Question

A shipping company handles rectangular boxes provided the sum of the height and the girth of the box does not exceed 96 in. (The girth is the perimeter of the smallest side of the box.) Find the dimensions of the box that meets this condition and has the largest volume.

Short answer

Answer: The dimensions of the box that maximize the volume are approximately 13.5 inches in length, 10.5 inches in width, and 22 inches in height.

Step-by-step solution

Define the variables

Let the length, width, and height of the rectangular box be denoted by l, w, and h, respectively.

Write the constraint equation

The constraint given is that the sum of the height and the girth does not exceed 96 in, which can be written as: h + 2(w + l) <= 96

Write the volume equation

The volume V of a rectangular box is given by the product of its length, width, and height. V = lwh

Solve for one variable in the constraint equation

From the constraint equation, we can write the equality for the upper limit of the sum (height and girth): h + 2(w + l) = 96 Solve for h: h = 96 - 2(w + l)

Substitute the expression for h in the volume equation

Replace h in the equation for V with the expression found in step 4: V(l, w) = lw(96-2(w+l))

Differentiate the volume equation with respect to length

To find the critical points, we can differentiate the volume equation V with respect to l. \[ \frac{dV}{dl} = w(96 - 4w - 4l) \]

Find the critical points

To find the critical points, set the derivative equal to zero and solve for w: \[ w(96 - 4w - 4l) = 0 \] Either \(w = 0\) or \(96 - 4w - 4l = 0\). When \(w = 0\), the volume is zero, which is not considered a valid solution. So, we need to focus on the second equation: \[96 - 4w - 4l = 0 \] Now, solve the equation for w: \[ w = 24 - l \]

Substitute the expression of w in the volume equation

Now, substitute the expression for w found in step 7 into the volume equation: \[ V(l) = l(24 - l)(96 - 2(24 - l) - 2l) \] Next, differentiate the volume equation V with respect to l. \[ \frac{dV}{dl} = (24 - l)(192 - 6l) - l(192 - 8l) \] To find the critical points, set the derivative equal to zero: \[ \frac{dV}{dl} = 0\]

Calculate the dimensions of the box

Solve the equation found in step 8 for l. Note that this requires numerical methods or using a calculator. After solving, we have l ≈ 13.5 in. Next, find the value of w using the expression found in step 7: \[w = 24 - l \] w ≈ 10.5 in. Finally, using the expression found in step 4: \[h = 96 - 2(w + l)\] h ≈ 22 in. So, the dimensions of the box that maximize the volume are: l ≈ 13.5 in, w ≈ 10.5 in, and h ≈ 22 in.

Key concepts

Constraint Optimization

Constraint optimization is a fundamental concept in calculus, particularly useful when dealing with real-world optimization problems where certain conditions must be met. This involves finding the maximum or minimum values of a function within a given set of constraints or boundaries.

For example, in the exercise provided, the shipping company's constraint is that the sum of the height and the girth of the box does not exceed 96 inches. This is the 'constraint equation' that must be satisfied. The art of constraint optimization is to express this equation in terms of the variables of interest—in this case, the dimensions of the box—and to incorporate it into the optimization process for the function we wish to maximize or minimize, which is the volume of the box in this exercise.

By setting up a constraint equation and substituting it into the volume function, one effectively reduces the number of variables, simplifying the problem. This allows for the application of methods to find critical points and determine the optimal values within the specified limits.

Volume Maximization

Volume maximization is a specific type of optimization problem in calculus that aims to find the largest possible volume of a geometric figure given some restrictions. The process involves modeling the volume as a mathematical function of its dimensions and then using calculus techniques to find the maximum value of this function.

In the context of our exercise, the objective is to maximize the volume V, expressed by the equation V = lwh, under the constraint that the sum of the box's height and girth is less than or equal to 96 inches. After manipulating the constraint to express one of the dimensions in terms of the others, we substitute this expression into the volume equation. Then, we differentiate the resulting function with respect to one variable and solve for critical points, which could potentially represent the maximum volume. Differentiation and solving for critical points are crucial steps in volume maximization problems.

Critical Points in Calculus

Critical points in calculus are locations on the graph of a function where its derivative is zero or undefined. These points are essential in identifying relative maxima, minima, and saddle points of the function. In optimization tasks, such as finding the box with the largest possible volume, we look for critical points since they can indicate where the volume is maximized.

To find these critical points for a volume function, we first differentiate the volume with respect to one of its variables and set the derivative equal to zero. This provides us with one or more critical points which we then analyze to identify which of these yield the maximum volume. In the given exercise, we differentiate V with respect to length, set the derivative equal to zero, and then solve the resulting equation to obtain the critical length l. Subsequently, we use this critical length to find the corresponding width w and height h, which will ultimately give us the dimensions for the box with the largest volume within the constraint.