Problem 10 - Thomas Calculus

Question

Catching rainwater \(A 1125 \mathrm{ft}^{3}\) open-top rectangular tank with a square base \(x\) ft on a side and \(y\) ft deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product \(x y .\) a. If the total cost is $$ c=5\left(x^{2}+4 x y\right)+10 x y $$ what values of \(x\) and \(y\) will minimize it? b. Give a possible scenario for the cost function in part (a).

Step-by-Step Solution

Answer

x = 15 ft, y = 5 ft. Cost function involves material and excavation costs.

Step 1: Understand the Problem

We need to minimize the cost \( c \) of constructing a tank with given volume, which depends on the dimensions \( x \) and \( y \). The formula for cost is \( c=5\left(x^{2}+4xy\right)+10xy \). The tank has a volume of \( 1125 \) cubic feet, given by \( x^2y = 1125 \).

Step 2: Set Up Constraints

Use the constraint from the problem: the volume of the tank is \( x^2y = 1125 \). Solve for one variable in terms of the other. For example, \( y = \frac{1125}{x^2} \).

Step 3: Substitute the Constraint into the Cost Function

Substitute \( y = \frac{1125}{x^2} \) into the cost equation: \[ c = 5(x^2 + 4x \cdot \frac{1125}{x^2}) + 10x \cdot \frac{1125}{x^2} \]. Simplifying this, we get: \[ c = 5x^2 + \frac{22500}{x} + \frac{11250}{x} \].

Step 4: Simplify the Cost Function

Combine like terms in the cost function: \( c = 5x^2 + \frac{33750}{x} \). Now, this is the simplified form of the cost function in terms of \( x \).

Step 5: Find Critical Points

To find the value of \( x \) that minimizes \( c \), take the derivative with respect to \( x \) and set it to zero. The derivative is \( \frac{dc}{dx} = 10x - \frac{33750}{x^2} \). Setting it to zero gives \( 10x = \frac{33750}{x^2} \).

Step 6: Solve for x

Solve \( 10x = \frac{33750}{x^2} \). Multiply both sides by \( x^2 \): \( 10x^3 = 33750 \). Solve for \( x \): \( x^3 = \frac{33750}{10} = 3375 \). So, \( x = \sqrt[3]{3375} = 15 \).

Step 7: Solve for y

Substitute \( x = 15 \) back into the volume constraint \( y = \frac{1125}{15^2} = 5 \). Thus, \( y = 5 \).

Step 8: Conclusion and Cost Function Context

The values of \( x \) and \( y \) that minimize the cost are \( x = 15 \) and \( y = 5 \). A possible scenario for the cost function: material costs \( 5\left(x^2 + 4xy\right) \) might include the base and sides, whereas excavation costs \( 10xy \) relate to the volume of earth moved.

Key Concepts

Cost Minimization

In optimization problems, particularly in calculus, cost minimization focuses on finding the most efficient way of achieving an objective at the lowest cost. For the problem of constructing a rainwater tank, we need to minimize the cost function given by the equation:
\( c = 5(x^2 + 4xy) + 10xy \).
This equation considers both the material required for the sides and base of the tank and the costs related to excavation. To minimize this cost effectively:

• Recognize that the cost function is influenced by variables related to the tank dimensions: \(x\) (the square base side) and \(y\) (depth).
• Use calculus tools, particularly derivatives, to find the values of these variables where the cost is minimized.

In this problem, once these variables are identified correctly, the total cost of the project can be reduced significantly.

Constraint Handling

Constraint handling is an essential skill in optimization as it helps to manage limitations or restrictions in a problem. In the context of this exercise, the tank's volume of 1125 cubic feet acts as a constraint given by:
\( x^2y = 1125 \).
To handle this constraint:

• Express one variable in terms of the other. For instance, solve for \(y\) as \( y = \frac{1125}{x^2} \).
• Substitute this expression into the cost function to eliminate one variable, simplifying the problem.

By using substitution, the complexity of the problem reduces, allowing for easier calculation of the optimal dimensions that satisfy both the volume constraint and cost minimization.

Derivatives for Extreme Values

Derivatives play a critical role in identifying extreme values, such as maxima or minima, in functions. In the context of this optimization problem, derivatives are used to pinpoint the base measurement \(x\) that minimizes cost.
To achieve this:

• Take the derivative of the cost function with respect to \(x\) to find the critical points: \( \frac{dc}{dx} = 10x - \frac{33750}{x^2} \).
• Set the derivative equal to zero to find potential minima: \( 10x = \frac{33750}{x^2} \).
• Solve this equation to determine the value of \(x\) that minimizes cost, which in this case yields \(x = 15\).

Calculating derivatives efficiently can lead you directly to the solution of finding the minimum cost configuration by identifying the points at which the cost function changes its increasing or decreasing trend.