Problem 21 - Thomas Calculus in SI Units

Question

a. Suppose that instead of having a box with square ends you have a box with square sides so that its dimensions are \(h\) by \(h\) by \(w\) and the girth is \(2 h+2 w\). What dimensions will give the box its largest volume now? b. Graph the volume as a function of \(h\) and compare what you see with your answer in part (a).

Step-by-Step Solution

Answer

The largest volume occurs at \( h = \frac{G}{3}, w = \frac{G}{6} \). The graph confirms this maximal point.

Step 1: Write the Volume Formula

For a box with square sides, the volume is given by the formula:\[ V = h^2 \times w \]where \( h \) is the height and width of the square sides, and \( w \) is the length of the box. The goal is to maximize the volume \( V \).

Step 2: Use Constraint for Girth

The girth is given by the sum of all side lengths excluding one of the width sides. Hence,\[ 2h + 2w = G \]where \( G \) is the specified girth. For simplicity, explore the case where \( G = 100 \), deriving \( w \) in terms of \( h \).\[ w = \frac{G - 2h}{2} \]Substitute this into the volume formula.

Step 3: Substitute Constraint into Volume

Substitute \( w \) from the previous equation into the volume formula:\[ V = h^2 \left( \frac{G - 2h}{2} \right) \]\[ V = \frac{h^2 \times (G - 2h)}{2} \]

Step 4: Derive with Respect to h

To maximize volume, take the derivative of \( V \) with respect to \( h \),\[ \frac{dV}{dh} = \frac{d}{dh} \left( \frac{h^2(G - 2h)}{2} \right) \]Using the product rule and simplifying gives:\[ \frac{dV}{dh} = \frac{G \cdot h - 3h^2}{2} \]

Step 5: Solve for Critical Points

Set the derivative equal to zero to find critical points:\[ \frac{G \cdot h - 3h^2}{2} = 0 \]Simplify and solve for \( h \):\[ G \cdot h - 3h^2 = 0 \]\[ h(G - 3h) = 0 \]This gives critical points \( h = 0 \) or \( h = \frac{G}{3} \).

Step 6: Evaluate Maximum Volume for h = G/3

Substitute \( h = \frac{G}{3} \) back into the volume equation:\[ w = \frac{G - 2\left(\frac{G}{3}\right)}{2} = \frac{G}{6} \]Thus, \( V = \left(\frac{G}{3}\right)^2 \times \frac{G}{6} = \frac{G^3}{54} \).

Step 7: Graph the Volume Function

Plot \( V(h) = \frac{h^2(G - 2h)}{2} \) using a graphing tool. Observe the peak in the graph, which should correspond with the critical point solution \( h = \frac{G}{3} \). The graph will confirm the calculation from the previous steps.

Key Concepts

Volume Maximization

In this optimization problem, our ultimate goal is to find the dimensions that maximize the volume of a box with square sides. The volume, represented by the formula \( V = h^2 \times w \), depends on two variables: \( h \) (the side of the square) and \( w \) (the length of the box). But we're not simply looking at varying all dimensions arbitrarily; instead, these dimensions must comply with a girth constraint.

To achieve volume maximization, we substitute \( w \) in terms of \( h \) using the girth constraint into the volume formula. This forms a new function of a single variable \( h \). Then, by taking the derivative with respect to \( h \) and setting it to zero, we find the critical points where the volume could be maximized. This step is essential because it pinpoints where the function's slope is flat, implying possible maximum volume.

Square Dimensions

In the given problem, the square dimensions refer to the fact that two sides of our box are equal, \( h \times h \). This simplification is vital when calculating the volume because it reduces the number of variables from three to two. So instead of dealing with width, height, and length independently, we account for just \( h \) and \( w \).

This symmetry makes the calculus easier to handle and allows us to focus on maximizing the box’s capacity by changing just one measurement of the square, \( h \), given the girth constraint. Square dimensions simplify the volume function and make it easier to derive and analyze.

Girth Constraint

The girth constraint is critical as it limits the possible dimensions of the box. Defined as \( 2h + 2w = G \), where \( G \) is the constant girth, it essentially binds \( h \) and \( w \) together.

Using this constraint, we can express \( w \) in terms of \( h \) as \( w = \frac{G - 2h}{2} \). By substituting this relationship into the volume formula, we tailor the formula according to the given conditions; allowing us eventually to set up a function dependent solely on \( h \). This resulting function can then be explored and maximized within the constraints to find the optimal solution.

Critical Points in Calculus

The concept of critical points is a cornerstone of calculus optimization problems, as it allows us to find where a function reaches its maximum or minimum point. In this exercise, after expressing the volume function in terms of only \( h \), the derivative \( \frac{dV}{dh} \) is calculated.

The critical points are identified by setting this derivative equal to zero: \( \frac{G \cdot h - 3h^2}{2} = 0 \). Solving this equation, we find possible values of \( h \) which maximize the volume. Testing these critical points \( h = 0 \) and \( h = \frac{G}{3} \), allows us to determine the dimensions that provide the largest volume.

The graphical representation of the volume function further supports these findings, illustrating where maximum values occur and validating our analytical solution.