Problem 10 - University Calculus: Early Transcendentals

Question

Catching rainwater \(\quad\) 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 tope 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

To minimize cost, use \( x \approx 13.35 \) and \( y \approx 6.31 \). The cost function represents material and excavation costs.

Step 1: Express Volume Condition

We know the volume of the tank must be 1125 ft³. With a square base, the volume can be represented as:\[ V = x^2y = 1125 \]Rearranging for \( y \) gives:\[ y = \frac{1125}{x^2} \]

Step 2: Substitute Volume in Cost Function

Now substitute \( y = \frac{1125}{x^2} \) into the cost function \( c \):\[ c = 5(x^2 + 4xy) + 10xy \]Substitute for \( y \):\[ c = 5(x^2 + 4x\left(\frac{1125}{x^2}\right)) + 10x\left(\frac{1125}{x^2}\right) \]Simplifying it results in:\[ c = 5x^2 + \frac{22500}{x} \]

Step 3: Differentiate Cost Function

To find the minimum cost, differentiate the cost function with respect to \( x \):\[ \frac{dc}{dx} = 10x - \frac{22500}{x^2} \]

Step 4: Solve for Critical Points

Set the derivative equal to zero to find critical points:\[ 10x - \frac{22500}{x^2} = 0 \]Multiplying the equation by \( x^2 \) gives:\[ 10x^3 = 22500 \]Solving for \( x \):\[ x^3 = 2250 \]\[ x = \sqrt[3]{2250} \approx 13.35 \]

Step 5: Determine Value of y

Substituting \( x = 13.35 \) in \( y = \frac{1125}{x^2} \):\[ y = \frac{1125}{(13.35)^2} \approx 6.31 \]

Step 6: Interpret Scenario of Cost Function

The cost function includes terms related to material costs and an excavation charge. The term \(5x^2\) represents the cost of the base, which depends on the area of the square. The terms involving \(4xy\) and \(10xy\) can reflect the costs of the sides and excavation based on the perimeter and depth, respectively. This function combines material cost and excavation in a proportional manner with depth \(y\).

Key Concepts

Cost Function

In optimization problems, a **cost function** is a mathematical expression that describes the total cost associated with a particular scenario. It can include variables dependent on different aspects of the system being considered.
For this specific exercise, the cost function is given by:\[ c = 5(x^2 + 4xy) + 10xy \]This equation breaks down into several components:

• 5\(x^2\): Represents the base material cost. Since the tank has a square base, this term calculates the cost in terms of how much material is needed for the base.
• 4xy and 10xy: These terms reflect the costs associated with the sides of the tank and excavation work, respectively. Here, the cost is tied directly to both the perimeter impacted by dimension \(x\) and depth \(y\).

Understanding how each part of the cost function is derived helps break down the total expenses, which is essential for finding the optimal dimensions that will minimize the cost.

Calculus

**Calculus** is a branch of mathematics that is used extensively in optimization problems to find both maximum and minimum values of functions. In this exercise, we apply calculus to minimize the cost function.
The first step involves creating a function for the cost with a given constraint, in this case, the tank's volume. Through calculus, we can rearrange terms and make substitutions to analyze the system's behavior at specific points.
Key techniques used here:

• **Differentiation**: This process involves calculating the derivative of the cost function in terms of \(x\). Differentiation helps us determine how the function changes, which is crucial for finding extremum points like minima or maxima.
• **Critical Points Detection**: By setting the derivative of the cost function to zero, we can find potential points at which the function's rate of change becomes zero, which may correspond to the minimum cost.

Calculus allows for both the precise calculation of these points and understanding how various aspects of the function contribute to either increases or decreases in cost.

Critical Points

**Critical points** are specific values in a function where the derivative is zero or undefined. These points are crucial in optimization as they often indicate local maxima, minima, or points of inflection.
In the exercise:

• The derivative of the cost function with respect to \(x\) is calculated as:

\[ \frac{dc}{dx} = 10x - \frac{22500}{x^2} \]

• Setting the derivative to zero, \(10x - \frac{22500}{x^2} = 0\), helps locate where the cost could be minimized. By solving this equation, we find the critical point \(x \approx 13.35\).
• This value of \(x\) is then used to determine the corresponding \(y\).

These steps confirm whether the identified critical points lead to a minimum or maximum for the cost function when interpreted with respect to the problem's context.

Volume Constraint

A **volume constraint** ensures that the solution adheres to the physical requirements of a problem. In this scenario, the volume constraint maintains the tank's capacity to 1125 cubic feet.
The volume of the tank with a square base \(x\) and depth \(y\) is expressed as:\[ x^2y = 1125 \]

• This equation is rearranged to solve for \(y\):

\[ y = \frac{1125}{x^2} \]

• The constraint formula helps substitute \(y\) back into the cost function to simplify the expression, demonstrating how the tank's dimensions influence the cost under real-world restrictions.

Balancing the volume constraint with minimizing cost is key to successfully solving optimization problems, ensuring solutions are both practically feasible and economically efficient.