Problem 57 - University Calculus: Early Transcendentals

Question

You are to construct an open rectangular box with a square base and a volume of \(48 \mathrm{ft}^{3} .\) If material for the bottom costs \(\$ 6 / \mathrm{ft}^{2}\) and material for the sides costs \(\$ 4 / \mathrm{ft}^{2},\) what dimensions will result in the least expensive box? What is the minimum cost?

Step-by-Step Solution

Answer

Optimal dimensions: 4 ft base, 3 ft height; Minimum cost: $288.

Step 1: Understand the Problem

We need to construct an open box with a square base, targeting a specific volume, and find dimensions that minimize the cost given the price per square foot of materials.

Step 2: Define Variables

Let the side length of the square base be \(x\) and the height of the box be \(h\). The volume of the box is given by \(x^2h = 48\).

Step 3: Express Height in Terms of Base Side Length

Using the volume equation \(x^2h = 48\), we solve for \(h\), giving \(h = \frac{48}{x^2}\).

Step 4: Determine Cost Functions

The cost of the bottom is \(6x^2\), and the cost for the four sides (each \(xh\)) is \(4(4xh) = 16xh\). Substitute \(h\) to get total cost \(C(x) = 6x^2 + 16x\left(\frac{48}{x^2}\right) = 6x^2 + \frac{768}{x}\).

Step 5: Optimize the Cost Function

Find the derivative of the cost function \(C(x)\) with respect to \(x\): \(C'(x) = 12x - \frac{768}{x^2}\). Set \(C'(x) = 0\) to find critical points: \(12x = \frac{768}{x^2}\).

Step 6: Solve for Critical Point

Multiply by \(x^2\): \(12x^3 = 768\). Simplify to find \(x^3 = 64\), resulting in \(x = 4\).

Step 7: Determine Height using Critical Point

Substitute \(x = 4\) into the height equation \(h = \frac{48}{x^2}\): \(h = \frac{48}{16} = 3\).

Step 8: Calculate Minimum Cost

Compute the cost using \(x = 4\) and \(h = 3\). Bottom cost is \(6(4^2) = 96\) and side cost is \(16 \cdot 4 \cdot 3 = 192\). Total cost is \(96 + 192 = 288\).

Step 9: Confirm Minimum by Second Derivative Test

Calculate \(C''(x) = 12 + \frac{2\cdot 768}{x^3}\). At \(x=4\), \(C''(4) > 0\), confirming a local minimum.

Key Concepts

Calculus

In optimization problems, calculus plays a crucial role by helping us find points on a curve where the value might be the lowest or highest, known as optimization.
This is where derivatives come in.

The derivative, often denoted as \(f'(x)\), measures how a function changes as its input changes. It's essentially finding the slope, or steepness, of the function at any given point.

• For cost minimization, we aim to find the point where the slope of the cost function is zero: the critical point.
• This critical point indicates a flat region, often a minimum or maximum.
• By setting the derivative \(C'(x)\) to zero, we solve for the critical point, which leads to the most economical dimensions of the box.

Once the critical points are found, the second derivative test helps confirm their nature. A positive result from the second derivative, \(C''(x)\), demonstrates a local minimum because it implies the curve smiles like a parabola at that point.
These concepts underline the importance of calculus in handling such optimization problems.

Cost minimization

Cost minimization is the main objective when constructing the box.
We are provided with different costs for the base and sides. This requires us to find a balance between dimensions and cost to use the least amount of funds.

• The base of the box has a cost of \(\\(6/\text{ft}^2\).
• The sides have a cost of \(\\)4/\text{ft}^2\). Each side needs to be built considering the height and base of the box.

The formula derived for the total cost, \(C(x) = 6x^2 + \frac{768}{x}\), incorporates these costs, and by optimizing it, we determine the most cost-effective dimensions.

After finding the critical point by setting \(C'(x) = 0\), using \(x = 4\) for the base side length, we simplify the costs:

• The base square costs \(6(4^2) = 96\) dollars.
• The four sides summed up cost \(16 \times 4 \times 3 = 192\) dollars.
• The total optimized cost is \(288\) dollars.

By considering both material costs separately and minimizing their sum, we ensure the overall expense is the lowest possible for the given constraints.

Volume constraints

Volume constraints are vital because we need our box to hold an exact volume of \(48 \text{ft}^3\).
This constraint directly influences the permissible dimensions of the box. Defined by the equation \(x^2h = 48\), this keeps the dimensions balanced while maintaining the required capacity.

• The volume constraint helps us express the height \(h\) in terms of the base side length \(x\): \(h = \frac{48}{x^2}\).
• Any dimension chosen should apply this constraint to ensure the box doesn't exceed or fail to meet the required volume.

By plugging back the determined \(x\) value from the cost optimization, \(x = 4\), into the volume equation, we solve for the height \(h\):

• Calculation yields \(h = \frac{48}{4^2} = 3\).

This ensures that while we minimize costs, we also respect the box's fundamental requirement of holding exactly \(48 \text{ft}^3\) efficiently. Compliance with volume constraints is key in optimization problems to maintain functionality while achieving minimum cost.