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

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 =

UpStudy AI · Accepted Answer

Let $$ x $$ be the side length of the square base and $$ h $$ be the height of the box. The box has an open top, so the total amount of material (surface area) is used for the base and the four sides. This gives the constraint

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

### Expressing $$ h $$ in Terms of $$ x $$

Solve the constraint for $$ h $$:

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

### Volume Expression

The volume $$ V $$ of the box is

$$
V = x^2h.
$$

Substitute $$ h $$ from the previous expression:

$$
V(x) = x^2 \cdot \frac{1600 - x^2}{4x} = \frac{x(1600 - x^2)}{4}.
$$

### Finding the Maximum Volume

To maximize $$ V $$, differentiate with respect to $$ x $$ and set the derivative equal to zero. First, rewrite the volume function:

$$
V(x) = \frac{1600x - x^3}{4}.
$$

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

$$
V'(x) = \frac{d}{dx} \left(\frac{1600x - x^3}{4}\right) = \frac{1600 - 3x^2}{4}.
$$

Set the derivative to zero to find the critical points:

$$
\frac{1600 - 3x^2}{4} = 0 \quad \Rightarrow \quad 1600 - 3x^2 = 0 \quad \Rightarrow \quad 3x^2 = 1600,
$$

$$
x^2 = \frac{1600}{3} \quad \Rightarrow \quad x = \frac{40}{\sqrt{3}}.
$$

### Determine the Height $$ h $$

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

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

Recall

$$
x^2 = \frac{1600}{3},
$$

so

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

Simplify:

$$
h = \frac{3200}{3} \cdot \frac{\sqrt{3}}{160} = \frac{20\sqrt{3}}{3}.
$$

### Maximum Volume

Now, substitute $$ x $$ and $$ h $$ into the volume formula:

$$
V = x^2h = \left(\frac{40}{\sqrt{3}}\right)^2 \cdot \frac{20\sqrt{3}}{3}.
$$

Since

$$
\left(\frac{40}{\sqrt{3}}\right)^2 = \frac{1600}{3},
$$

we have

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

Thus, the largest possible volume of the box is

$$
\boxed{\frac{32000\sqrt{3}}{9}} \quad \text{cubic centimeters}.
$$
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 =

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