Problem 21 - Calculus Early Transcendentals

Question

An open box is to be made from a \(3 \mathrm{ft}\) by 8 ft rectangular piece of sheet metal by cutting out squares of equal size from the four corners and bending up the sides. Find the maximum volume that the box can have.

Step-by-Step Solution

Answer

The maximum volume of the box is approximately 11.85 cubic feet.

Step 1: Define the Variables

Let the side length of the squares cut out from each corner be \(x\) feet. After the squares are cut and the sides are folded up, the dimensions of the box become \((3-2x)\) feet in width, \((8-2x)\) feet in length, and \(x\) feet in height.

Step 2: Write the Volume Function

The volume of the box \(V\) can be expressed as a function of \(x\): \[ V(x) = (3-2x)(8-2x)x \].On expanding this, we get the volume function:\[ V(x) = 4x^3 - 22x^2 + 24x \].

Step 3: Determine the Domain

The domain of \(x\) is determined by the physical constraints that the length and width must be positive after cutting, which means:\(0 < x < \frac{3}{2}\).Hence, the domain of \(x\) is \(0 < x < 1.5\).

Step 4: Find the Critical Points

To find the critical points, take the derivative of \(V(x)\):\[ V'(x) = 12x^2 - 44x + 24 \].Set \(V'(x) = 0\) to solve for critical points:\[ 12x^2 - 44x + 24 = 0 \].This can be factored as:\[ (6x - 4)(2x - 6) = 0 \], leading to solutions \(x = \frac{2}{3}, 2\). Given the domain constraint, only \(x = \frac{2}{3}\) is within the domain.

Step 5: Evaluate the Volume at Critical Points and Boundaries

Evaluate \(V(x)\) at \(x = \frac{2}{3}\) and the boundary values \(x = 0\) and \(x = 1.5\).- \(V\left(0\right) = 0\)- \(V\left(\frac{2}{3}\right) = 4\left(\frac{2}{3}\right)^3 - 22\left(\frac{2}{3}\right)^2 + 24\left(\frac{2}{3}\right) = \frac{320}{27} \approx 11.85 \, \text{ft}^3\)- \(V\left(1.5\right) = 0\)

Step 6: Conclude with the Maximum Volume

From the evaluations, the maximum volume occurs at \(x = \frac{2}{3}\) with the volume approximately \(11.85 \, \text{ft}^3\).

Key Concepts

Volume Optimization

Volume optimization is about finding the maximum or minimum volume of an object. In the given problem, we aim to determine the size of squares to cut from the corners of a rectangular sheet to form a box with maximum volume. To start, we express the volume of the box as a function of variable size cuts. Here, after cutting squares of side length \(x\) feet from each corner, the box's volume \(V(x)\) can be expressed as:

• The width of the box becomes \(3 - 2x\) feet
• The length becomes \(8 - 2x\) feet
• The height is \(x\) feet

Thus, the volume \(V\) is given by:\[ V(x) = (3-2x)(8-2x)x \] Expanding this helps us better handle and optimize the function. The goal is to find the value of \(x\) that maximizes this volume. With \(x = \frac{2}{3}\) leading to the maximum volume, we are using this as a practical example of volume optimization.

Critical Points

Critical points are potential locations where a function reaches a maximum or minimum value. Determining these points involves finding where the derivative of the function is zero—indicating a slope of zero, or flatness, at that point. For this problem, the function \(V(x)\) represents the volume of the box, and we compute its derivative to locate where the maximum volume might occur.We calculate the derivative \(V'(x)\) to explore where the rate of change of volume is zero:\[ V'(x) = 12x^2 - 44x + 24 \] Setting \(V'(x) = 0\), we solve the equation:\[ 12x^2 - 44x + 24 = 0 \] Factoring yields solutions \(x = \frac{2}{3}, 2\). However, with constraints (considering physical properties of the box), only \(x = \frac{2}{3}\) lies within the feasible domain. Evaluating the function at this critical point confirms it is a maximum point for volume.

Derivatives

Derivatives help us understand how functions change. They are crucial in optimization problems because they facilitate finding critical points—where function values may be maximized or minimized.For our box volume, \(V(x)\), the derivative \(V'(x)\) was derived from:\[ V(x) = 4x^3 - 22x^2 + 24x \] Taking the derivative gives us:\[ V'(x) = 12x^2 - 44x + 24 \] The derivative tells us the rate at which the volume changes as we tweak \(x\). Setting the derivative to zero lets us solve for critical points, essentially finding spots where the volume plateaus, and potentially peaks or valleys.

Domain and Range

In optimization, the domain and range define the scope within which the problem operates. The domain refers to the set of possible values for \(x\), here constrained by the physical limits of constructing the box. From the problem, constraint conditions give us:

• Width: \(3 - 2x > 0\)
• Length: \(8 - 2x > 0\)

These inequalities imply the domain for \(x\) is:\[ 0 < x < \frac{3}{2} \]The range, however, is the potential outputs from the volume function within this domain, specifically how the volume changes as \(x\) varies between 0 and 1.5. Understanding domain and range ensures we limit our problem-solving to realistic values that maintain the structural integrity of the box.