Problem 40 - Calculus Early Transcendentals

Question

A commercial cattle ranch currently allows 20 steers per acre of grazing land; on the average its steers weigh 2000 lb at market. Estimates by the Agriculture Department indicate that the average market weight per steer will be reduced by 50 lb for each additional steer added per acre of grazing land. How many steers per acre should be allowed in order for the ranch to get the largest possible total market weight for its cattle?

Step-by-Step Solution

Answer

30 steers per acre maximize total market weight.

Step 1: Understand the Relationship

We start by understanding that the total weight depends on the number of steers per acre, denoted as \(x\). With 20 steers per acre, each steer weighs 2000 lb. For every additional steer, the weight of each steer is reduced by 50 lb.

Step 2: Set up the Expression for Total Weight

We express the total weight, \(W(x)\), as a function of steers per acre, \(x\). If \(x\) steers are allowed, each will weigh \(2000 - 50(x - 20)\) lb because each additional steer reduces the weight by 50 lb. Thus, the total weight is \(W(x) = x[2000 - 50(x - 20)]\).

Step 3: Simplify the Expression

Simplify the expression: \[ W(x) = x \times (2000 - 50(x - 20)) \] \(= x \times (2000 - 50x + 1000)\) \(= x \times (3000 - 50x)\) \(= 3000x - 50x^2\).

Step 4: Find the Critical Points

To find the value of \(x\) that maximizes \(W(x)\), take the derivative of the expression with respect to \(x\) and set it to zero: \[ \frac{dW}{dx} = 3000 - 100x = 0 \] Solving for \(x\) gives \(x = 30\).

Step 5: Verify Maximum Value

We verify that \(x = 30\) gives a maximum by checking the second derivative: \[ \frac{d^2W}{dx^2} = -100 \] Since the second derivative is negative, \(x = 30\) is indeed a maximum.

Key Concepts

Critical Points

In the world of optimization in calculus, a critical point is where we find potential minimum or maximum values of a function. These points occur where the derivative of the function is zero or undefined. In simpler terms, it's like a special spot on the graph of the function where it "flattens out". In the context of our exercise, critical points help us determine the number of steers that should be allowed per acre to maximize the total market weight.

To identify critical points, we look for where the derivative of our function equals zero. Once we find these points, we can decide whether they are minimums, maximums, or neither. This is crucial in solving problems like optimizing the weight of cattle in a ranch setting. Understanding these points helps us make decisions that optimize a target outcome, such as total weight, cost, or efficiency.

Derivative Test

The derivative test is a powerful tool to discern what critical points mean. Specifically, we often use the first derivative test to determine local maxima or minima in a function.

Here's how it works in our scenario: We take the derivative of the function that represents the total weight of steers per acre, denoted as \( W(x) \). By finding when the derivative \( \frac{dW}{dx} \) is zero, we find our critical points. Remember, in our exercise, this resulted in \( 3000 - 100x = 0 \), leading to the solution \( x = 30 \).

• If the function changes from increasing to decreasing at a critical point, it's a local maximum.
• If the function changes from decreasing to increasing, it's a local minimum.

By using the derivative test, we establish the point's behavior, guiding us towards optimizing outcomes in real-life applications, such as maximizing cattle weight.

Second Derivative Test

The second derivative test provides additional assurance about whether a critical point is a maximum or a minimum. It involves taking the second derivative of the function and evaluating it at the critical point.

For our total weight function \( W(x) \), once we identified the critical point \( x = 30 \) from the first derivative, we calculated the second derivative \( \frac{d^2W}{dx^2} \). In our case, this resulted in \( \frac{d^2W}{dx^2} = -100 \), which is a negative value.

• If the second derivative is negative at the critical point, then it's a local maximum (the graph is concave down).
• If it's positive, then it's a local minimum (the graph is concave up).
• If it equals zero, the second derivative test is inconclusive.

In the cattle ranch problem, since the second derivative is negative, it confirms that \( x = 30 \) indeed results in the maximum total market weight, allowing us to make sound and optimal decisions.