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?
Understand the Problem
The question is asking how large the squares should be that are cut from the corners of a rectangular sheet in order to maximize the volume of the resulting open-top box.
Answer
The size of the squares cut from the corners to maximize the box's volume is $x = 2 \text{ cm}$.
Answer for screen readers
The size of the squares cut from the corners to maximize the box's volume is $x = 2 \text{ cm}$.
Steps to Solve
- 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: $x$
- Write the volume function
The volume $V$ of the box can be expressed in terms of $x$: $$ V = (12 - 2x)(12 - 2x)(x) = x(12 - 2x)^2 $$
- Expand the volume function
Now, let's expand the volume expression: $$ V = x(12 - 2x)^2 = x(144 - 48x + 4x^2) $$ So, $$ V = 144x - 48x^2 + 4x^3 $$
- 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: $$ 12x^2 - 96x + 144 = 0 $$
- Solve the quadratic equation
Using the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Here, $a = 12$, $b = -96$, and $c = 144$: $$ x = \frac{96 \pm \sqrt{(-96)^2 - 4 \cdot 12 \cdot 144}}{2 \cdot 12} $$
- 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:
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$$ x = \frac{144}{24} = 6 $$
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$$ x = \frac{48}{24} = 2 $$
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Determine the feasible solution
Since $x$ must be less than half the dimensions of the box, the feasible value is: $$ x = 2 $$
The size of the squares cut from the corners to maximize the box's volume is $x = 2 \text{ cm}$.
More Information
This problem involves finding the maximum volume of an open-top box made from a flat sheet by strategically cutting out squares from the corners. This optimization is crucial in various real-world applications, including packaging and design.
Tips
- Skipping the step to determine feasible values for $x$. Remember that $x$ must be less than half the shorter side of the rectangle.
- Incorrectly applying the quadratic formula or making arithmetic mistakes when calculating the discriminant.
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