Question

All boxes with a square base and a volume of \(50 \mathrm{ft}^{3}\) have a surface area given by \(S(x)=2 x^{2}+200 / x\) where \(x\) is the length of the sides of the base. Find the absolute minimum of the surface area function on the interval \((0, \infty) .\) What are the dimensions of the box with minimum surface area?

Short answer

Answer: The box with minimum surface area has dimensions of \(\sqrt[3]{50}\) ft by \(\sqrt[3]{50}\) ft by \(\frac{50}{(\sqrt[3]{50})^2}\) ft.

Step-by-step solution

Find the first derivative

We have the given function \(S(x) = 2x^2 + \frac{200}{x}\). To find its critical points, we need to take the first derivative of \(S(x)\) with respect to \(x\). Using the Power Rule and Sum Rule, we have: \(S'(x) = \frac{d}{dx}(2x^2) + \frac{d}{dx}(\frac{200}{x})\) \(S'(x) = 4x - \frac{200}{x^2}\) Now that we have the first derivative, we can proceed to find the critical points.

Find the critical points

To find the critical points, set \(S'(x)\) equal to zero and solve for x. \(4x - \frac{200}{x^2} = 0\) To solve this equation for x, we can multiply both sides by \(x^2\) to eliminate the fraction: \(4x^3 - 200 = 0\) Now, solve this equation for \(x\): \(4x^3 = 200\) \(x^3 = \frac{200}{4}\) \(x^3 = 50\) Therefore, \(x = \sqrt[3]{50}\) This gives us a single critical point, \(x = \sqrt[3]{50}\).

Test the critical point

The function \(S(x) = 2x^2 + \frac{200}{x}\) is defined on the interval \((0, \infty)\). Since there is only one critical point, we know that this point represents the absolute minimum. However, it is still useful to confirm that the second derivative is positive for this value of \(x\). First, let's find the second derivative, \(S''(x)\): \(S''(x) = \frac{d}{dx}(4x - \frac{200}{x^2})\) \(S''(x) = 4 + \frac{400}{x^3}\) Now, let's check if \(S''(\sqrt[3]{50}) > 0\): \(S''(\sqrt[3]{50}) = 4 + \frac{400}{(\sqrt[3]{50})^3}\) \(S''(\sqrt[3]{50}) = 4 + \frac{400}{50}\) \(S''(\sqrt[3]{50}) = 4 + 8\) \(S''(\sqrt[3]{50}) = 12\) Since \(S''(\sqrt[3]{50}) > 0\), the critical point represents a local minimum and, in this case, an absolute minimum.

Determine the dimensions of the box

Now that we have found the value of \(x\) where the surface area is minimized, we can determine the dimensions of the box with minimum surface area. The volume of the box is 50 cubic feet, and the box has a square base with side length \(x\). Thus, we can write its dimensions as \(x\), \(x\), and \(\frac{50}{x^2}\). Since \(x = \sqrt[3]{50}\), the dimensions of the box with minimum surface area are: Length: \(\sqrt[3]{50} \mathrm{ft}\) Width: \(\sqrt[3]{50} \mathrm{ft}\) Height: \(\frac{50}{(\sqrt[3]{50})^2} \mathrm{ft}\) In conclusion, the box with minimum surface area has dimensions \(\sqrt[3]{50}\) ft by \(\sqrt[3]{50}\) ft by \(\frac{50}{(\sqrt[3]{50})^2}\) ft.

Key concepts

Critical Points

In calculus, finding critical points is an essential step in optimization problems. These points are where a function's derivative is zero, or the derivative does not exist. They help identify possible extrema (i.e., maximum and minimum values) on a given interval.
To find the critical points of a function, follow these steps:

• Calculate the first derivative of the function.
• Set the derivative equal to zero and solve for the variable.
• Consider the points where the derivative does not exist, if such points are within the interval.

For the given function, the first derivative is found using the derivative rules, leading to solving the equation derived from setting it to zero. In our problem, this led to a single critical point at \( x = \sqrt[3]{50} \). This critical point helps us determine where the surface area of the box is optimized.

Derivative

The derivative of a function describes how the function's value changes as its input changes. For optimization, it helps identify where a function increases or decreases. In this problem, we are given a surface area function \( S(x) = 2x^2 + \frac{200}{x} \).
To find the first derivative, apply calculus rules:

• The Power Rule: For \( x^n \), the derivative is \( nx^{n-1} \).
• The Quotient Rule: For \( \frac{a}{x} \), the derivative is \( -\frac{a}{x^2} \).

Applying these, the derivative \( S'(x) = 4x - \frac{200}{x^2} \) emerges, indicating the rate at which the surface area changes with the length of the base side \( x \). This derivative is crucial for locating critical points and understanding the behavior of \( S(x) \).

Second Derivative

The second derivative of a function provides insight into the concavity and points of inflection of the function. It is also used in the second derivative test to determine whether a critical point is a maximum, minimum, or a point of inflection.
To find the second derivative of a function, take the derivative of the first derivative. For our function:

• The first derivative is \( S'(x) = 4x - \frac{200}{x^2} \).
• Then the second derivative is \( S''(x) = 4 + \frac{400}{x^3} \).

In our problem, evaluating the second derivative at the critical point \( x = \sqrt[3]{50} \) gives \( S''(\sqrt[3]{50}) = 12 \), which is positive. A positive second derivative at this point confirms that \( x = \sqrt[3]{50} \) is a minimum for the surface area function.

Volume of a Box

Understanding the volume of a box with a square base is key in solving this optimization problem. The volume \( V \) of such a box is calculated by the formula \( V = x^2 \times h \), where \( x \) is the side length of the square base, and \( h \) is the height.
Given a fixed volume \( V = 50 \) for the box, the relationship \( x^2 \times h = 50 \) allows us to express the height \( h \) in terms of the base side length \( x \):

• \( h = \frac{50}{x^2} \)

This height expression is essential for determining the dimensions of the box when optimization conditions (minimum surface area) are met. As found earlier, by deriving and solving for \( x \), the optimized box dimensions become \( x = \sqrt[3]{50} \), and the height is \( \frac{50}{(\sqrt[3]{50})^2} \). This ensures that the box uses the minimum surface area possible while maintaining the given volume.