How large should the squares cut from the corners of a 12-cm-by-12-cm sheet of tin be to make the box hold as much as possible?

Steps to Solve

1. Define the dimensions of the box

Let $x$ be the length of a side of the square cut from each corner. After cutting the squares, the new dimensions of the box will be:

• Length: $12 - 2x$ (cutting $x$ from both sides)
• Width: $12 - 2x$
• Height: $x$

2. Write the volume function

The volume $V$ of the box can be expressed in terms of $x$: $$ V = (12 - 2x)(12 - 2x)(x) = x(12 - 2x)^2 $$

3. Expand the volume function

Now, let's expand the volume expression: $$ V = x(12 - 2x)^2 = x(144 - 48x + 4x^2) $$ So, $$ V = 144x - 48x^2 + 4x^3 $$

4. Find the critical points

To maximize the volume, take the derivative of $V$ with respect to $x$ and set it to zero: $$ V' = 144 - 96x + 12x^2 $$ Set the derivative to zero: $$ 12x^2 - 96x + 144 = 0 $$

5. Solve the quadratic equation

Using the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Here, $a = 12$, $b = -96$, and $c = 144$: $$ x = \frac{96 \pm \sqrt{(-96)^2 - 4 \cdot 12 \cdot 144}}{2 \cdot 12} $$

6. Calculate the value of x

First, calculate the discriminant: $$ (-96)^2 - 4 \cdot 12 \cdot 144 = 9216 - 6912 = 2304 $$ Then find the square root: $$ \sqrt{2304} = 48 $$ Now substitute back into the quadratic formula: $$ x = \frac{96 \pm 48}{24} $$

Thus, the two solutions are:

1. $$ x = \frac{144}{24} = 6 $$

2. $$ x = \frac{48}{24} = 2 $$

3. Determine the feasible solution

Since $x$ must be less than half the dimensions of the box, the feasible value is: $$ x = 2 $$