Problem 5 - Calculus: Early Transcendentals

Question

A box with square base and no top is to hold a volume V. Find (in terms of V) the dimensions of the box that requires the least material for the five sides. Also find the ratio of height to side of the base. (This ratio will not involve V.)

Step-by-Step Solution

Answer

The base side is \(x = \sqrt[3]{2V}\); the ratio of height to base side is \(\frac{1}{2\sqrt[3]{2}}\).

Step 1: Define the Variables and Objective

First, define the variables. Let the side of the base be \(x\) and the height of the box be \(h\). The volume of the box, \(V\), is constant and given by the formula \(V = x^2h\). The objective is to minimize the material used, which corresponds to minimizing the surface area of the four sides and bottom, since the top is open.

Step 2: Express Surface Area in Terms of Variables

The surface area \(A\) of the box with a square base involves the base and the four vertical sides. The area of the base is \(x^2\), and the area of the four sides is \(4xh\). Thus, the total area is given by the equation \(A = x^2 + 4xh\).

Step 3: Substitute Height from Volume Formula

Using the volume formula \(V = x^2h\), solve for \(h\): \[ h = \frac{V}{x^2} \]Substitute \(h\) in the expression for the area:\[ A = x^2 + 4x\left(\frac{V}{x^2}\right) = x^2 + \frac{4V}{x} \]

Step 4: Optimization - Find Critical Points

To find the dimensions for minimal material, take the derivative of \(A\) with respect to \(x\) and set it to zero:\[ \frac{dA}{dx} = 2x - \frac{4V}{x^2} \]Set \(\frac{dA}{dx} = 0\) and solve:\[ 2x = \frac{4V}{x^2} \] \[ 2x^3 = 4V \] \[ x^3 = 2V \] \[ x = \sqrt[3]{2V} \]

Step 5: Calculate Height using Optimal Base Side

Substitute \(x\) back into the formula for \(h = \frac{V}{x^2}\):\[ h = \frac{V}{(\sqrt[3]{2V})^2} = \frac{V}{(2V)^{2/3}} \]\[ h = \frac{V}{4V^{2/3}} = \frac{1}{4}V^{1/3} \]

Step 6: Determine the Ratio of Height to Side of the Base

The ratio of the height \(h\) to the side of the base \(x\) is given by:\[ \text{Ratio} = \frac{h}{x} = \frac{\frac{1}{4}V^{1/3}}{\sqrt[3]{2V}} = \frac{1/4}{\sqrt[3]{2}} \]This simplifies to approximately \(\frac{1}{2\sqrt[3]{2}}\), which does not involve \(V\).

Key Concepts

Volume of a Box

When dealing with the volume of a box, you often encounter formulas and concepts that help define its capacity. For a box with a square base and no top, the volume can be expressed as \( V = x^2h \), where \( x \) is the side length of the square base and \( h \) is the height of the box. The formula implies that the volume is a product of the area of the base and the height. In problems involving optimization in calculus, a constant volume is given, and your task may involve finding other dimensions that meet a specific condition such as minimizing the material needed.

Surface Area Minimization

Minimizing the surface area of an open box with a square base is a classic problem in optimization. The surface area \( A \) of such a box includes the area of the base and the areas of the four sides. It is calculated using:

• The area of the base: \( x^2 \)
• The combined area of the four sides: \( 4xh \)

These combine to give \( A = x^2 + 4xh \). The challenge is to find the dimensions that require the least material, i.e., the minimal surface area, given a constant volume. This involves expressing the surface area in terms of one variable and using calculus to find the necessary critical points.

Critical Points

To optimize a function, finding its critical points is essential. Critical points are where the derivative of the function is zero or undefined, indicating potential local maxima or minima. In our box problem, we express the surface area as \( A = x^2 + \frac{4V}{x} \) and then calculate its derivative with respect to \( x \). The derivative, \( \frac{dA}{dx} = 2x - \frac{4V}{x^2} \), is set to zero to find critical values of \( x \). Solving \( 2x = \frac{4V}{x^2} \) results in the ideal base side length \( x = \sqrt[3]{2V} \). This ensures that our conditions for minimal material usage are met when constructing the box.

Derivative Applications

Derivatives are incredibly useful in optimization problems as they help find points of interest like local extrema. In this box problem, we used derivatives to calculate where the surface area is minimized. The key derivative expression \( \frac{dA}{dx} \) helped determine the ideal side length \( x \) of the base. By applying the concept of derivatives, we were then able to solve for \( x \), and subsequently, calculate the height \( h \) by substituting back into \( h = \frac{V}{x^2} \). Lastly, the derivative technique enabled us to find the ratio of the height to the base side. By solving \( \text{Ratio} = \frac{h}{x} = \frac{1}{4\sqrt[3]{2}} \), we confirmed that the solution was independent of volume, a common objective in optimization problems.