Maximizing Volume of an Open Box

This question asks us to find the maximum volume of an open box with a square base, given its surface area is 36 square units. This is a classic optimization problem involving calculus.

Setting up the Problem

Let the side length of the square base be \( x \) units and the height of the box be \( h \) units.

• The area of the square base is \( x^2 \).
• The box is open, so there is no top face.
• The areas of the four side faces are \( x \times h \) each.

The total surface area (\( A \)) of the open box is the sum of the area of the base and the areas of the four sides:

\( A = x^2 + 4xh \)

We are given that the surface area is 36 units. So, we have the constraint equation:

\( x^2 + 4xh = 36 \)

The volume (\( V \)) of the box is given by the area of the base multiplied by the height:

\( V = x^2 h \)

Our goal is to maximize this volume \( V \) subject to the surface area constraint.

Expressing Volume as a Single Variable Function

To maximize \( V \), we need to express it as a function of a single variable, either \( x \) or \( h \). From the surface area constraint, we can solve for \( h \):

\( 4xh = 36 - x^2 \)

Since \( x \) must be a length, \( x > 0 \). Also, the height \( h \) must be positive, which implies \( 36 - x^2 > 0 \), so \( x^2 < 36 \). This means \( 0 < x < 6 \).

Now, solve for \( h \):

\( h = \frac{36 - x^2}{4x} \)

Substitute this expression for \( h \) into the volume formula \( V = x^2 h \):

\( V(x) = x^2 \left( \frac{36 - x^2}{4x} \right) \)

Simplify the expression for \( V(x) \):

\( V(x) = \frac{x(36 - x^2)}{4} = \frac{36x - x^3}{4} \)

Finding the Maximum Volume using Calculus

To find the maximum volume, we need to find the critical points of \( V(x) \) by taking the derivative with respect to \( x \) and setting it to zero.

\( V'(x) = \frac{d}{dx} \left( \frac{36x - x^3}{4} \right) \)

\( V'(x) = \frac{1}{4} \frac{d}{dx} (36x - x^3) \)

\( V'(x) = \frac{1}{4} (36 - 3x^2) \)

Set \( V'(x) = 0 \) to find critical points:

\( \frac{1}{4} (36 - 3x^2) = 0 \)

\( 36 - 3x^2 = 0 \)

\( 3x^2 = 36 \)

\( x^2 = 12 \)

\( x = \sqrt{12} \) or \( x = -\sqrt{12} \)

Since \( x \) must be positive (length of a side), we take \( x = \sqrt{12} \). Simplify \( \sqrt{12} \):

\( x = \sqrt{4 \times 3} = 2\sqrt{3} \)

This value \( x = 2\sqrt{3} \) is within our domain \( 0 < x < 6 \) because \( \sqrt{12} \) is between \( \sqrt{9}=3 \) and \( \sqrt{36}=6 \).

Verifying Maximum Volume

We can use the second derivative test to confirm that \( x = 2\sqrt{3} \) corresponds to a maximum volume. Calculate the second derivative \( V''(x) \):

\( V''(x) = \frac{d}{dx} \left( \frac{1}{4} (36 - 3x^2) \right) \)

\( V''(x) = \frac{1}{4} (-6x) = -\frac{3x}{2} \)

Evaluate \( V''(x) \) at \( x = 2\sqrt{3} \):

\( V''(2\sqrt{3}) = -\frac{3(2\sqrt{3})}{2} = -3\sqrt{3} \)

Since \( V''(2\sqrt{3}) < 0 \), the volume is indeed maximum at \( x = 2\sqrt{3} \).

Calculating the Maximum Volume

Substitute the value of \( x = 2\sqrt{3} \) back into the volume formula \( V(x) = \frac{36x - x^3}{4} \):

\( V_{max} = V(2\sqrt{3}) = \frac{36(2\sqrt{3}) - (2\sqrt{3})^3}{4} \)

Calculate \( (2\sqrt{3})^3 \):

\( (2\sqrt{3})^3 = (2^3) \times (\sqrt{3})^3 = 8 \times (3\sqrt{3}) = 24\sqrt{3} \)

Now substitute back into the volume formula:

\( V_{max} = \frac{72\sqrt{3} - 24\sqrt{3}}{4} \)

\( V_{max} = \frac{48\sqrt{3}}{4} \)

\( V_{max} = 12\sqrt{3} \)

The maximum volume of the open box is \( 12\sqrt{3} \) cubic units.

Summary of Steps

• Defined variables for base side and height.
• Wrote formulas for surface area and volume.
• Used the surface area constraint to express height in terms of base side.
• Substituted height into the volume formula to get volume as a function of the base side.
• Found the derivative of the volume function.
• Set the derivative to zero to find critical points.
• Selected the valid critical point based on physical constraints.
• Used the second derivative test to confirm it's a maximum.
• Calculated the maximum volume by substituting the base side value back into the volume formula.

Revision Table: Box Optimization Concepts

Additional Information: Related Optimization Problems

Optimization problems using calculus appear in many contexts. Here are a few examples:

• Finding the dimensions of a rectangle with a fixed perimeter that maximize its area. This involves setting up area as a function of one side and using the perimeter constraint.
• Finding the shortest distance from a point to a curve. This involves minimizing the distance function squared.
• Minimizing the cost of materials to construct a container of a given volume. Similar to the box problem, but minimizing surface area for a fixed volume.

These problems generally involve setting up a function to be optimized (like volume or area) and a constraint equation (like fixed surface area or perimeter). The constraint is used to reduce the function to a single variable, and then calculus (derivatives) is applied to find the optimal value.