Problem 315 - Calculus Volume 1

Question

For the following exercises, 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 at height \( h = 31 \), maximizing box dimensions.

Step 1: Define the Variables and Constraints

Given that the total of the length, width, and height must not exceed 62 inches, we set \[ l + w + h \leq 62 \] where \( l \) is the length, \( w \) is the width, and \( h \) is the height. Assuming the height \( h \) is fixed, we will explore how to maximize the volume \( V = l \times w \times h \).

Step 2: Express Dimensions in Terms of Height

To fit the constraint, express the dimensions \( l \) and \( w \) in terms of \( h \): \[ l + w = 62 - h \]. Since the problem asks for an expression for the maximum volume as a function of \( h \), set \( l=w \), then:\[ 2l = 62 - h \] \[ l = \frac{62 - h}{2} \] Thus, the width \( w \) can also be expressed as \( \frac{62 - h}{2} \).

Step 3: Formulate the Volume Equation

Substitute the expressions for \( l \) and \( w \) into the volume formula to get:\[ V = l \times w \times h = \left( \frac{62 - h}{2} \right)^2 \times h \]Which simplifies to:\[ V = h \left( \frac{62 - h}{2} \right)^2 \]

Step 4: Simplify the Volume Expression

The expression can be simplified further:\[ V = h \left( \frac{62 - h}{2} \right)^2 \]Expanding the term \( \left( \frac{62 - h}{2} \right)^2 \) gives:\[ V = h \cdot \frac{(62 - h)^2}{4} \]\[ V = \frac{h}{4} \cdot (62 - h)^2 \]Simplifying, this is equivalent to:\[ V = h \left(31 - \frac{1}{2}h \right)^2 \]

Step 5: Find the Derivative to Maximize the Volume

To find the maximum volume, take the derivative of the volume function with respect to \( h \):\[ V'(h) = \left(31 - \frac{1}{2}h \right)^2 + 2h(-\frac{1}{2})(31-\frac{1}{2}h) \]\[ V'(h) = 3(31 - \frac{1}{2}h)(1-h) \]

Step 6: Solve for Critical Points

Set the first derivative to zero to find the critical points:\[ 3(31 - \frac{1}{2}h)(1-h) = 0 \]Solving the equation for \( h \), we get:\[ 31 - \frac{1}{2} h = 0 \quad or \quad 1 - h = 0 \]Thus, \( h = 62 \) or \( h = 1 \). Since the height cannot be 62 (no room for other dimensions), choose logical solution \( h eq 1 \) as feasible.

Key Concepts

Maximum Volume Calculation

Optimization is all about finding the best solution within set constraints. In this exercise, we aim to maximize the volume of a suitcase given its dimensional constraint: the sum of length, width, and height should not exceed 62 inches. Knowing this, our first task is to express the suitcase's volume in terms of one variable.

Given that the suitcase height, denoted as \( h \), is fixed, an efficient approach is to express both length \( l \) and width \( w \) in terms of height. The equation \( l + w + h \leq 62 \) ensures that these dimensions fit the air travel regulation.

• If we set \( l = w \), both dimensions become equal parts of the remaining allowance after accounting for \( h \): \( l + w = 62 - h \).
• Thus, each dimension is \( \frac{62 - h}{2} \), and the volume formula transforms to \( V = l \cdot w \cdot h = \left(\frac{62-h}{2}\right)^2 \cdot h \).

So, the expression for volume becomes \( V = h\left(31-\frac{1}{2}h\right)^2 \). This calculation sets the stage for finding the maximum volume the suitcase can have, based on varying height within constraints.

Derivatives in Optimization

Derivatives are powerful tools in calculus, extensively used for optimization problems such as this one. They help us find where functions reach their maximum or minimum values. Here, our goal is to maximize the suitcase volume by utilizing the derivative of the volume expression.

• To start, derive the volume function \( V(h) = h\left(31-\frac{1}{2}h\right)^2 \) with respect to \( h \).
• This derivative indicates how volume changes as the height changes.

The derived function is \( V'(h) = \left(31-\frac{1}{2}h\right)^2 + 2h\left(-\frac{1}{2}\right)(31-\frac{1}{2}h) \). Here, mathematical manipulation assists in finding how shape changes impact volume.

By understanding this derivative, we can locate points where the slope is zero, indicating potential maxima or minima. Taking the derivative is a crucial step toward understanding not just *where* the maximum volume occurs, but *why* it occurs at that height.

Critical Points Analysis

Critical points are where an important part of the optimization journey occurs. These are the spots in the derivative function where the slope is zero or undefined, offering clues about optimal dimensions.

• Setting \( V'(h) \) to zero, we find the critical points: \( 3(31-\frac{1}{2}h)(1-h) = 0 \).
• This breaks down into two conditions: either \( 31 - \frac{1}{2}h = 0 \) or \( 1-h = 0 \).

Solving these gives potential heights \( h = 62 \) and \( h = 1 \).

We dismiss \( h = 62 \) as infeasible because it leaves no room for length and width (a height alone can't form a suitcase!). Therefore, \( h = 1 \) is a sensible choice. It implies the largest volume occurs when height remains minimal but accommodates both width and length efficiently. Critical points guide us to such practical insights, linking calculus to real-world solutions.