Key Concepts and Formulas

• Volume of a Rectangular Prism: $V = lwh$, where $l$ is the length, $w$ is the width, and $h$ is the height.
• Optimization using Derivatives: To find the maximum or minimum of a function $f(x)$, find the critical points by solving $f'(x) = 0$. Then, use the second derivative test: if $f''(x) > 0$, the point is a local minimum; if $f''(x) < 0$, the point is a local maximum.
• Quadratic Formula: The solutions to the quadratic equation $ax^2 + bx + c = 0$ are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

Step 1: Express the volume of the box in terms of x.

We are cutting squares of side $x$ from each corner of the rectangular sheet. After folding, the dimensions of the box will be:

• Length: $l = a - 2x$
• Width: $w = b - 2x$
• Height: $h = x$

Therefore, the volume of the box, $V$, is given by:

$V(x) = (a - 2x)(b - 2x)x = (ab - 2ax - 2bx + 4x^2)x = 4x^3 - 2(a+b)x^2 + abx$

Step 2: Find the first derivative of the volume with respect to x.

To find the critical points, we need to find the derivative of $V(x)$ with respect to $x$:

$\frac{dV}{dx} = \frac{d}{dx}(4x^3 - 2(a+b)x^2 + abx) = 12x^2 - 4(a+b)x + ab$

Step 3: Set the first derivative equal to zero and solve for x.

To find the critical points, we set $\frac{dV}{dx} = 0$:

$12x^2 - 4(a+b)x + ab = 0$

This is a quadratic equation in $x$. We can use the quadratic formula to solve for $x$:

$x = \frac{-(-4(a+b)) \pm \sqrt{(-4(a+b))^2 - 4(12)(ab)}}{2(12)} = \frac{4(a+b) \pm \sqrt{16(a+b)^2 - 48ab}}{24}$

Simplify the expression under the square root:

$16(a^2 + 2ab + b^2) - 48ab = 16a^2 + 32ab + 16b^2 - 48ab = 16a^2 - 16ab + 16b^2 = 16(a^2 - ab + b^2)$

Substitute this back into the equation for $x$:

$x = \frac{4(a+b) \pm \sqrt{16(a^2 - ab + b^2)}}{24} = \frac{4(a+b) \pm 4\sqrt{a^2 - ab + b^2}}{24} = \frac{a+b \pm \sqrt{a^2 - ab + b^2}}{6}$

So we have two possible values for $x$:

$x_1 = \frac{a+b + \sqrt{a^2 - ab + b^2}}{6}$ $x_2 = \frac{a+b - \sqrt{a^2 - ab + b^2}}{6}$

Step 4: Determine which value of x maximizes the volume using the second derivative test.

Find the second derivative of $V(x)$:

$\frac{d^2V}{dx^2} = \frac{d}{dx}(12x^2 - 4(a+b)x + ab) = 24x - 4(a+b)$

Now, evaluate the second derivative at both critical points. First, consider $x_1 = \frac{a+b + \sqrt{a^2 - ab + b^2}}{6}$:

$\frac{d^2V}{dx^2}(x_1) = 24\left(\frac{a+b + \sqrt{a^2 - ab + b^2}}{6}\right) - 4(a+b) = 4(a+b) + 4\sqrt{a^2 - ab + b^2} - 4(a+b) = 4\sqrt{a^2 - ab + b^2}$

Since $a^2 - ab + b^2$ is always positive for real $a$ and $b$ (except when $a=b=0$, which is not a valid case for this problem), the second derivative at $x_1$ is positive. This means that $x_1$ corresponds to a local minimum of the volume.

Now, consider $x_2 = \frac{a+b - \sqrt{a^2 - ab + b^2}}{6}$:

$\frac{d^2V}{dx^2}(x_2) = 24\left(\frac{a+b - \sqrt{a^2 - ab + b^2}}{6}\right) - 4(a+b) = 4(a+b) - 4\sqrt{a^2 - ab + b^2} - 4(a+b) = -4\sqrt{a^2 - ab + b^2}$

The second derivative at $x_2$ is negative, which means that $x_2$ corresponds to a local maximum of the volume.

Step 5: Check for validity of the solution

We must ensure that $x > 0$, $a - 2x > 0$ and $b - 2x > 0$. This implies $0 < x < a/2$ and $0 < x < b/2$. The solution $x_1$ is greater than both $a/2$ and $b/2$, hence it can be discarded. Therefore, the correct solution is $x_2$.

Therefore, the value of $x$ that maximizes the volume is:

$x = \frac{a+b - \sqrt{a^2 - ab + b^2}}{6}$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when differentiating and applying the quadratic formula. A small sign error can lead to a completely incorrect answer.
• Validity of Solutions: Always check if the solutions you obtain are physically meaningful. In this case, $x$ must be positive and less than half of both $a$ and $b$. Discard any solutions that don't satisfy these conditions.
• Second Derivative Test: Remember to use the second derivative test to determine whether a critical point is a maximum or minimum.

Summary

To find the value of $x$ that maximizes the volume of the open box, we first expressed the volume as a function of $x$. Then, we found the critical points by taking the first derivative and setting it equal to zero. Finally, we used the second derivative test to determine which critical point corresponded to a maximum volume. The value of $x$ that maximizes the volume is $x = \frac{a+b - \sqrt{a^2 - ab + b^2}}{6}$.

Final Answer

The final answer is $\boxed{\frac{a + b - \sqrt{a^2 + b^2 - ab}}{6}}$, which corresponds to option (C).