If 1600 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.
Volume
Question Help: cubic centimeters.

Question

If 1600 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box. Volume $$ = $$ Question Help: cubic centimeters.

UpStudy AI · Accepted Answer

Let the side length of the square base be $$ x $$ and the height of the box be $$ y $$. Since the box has an open top, the surface area is used for the base and the four sides. The constraint on the surface area is

$$
x^2 + 4xy = 1600.
$$

The volume of the box is given by

$$
V = x^2 y.
$$

**Step 1. Express $$ y $$ in terms of $$ x $$:**

From the surface area constraint, solve for $$ y $$:

$$
4xy = 1600 - x^2 \quad \Longrightarrow \quad y = \frac{1600 - x^2}{4x}.
$$

**Step 2. Write the volume in terms of $$ x $$ only:**

Substitute $$ y $$ into the volume formula:

$$
V(x) = x^2 \left(\frac{1600 - x^2}{4x}\right) = \frac{x\,(1600 - x^2)}{4} = 400x - \frac{x^3}{4}.
$$

**Step 3. Optimize the volume by differentiating with respect to $$ x $$:**

Differentiate $$ V(x) $$ with respect to $$ x $$:

$$
V'(x) = 400 - \frac{3x^2}{4}.
$$

Set the derivative equal to zero to find the critical point:

$$
400 - \frac{3x^2}{4} = 0.
$$

Solve for $$ x $$:

$$
\frac{3x^2}{4} = 400 \quad \Longrightarrow \quad 3x^2 = 1600 \quad \Longrightarrow \quad x^2 = \frac{1600}{3}.
$$

Taking the positive square root (since $$ x $$ must be positive):

$$
x = \sqrt{\frac{1600}{3}} = \frac{40}{\sqrt{3}}.
$$

**Step 4. Find $$ y $$ corresponding to $$ x = \frac{40}{\sqrt{3}} $$:**

Substitute $$ x = \frac{40}{\sqrt{3}} $$ into the expression for $$ y $$:

$$
y = \frac{1600 - x^2}{4x}.
$$

We already have $$ x^2 = \frac{1600}{3} $$, so:

$$
y = \frac{1600 - \frac{1600}{3}}{4 \cdot \frac{40}{\sqrt{3}}} = \frac{1600\left(1 - \frac{1}{3}\right)}{160 \cdot \frac{1}{\sqrt{3}}} = \frac{1600\left(\frac{2}{3}\right)}{160 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{3200}{3}}{160/\sqrt{3}} = \frac{3200}{3} \cdot \frac{\sqrt{3}}{160} = \frac{20\sqrt{3}}{3}.
$$

**Step 5. Calculate the maximum volume:**

Substitute $$ x = \frac{40}{\sqrt{3}} $$ and $$ y = \frac{20\sqrt{3}}{3} $$ back into the volume formula:

$$
V = x^2 y.
$$

Since $$ x^2 = \frac{1600}{3} $$:

$$
V = \frac{1600}{3} \cdot \frac{20\sqrt{3}}{3} = \frac{32000\sqrt{3}}{9}.
$$

Thus, the largest possible volume of the box is

$$
\boxed{\frac{32000\sqrt{3}}{9}} \text{ cubic centimeters}.
$$
The largest possible volume of the box is $$ \frac{32000\sqrt{3}}{9} $$ cubic centimeters.

If 1600 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.
Volume
Question Help: cubic centimeters.

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If 1600 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box. Volume Question Help: cubic centimeters.

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If 1600 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.
Volume
Question Help: cubic centimeters.