2. ( 2 points) A box with an open top is to be constructed from a square piece of cardboard with dimensions $$ 4 \mathrm{~cm} imes 4 \mathrm{~cm} $$. This box is constructed by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have.

Question

2. ( 2 points) A box with an open top is to be constructed from a square piece of cardboard with dimensions $$ 4 \mathrm{~cm} 	imes 4 \mathrm{~cm} $$. This box is constructed by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have.

UpStudy AI · Accepted Answer

Let $$ x $$ be the side length of the square cut from each corner. After cutting, the dimensions of the base of the box will be
$$
4 - 2x \quad \text{by} \quad 4 - 2x.
$$
The height of the box is $$ x $$. Thus, the volume $$ V $$ of the box is given by
$$
V = x(4-2x)^2.
$$

## Step 1. Express the Volume Function

Expanding the function:
$$
V = x(4-2x)^2 = x(16 - 16x + 4x^2) = 16x - 16x^2 + 4x^3.
$$

## Step 2. Find the Critical Points

Differentiate $$ V $$ with respect to $$ x $$:
$$
\frac{dV}{dx} = 16 - 32x + 12x^2.
$$
Set the derivative equal to zero:
$$
16 - 32x + 12x^2 = 0.
$$
Divide the equation by 2 to simplify:
$$
8 - 16x + 6x^2 = 0.
$$
Rearrange the terms:
$$
6x^2 - 16x + 8 = 0.
$$

## Step 3. Solve the Quadratic Equation

Using the quadratic formula, 
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
$$
where $$ a = 6 $$, $$ b = -16 $$, and $$ c = 8 $$:
$$
x = \frac{16 \pm \sqrt{(-16)^2 - 4(6)(8)}}{2 \times 6} = \frac{16 \pm \sqrt{256 - 192}}{12} = \frac{16 \pm \sqrt{64}}{12} = \frac{16 \pm 8}{12}.
$$

This gives two solutions:
$$
x = \frac{16 + 8}{12} = \frac{24}{12} = 2 \quad \text{or} \quad x = \frac{16 - 8}{12} = \frac{8}{12} = \frac{2}{3}.
$$

Since $$ x = 2 $$ leads to zero base dimensions (because $$ 4-2(2)=0 $$), it must be rejected. The feasible solution is:
$$
x = \frac{2}{3}.
$$

## Step 4. Calculate the Maximum Volume

Substitute $$ x = \frac{2}{3} $$ into the volume function:
$$
V = \frac{2}{3}\left(4 - 2\left(\frac{2}{3}\right)\right)^2.
$$

Simplify:
$$
4 - 2\left(\frac{2}{3}\right) = 4 - \frac{4}{3} = \frac{12}{3} - \frac{4}{3} = \frac{8}{3}.
$$

Thus,
$$
V = \frac{2}{3} \left(\frac{8}{3}\right)^2 = \frac{2}{3} \cdot \frac{64}{9} = \frac{128}{27}.
$$

## Final Answer

The largest volume that the box can have is
$$
\boxed{\frac{128}{27}} \text{ cubic centimeters}.
$$
The largest volume of the box is $$ \frac{128}{27} $$ cubic centimeters.

( 2 points) A box with an open top is to be constructed from a square piece of cardboard with dimensions
. This box is constructed by cutting out a square from each of the four corners and bending up
the sides. Find the largest volume that such a box can have.

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Solución

Step 1. Express the Volume Function

Step 2. Find the Critical Points

Step 3. Solve the Quadratic Equation

Step 4. Calculate the Maximum Volume

Final Answer

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( 2 points) A box with an open top is to be constructed from a square piece of cardboard with dimensions . This box is constructed by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have.