Problem 43 - College Algebra and Calculus: An Applied Approach

Question

Find the dimensions of a rectangular package of maximum volume that may be sent by a shipping company assuming that the sum of the length and the girth (perimeter of a cross section) cannot exceed 96 inches.

Step-by-Step Solution

Answer

The dimensions that will give the maximum volume for the rectangular package are: Length = 48 inches, Width = 16 inches and Height = 16 inches

Step 1: Convert the function of three variables to one variable

Let's convert the problem into a function of one variable by solving the constraint equation for L: \[ L = 96 - 2W - 2H \]

Step 2: Substitute the expression for L into the volume function

Substitute this result into the volume function: \[ V = (96 - 2W - 2H)*W*H = 96WH - 2W^2H - 2WH^2 \]

Step 3: Differentiate the volume function

We can't resolve the equation for maximum volume as it stands, because it has two variables, W and H. Let's assume a square cross section, so W = H. The volume function becomes \[ V = 96W^2 - 4W^3 \] Differentiate this function with respect to W: \[ V' = 192W - 12W^2 \]

Step 4: Solve V' = 0 to find the critical points

To find the critical points of the volume function, solve the equation \[ V' = 192W - 12W^2 = 0 \] This gives us W = 0 or W = 16. 0 is a trivial solution that will give us a zero volume, and so our possible solution is W=H=16

Step 5: Substitute W=16 into the constraint equation to find L

Substitute W=16 into the equation from Step 1 to find L: \[ L = 96 - 2W - 2H = 96 - 2*16 - 2*16 = 48\]

Key Concepts

Rectangular Package Dimensions

Maximizing the volume of a rectangular package, while adhering to shipping constraints, is a practical example of an applied mathematics problem. For students and professionals alike, it's essential to understand how to find the optimal dimensions that utilize the maximum allowed space.

Consider a scenario where a shipping company limits the combined length and girth of a package to 96 inches. The girth is the perimeter of the cross-section, which for a rectangular package means twice its width plus twice its height. Gaining a thorough understanding of these dimensions and how they relate to each other under given constraints can aid in a variety of real-world applications from packaging to construction and beyond.

To make these concepts easier to visualize, consider a box where you measure the girth around its width and height. The sum of this measure plus the length of the box cannot exceed the shipping limit. It's about finding a balance between all sides of the box to ensure not a single inch goes to waste.

Applied Calculus

Calculus is often perceived as an abstract field of mathematics, but its principles are crucial in solving practical optimization problems. Applied calculus involves using differential calculus to find the maximum or minimum values of functions pertaining to real-world scenarios. In this case, we're using calculus to maximize the volume of a rectangular package within certain constraints.

By formulating the problem as a mathematical function, calculus enables us to translate physical restrictions into equations we can differentiate and solve. Taking the derivative of the volume equation with respect to a variable allows us to locate the maximum volume efficiently rather than testing endless combinations of dimensions, which highlights the power and utility of calculus in designing and decision-making processes.

Constrained Optimization

Constrained optimization is a fundamental concept in calculus where we maximize or minimize a function subject to certain conditions. In the textbook problem, the shipping constraint acts as the limit within which the package dimensions must be optimized.

It involves expressing one or more constraints in equation form and incorporating them into our function in a way that reduces the number of variables. By applying this technique, complicated problems become more manageable, allowing for solving them with ordinary differential calculus methods. For practical circumstances such as packaging, this approach is invaluable for creating solutions that adhere to regulatory or physical limits. In essence, constrained optimization is the balancing act of achieving the best possible outcome under a set of given rules or limits.

Differentiation Applications

Differentiation, a core operation in calculus, is widely used for finding the rate at which something changes. When applied to optimization problems, differentiation helps in determining the points at which the quantity we're optimizing (like volume) reaches its maximum or minimum value.

In the given problem, differentiating the volume function with respect to the width and then setting the derivative equal to zero allows us to find the critical points. Critical points indicate where the function's slope changes, which, in optimization problems, can signal an optimum. By comprehensively understanding how to apply differentiation to practical problems, students can unlock the potential to solve complex issues in engineering, economics, physics, and beyond, turning mathematical concepts into tools for innovation and efficiency.