Question

Minimum-surface-area box All boxes with a square base and a volume of \(50 \mathrm{ft}^{3}\) have a surface area given by \(S(x)=2 x^{2}+\frac{200}{x}\) 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 dimensions of the box with minimum surface area are base length and width equal to \(\sqrt[3]{50}\) feet and height equal to \(\frac{50}{(\sqrt[3]{50})^2}\) feet.

Step-by-step solution

Find the first derivative

To find the critical points, we'll first find the first derivative of the function \(S(x)\). Use the power rule and the chain rule to find the derivative: \(\frac{d}{dx}(2x^2) + \frac{d}{dx}(\frac{200}{x})\) \(S'(x) = 4x - \frac{200}{x^2}\)

Find the critical points

Next, we need to find the critical points by setting the first derivative equal to 0 and solving for \(x\): \(4x - \frac{200}{x^2} = 0\) To solve this equation, first multiply both sides by \(x^2\) to eliminate the fraction: \(4x^3 - 200 = 0\) Now add 200 to both sides and divide by 4 to isolate the term with \(x\): \(x^3 = \frac{200}{4}\) \(x^3 = 50\) Now, take the cube root of both sides to find the critical point: \(x = \sqrt[3]{50}\) So, the critical point is \(x = \sqrt[3]{50}\).

Find the second derivative

To check if the critical point yields a minimum surface area, we'll find the second derivative of \(S(x)\): \(\frac{d^2}{dx^2}(4x - \frac{200}{x^2})\) \(S''(x) = 4 + \frac{400}{x^3}\)

Check for concavity at the critical point

Now, we'll evaluate the second derivative at the critical point \(x = \sqrt[3]{50}\): \(S''(\sqrt[3]{50}) = 4 + \frac{400}{(\sqrt[3]{50})^3}\) \(S''(\sqrt[3]{50}) = 4 + \frac{400}{50}\) \(S''(\sqrt[3]{50}) = 4 + 8\) \(S''(\sqrt[3]{50}) = 12\) Since the second derivative is positive at the critical point, this proves that the surface area function \(S(x)\) exhibits a minimum value at \(x = \sqrt[3]{50}\).

Find the dimensions of the box

Finally, we will calculate the dimensions of the box: Base length: \(x = \sqrt[3]{50}\) Base width: \(x = \sqrt[3]{50}\) Height: To find the height, use the volume formula: \(Volume = length \cdot width \cdot height\) \(50 = (\sqrt[3]{50})(\sqrt[3]{50}) \cdot height\) Divide both sides by \((\sqrt[3]{50})^2\): \(height = \frac{50}{(\sqrt[3]{50})^2}\) So, the dimensions of the box with minimum surface area are base length and width equal to \(\sqrt[3]{50}\) feet and height equal to \(\frac{50}{(\sqrt[3]{50})^2}\) feet.

Key concepts

Critical Points

Critical points are where a function's derivative either equals zero or is undefined. These points are vital in optimization problems since they help identify potential maximum or minimum values of a function. For the surface area function given, we identified the critical point by setting the first derivative equal to zero:

• We started by finding the derivative of the function.
• Set the derivative equal to zero: \(4x - \frac{200}{x^2} = 0\).
• Solving this equation led to the critical point: \(x = \sqrt[3]{50}\).

This critical point is an essential step in determining where the minimum surface area occurs, as it provides the possible solution within a given interval.

Derivative

The derivative is a fundamental tool in calculus that describes the rate at which a function is changing at any given point. It is paramount in finding critical points. In this problem:

• The surface area function is \(S(x) = 2x^2 + \frac{200}{x}\).
• We calculated the first derivative \(S'(x) = 4x - \frac{200}{x^2}\).

This derivative helps us determine where the slope of the tangent line to the function is zero, indicating potential places for minima or maxima. Using rules like the power rule and chain rule, we easily find the expression that guides us towards optimization.

Concavity

Concavity tells us how a function curves, which is useful to confirm whether a critical point is indeed a minimum or maximum. We determine this by using the second derivative. In this exercise:

• The second derivative is \(S''(x) = 4 + \frac{400}{x^3}\).
• By evaluating the second derivative at our critical point \(x = \sqrt[3]{50}\), we found \(S''(\sqrt[3]{50}) = 12\).

Since the second derivative was positive, this indicated the graph is concave up at the critical point, confirming a local minimum. Understanding concavity ensures accurate conclusions about the function's behavior at critical points.

Volume Calculation

Volume calculations in geometry often connect various dimensions of an object. For a box with a square base, the volume formula is:

• Volume = length \(\times\) width \(\times\) height.
• Given the volume \(= 50 \, \text{ft}^3\), both length and width are \(x = \sqrt[3]{50}\).
• This leads us to find the height using: \(height = \frac{50}{(\sqrt[3]{50})^2}\).

These calculations determine the specific dimensions that give the minimum surface area while satisfying the volume constraint. This approach highlights the relationship between algebraic functions and geometric properties, providing a deeper understanding of volume optimization.