Question

Rectangular boxes with a volume of \(10 \mathrm{m}^{3}\) are made of two materials. The material for the top and bottom of the box costs \(\$ 10 / \mathrm{m}^{2}\) and the material for the sides of the box costs \(\$ 1 / \mathrm{m}^{2}\). What are the dimensions of the box that minimize the cost of the box?

Short answer

Answer: The approximate dimensions of the rectangular box that minimize the cost are x = 0.829 m, y = 0.207 m, and z = 6.073 m.

Step-by-step solution

Define the Variables

Let x, y, and z be the length, width, and height of the box, respectively. Given that the volume of the box is 10 m³, we can write that: volume (V) = x * y * z = 10 m³ Additionally, the top and bottom of the box have an area of x * y each and cost \(10/m², and the four sides of the box have areas y * z, x * z, y * z, and x * z, each costing \)1/m².

Define the Cost Function

Now, we must express the cost of the box in terms of its dimensions. The cost of the top and bottom is 10(x * y), while the cost of the sides is (2x * z) + (2y * z). Thus, the total cost (C) can be represented as: C(x, y, z) = 10(x * y) + (2x * z) + (2y * z)

Convert the Cost Function to Two Variables

Since the volume is given as 10 m³, we can express z in terms of x and y, using the equation: z = 10 / (x * y) Substitute z into the cost function: C(x, y) = 10(x * y) + 2x(10 / (x * y)) + 2y(10 / (x * y))

Simplify the Cost Function

Simplify C(x, y): C(x, y) = 10(x * y) + (20 / y) + (20 / x)

Find the Partial Derivatives

To find the minimum cost, we need to find the critical points. To do this, we must find the first partial derivatives of C(x, y) with respect to x and y, and set them equal to zero: ∂C/∂x = 10*y - 20/(x²) = 0 ∂C/∂y = 10*x - 20/(y²) = 0

Solve the System of Equations

To find the critical points, solve the system of equations formed by the partial derivatives: 10*y = 20 / (x²) 10*x = 20 / (y²)

Find the Dimensions

Multiply both equations by x² * y² to eliminate the fractions: 20*x²*y² = 10*x²*y 20*x²*y² = 10*x*y² Divide both equations by 10x and 10y, respectively: 2*y² = x 2*x² = y Substitute the value of x from the second equation into the first: 2 * (2*y²)² = y Simplify and solve for y: 16*y^4 = y^3 16*y = 1 y ≈ 0.207 (approx.) x ≈ 0.829 (approx.) Then, z can be calculated using the volume equation: z = 10 / (x * y) z ≈ 6.073 (approx.)

Conclusion

The dimensions of the rectangular box that minimize the cost, while maintaining a volume of 10 m³, are approximately x = 0.829 m, y = 0.207 m, and z = 6.073 m.

Key concepts

Calculus

Calculus is a branch of mathematics that deals with rates of change and the accumulation of quantities. It includes concepts such as derivatives, integrals, limits, and infinite series. In this exercise, we are utilizing calculus to optimize the cost of constructing a box by determining the best dimensions.
When we talk about optimization in calculus, we refer to finding the maximum or minimum values of a function. This is key in many real-world applications, including minimizing costs, as shown here. The main tools we use in this context are derivatives and critical points.

Cost Function

A cost function represents the cost of producing a certain number of goods or, in this case, constructing a box. It is a mathematical expression of how various factors affect the total cost.

• The cost function in our problem is impacted by the areas of the box's surfaces, as the material costs are different for various surfaces.
• For our box, the cost function is expressed as: \[ C(x, y, z) = 10(x \cdot y) + (2x \cdot z) + (2y \cdot z) \]
• This function combines the cost of the top and bottom materials with that of the side materials.

By simplifying this cost function based on the volume constraint, we find a simpler function of two variables. This makes the problem easier to solve.

Partial Derivatives

Partial derivatives are a concept from calculus used to find the rate of change of a multivariable function with respect to one variable while keeping other variables constant. They are essential for finding critical points in optimization problems.
In this exercise, we calculate partial derivatives of the cost function: \[ \text{\( \frac{\partial C}{\partial x} = 10 \cdot y - \frac{20}{x^2} \)} \] \[ \text{\( \frac{\partial C}{\partial y} = 10 \cdot x - \frac{20}{y^2} \)} \]

• These derivatives help find where the cost function has a potential minimum, indicating the dimensions at which the cost is optimized.
• Solving these equations by setting them equal to zero leads us to the critical points.

Critical Points

Critical points are values of variables at which the first derivative of a function is zero or undefined, and they represent potential maxima or minima. In optimization problems, these points are the candidates for the locations where a function reaches an extreme value.
In our exercise, finding the critical points of the cost function is crucial in determining the optimum box dimensions to minimize cost.

• We set partial derivatives equal to zero to find critical points: \[ 10 \cdot y - \frac{20}{x^2} = 0 \] \[ 10 \cdot x - \frac{20}{y^2} = 0 \]
• By solving these equations, we find the optimal values of \( x \) and \( y \) that provide the lowest construction cost under the given volume constraint.

Once the critical points are found, checking them ensures that it corresponds to a minimum and not a maximum or a saddle point, which ultimately confirms the solution to the problem.