Question

The material for the bottom of a rectangular box costs three times as much per square foot as the material for the sides and top. Find the greatest volume that such a box can have if the total amount of money available for material is \(\$ 12\) and the material for the bottom costs \(\$ 0.60\) per square foot.

Short answer

Maximize volume by substituting variables using the cost constraint and solve for dimensions.

Step-by-step solution

Define Variables and Costs

Let's denote the dimensions of the box as length \( l \), width \( w \), and height \( h \). The cost of the bottom material is \( \\(0.60 \) per square foot, while the top and sides cost \( \\)0.20 \) per square foot. The bottom area is \( lw \), the top is also \( lw \), and the sides together are \( 2lh + 2wh \).

Formulate Total Cost Equation

The total cost to build the box is given by the cost of the bottom (\(0.60\times lw\)) plus the cost of the top and sides (\(0.20\times(2lh + 2wh + lw)\)). The total budget is \(\$12\). We have:\[0.60lw + 0.20(2lh + 2wh + lw) = 12\]

Simplify the Cost Equation

Expanding and simplifying, we obtain:\[0.60lw + 0.40lh + 0.40wh + 0.20lw = 12\]This simplifies to:\[0.80lw + 0.40lh + 0.40wh = 12\]

Maximize the Volume

The volume \(V\) of the box is \(lwh\). We need to maximize \(V = lwh\) under the constraint:\[0.80lw + 0.40lh + 0.40wh = 12\]

Use the Constraint to Express Variables

Solve the constraint for one variable, e.g., \(h\):\[h = \frac{12 - 0.80lw}{0.40(l+w)}\]

Substitute and Differentiate

Substitute \(h\) back into the volume expression. Differentiate \(V\) with respect to another variable (say \(w\)) and set the derivative to zero to find critical points.

Solve for Dimensions

After finding \(w\), compute corresponding \(l\) and \(h\) using the equations. Check constraints and ensure that all values are valid dimensions.

Calculate Maximum Volume

Using the dimensions found, calculate the maximum volume using the volume formula.

Key concepts

Cost Constraints

In optimization problems, especially ones involving physical objects like a rectangular box, constraints on cost play an essential role. For the given problem, cost constraints arise due to the limited budget of $12, which dictates how much material can be purchased.

- The bottom of the box uses a material that is thrice as costly as the sides and top. Therefore, the choice of materials and how they are used directly impacts the maximum dimensions of the box. - Understanding these constraints helps ensure that any solution is feasible within the given financial limits.
By formulating the total cost equation, combining the different costs, and aligning them with the budget, we can manage resources effectively. This step lays the groundwork for subsequent optimization analysis.

Maximizing Volume

Maximizing the volume of a structure under specific constraints is a classic optimization problem. Here, the objective is to find the greatest volume that the rectangular box can have, given the cost constraint.

- Volume, represented mathematically as \(V = lwh\), where \(l\), \(w\), and \(h\) are the length, width, and height of the box, acts as the primary function that needs optimization.- To achieve maximization, it's essential to derive a constraint equation from the cost conditions, allowing us to express one variable in terms of the others.
By solving the volume maximization, we identify the optimal dimensions that can be obtained without exceeding the budget.

Rectangular Box Problem

The rectangular box problem is a vivid application of optimization in calculus. It involves finding the ideal dimensions of a box that maximizes or minimizes certain characteristics, under given conditions.

- In this case, the problem requires balancing the budget with the desired box volume. - Critical thinking and strategic planning are necessary to define the way materials are distributed across the box while respecting the cost boundaries.
This type of problem highlights how an understanding of geometry and cost analysis leads to practical real-world solutions.

Differentiation in Optimization

Differentiation plays a crucial role in solving optimization problems. Once an expression for volume is derived considering the constraints, differentiation helps find the values of variables that optimize the function.

- After substituting one variable into the volume equation, differentiation with respect to another variable allows us to find points where the volume reaches its maximum or minimum. - By setting the derivative equal to zero, we locate critical points, which are then tested for validity within the constraints.
Differentiation in this problem thus not only simplifies the volume equation but also guides towards the most efficient configuration of the box.