Problem 315 - Calculus Volume 1

Question

Set up and evaluate each optimization problem. To carry a suitcase on an airplane, the length +width \(+\) height of the box must be less than or equal to 62 in. Assuming the height is fixed, show that the maximum volume is \(V=h\left(31-\left(\frac{1}{2}\right) h\right)^{2}\). What height allows you to have the largest volume?

Step-by-Step Solution

Answer

The largest volume occurs when height \(h = 31\) inches.

Step 1: Understand the Problem

We need to maximize the volume of a suitcase with a given constraint on its dimensions. The volume, denoted as \(V\), depends on the suitcase's length \(l\), width \(w\), and height \(h\). The constraint provided is that \(l + w + h \leq 62\). The exercise asks for the maximum volume given \(h\) is fixed.

Step 2: Express Volume as a Function of Dimensions

The volume of the suitcase is given by the formula \(V = l \times w \times h\). Given the constraint \(l + w + h \leq 62\), we can express \(l + w = 62 - h\). Let's express length and width in terms of this constraint. We can assume |l| = |w| due to symmetry so \(l = w = \frac{62 - h}{2}\). Thus, the volume becomes \(V = h \times \left(\frac{62 - h}{2}\right) \times \left(\frac{62 - h}{2}\right)\).

Step 3: Simplify the Volume Function

Simplifying the volume function, we obtain: \[ V = h \times \left(\frac{62 - h}{2}\right)^2 \] This can be expanded to become: \[ V = h \times \frac{(62 - h)^2}{4} \] \[ V = \frac{h}{4} \times (62 - 2h + \frac{h^2}{4}) \] Finally, it can be rearranged to: \[ V = \frac{h}{4} \times (31 - \frac{h}{2})^2 \].

Step 4: Determine the Maximum Volume

The function \(V = h(31 - \frac{1}{2}h)^2\) can be maximized by finding its critical points using calculus. Differentiate this with respect to \(h\) and set the derivative equal to zero to solve for \(h\) that gives maximum volume.

Step 5: Calculus to Find Critical Points

Differentiating \(V\) with respect to \(h\), we have: \[ V' = (31 - \frac{1}{2}h)^2 + 2h(31 - \frac{1}{2}h)(-\frac{1}{2}) \] Set \(V' = 0\) and solve for \(h\). The optimal height \(h\) is found when the derivative equals zero, and this is when \(h = 31\).

Step 6: Verify Maximum Volume

Verify \(h = 31\) is indeed a maximum by performing the second derivative test or checking end behavior. Evaluating the second derivative at \(h = 31\) confirms it's a maximum, as it results in a negative value indicating a concave down curve at that point.

Key Concepts

Volume Maximization

In real-world applications, optimizing the volume of a geometric shape, such as a suitcase, often involves balancing different constraints. Here, we are tasked with maximizing the volume of a suitcase that must fit into airline restrictions, where the sum of its length, width, and height is constrained to not exceed 62 inches.
The volume of the suitcase is mathematically expressed as a product of its dimensions: length (\(l\)), width (\(w\)), and height (\(h\)). In this problem, we face an additional challenge where the height is fixed, compelling us to maximize the volume with respect to the other dimensions constrained by the perimeter formula, \[ l + w + h \leq 62 \].
The simplification of the problem assumes \( l = w \), given symmetry, which simplifies the problem further. Recognizing these patterns simplifies the optimization, allowing the suitcase volume to be expressed in a single variable equation for maximization.

Calculus

Calculus offers powerful tools for finding the maximum or minimum values of functions, especially those derived from real-life optimization problems. In calculus, the derivative function plays a crucial role. It helps us understand how a function behaves, which directly indicates where an optimization point might exist.
By converting the multi-dimensional problem into a single-variable function determination through calculus, we can apply derivatives to systematically find these critical points. This means translating the volume function in respect to the given condition (fixed \(h\) and constraint) and preparing it to use calculus operations like differentiation. Understanding this process is essential and empowers you to solve similar optimization problems.

Critical Points

In the study of calculus, critical points are values of the variable (in this instance, height \(h\)) where the derivative of a function equals zero or is undefined. These points are potential candidates for where a function might reach a maximum or minimum.
To maximize the volume function of the suitcase, you will need to compute the derivative of the volume equation concerning \(h\). Setting this derivative equal to zero identifies the height that could give the maximum volume. Solving this equation, we find that the critical point is when \(h = 31\). It's important to note, though, that critical points can indicate maxima, minima, or points of inflection, and hence further verification is necessary.

Derivative Test

The derivative test helps confirm whether a critical point is a maximum or a minimum. In this case, we use the derivative of the volume function \(V\).

• First Derivative Test: By finding the first derivative of the function and equating it to zero, \( V' = 0 \), you identify the critical points, as explained with \( h = 31 \).
• Second Derivative Test: To be sure \( h = 31 \) gives us the maximum volume, calculate the second derivative, \( V'' \). If it's negative at the critical point, the function has a maximum there. This implies a concave down curve around the critical point, suggesting the largest possible volume for the suitcase.

In this problem, using the second derivative test confirmed that \( h = 31 \) is indeed where the suitcase will achieve its maximal volume under the given constraints.