formula

After cutting the squares and folding, the resulting tray has:

• Height = $x$ cm
• Base Length = $(12 - 2x)$ cm
• Base Width = $(12 - 2x)$ cm

The volume $V$ of the tray is given by the formula:

$V(x) = \text{Height} \times \text{Base Length} \times \text{Base Width}$
$V(x) = x (12 - 2x) (12 - 2x)$
$V(x) = x (12 - 2x)^2$

Finding the Maximum Volume using Calculus

To find the maximum volume, we need to find the value of $x$ that maximizes $V(x)$. This typically involves finding the derivative of $V(x)$ and setting it to zero.

1. Expand the volume function:
   $V(x) = x (144 - 48x + 4x^2)$
   $V(x) = 4x^3 - 48x^2 + 144x$
2. Find the first derivative $V'(x)$:
   $V'(x) = \frac{dV}{dx} = 12x^2 - 96x + 144$
3. Set the derivative to zero to find critical points:
   $12x^2 - 96x + 144 = 0$
   Divide by 12:
   $x^2 - 8x + 12 = 0$
4. Solve the quadratic equation:
   Factor the equation:
   $(x - 2)(x - 6) = 0$
   This gives two possible values for $x$: $x = 2$ or $x = 6$.

Determining the Maximum

We need to consider the physical constraints of the problem. The side of the cutout square, $x$, must be positive ($x > 0$). Also, the dimensions of the base must be positive, so $12 - 2x > 0$, which implies $2x < 12$, or $x < 6$. Therefore, the valid range for $x$ is $0 < x < 6$.

The critical point $x = 6$ is outside the valid range for a tray with positive base dimensions (it would result in a base of zero area and zero volume). The critical point $x = 2$ lies within the valid range.

To confirm that $x = 2$ yields a maximum volume, we can use the second derivative test:

1. Find the second derivative $V''(x)$:
   $V''(x) = \frac{d^2V}{dx^2} = 24x - 96$
2. Evaluate $V''(x)$ at the critical point $x=2$:
   $V''(2) = 24(2) - 96 = 48 - 96 = -48$

Since $V''(2) < 0$, the volume function $V(x)$ has a local maximum at $x = 2$.

Conclusion

The value of $x$ that maximizes the tray's volume is $2\text{ cm}$.