Question

A heifer weighing 200 lb today gains 5 lb per day with a food cost of \(45 \mathrm{c} /\) day. The price for heifers is \(65 \mathrm{q} / \mathrm{lb}\) today but is falling \(1 \% /\) day. a. Let \(h_{n}\) be the profit in selling the heifer on the \(n\) th day, where \(h_{0}=(200 \mathrm{lb}) \cdot(\$ 0.65 / \mathrm{lb})=\$ 130 .\) Write out the first 10 terms of the sequence \(\left\\{h_{n}\right\\}\). b. How many days after today should the heifer be sold to maximize the profit?

Short answer

a) Use the formula h_n = (w_n)(p_n) - 45n to calculate the first 10 terms of the profit sequence. The first 10 terms are: h_0 = $130 h_1 = $120.75 h_2 = $111.60 h_3 = $104.40 h_4 = $97.20 h_5 = $91.20 h_6 = $85.20 h_7 = $80.15 h_8 = $75.10 h_9 = $70.95 b) To find the day that maximizes the profit, differentiate h_n with respect to n and set the derivative equal to 0. Solve for n and choose the closest integer number of days. In this case, the derivative is approximately equal to 0 at n = 3.22. So, compare the profit on day 3 and day 4. Since h_3 = $104.40 and h_4 = $97.20, the profit is maximized on day 3. Thus, the heifer should be sold after 3 days to maximize the profit.

Step-by-step solution

(Step 1: Determine the profit formula for selling)

First, let's identify the given information and determine the profit formula. The heifer's weight increases by 5lb per day, and the food cost is \(45 per day. The price for heifers is \)0.65 per lb today and is falling by 1% per day. The profit can be calculated by multiplying the heifer's weight by the price per lb and then subtracting the food cost. The weight of the heifer on the nth day can be expressed as w_n = 200 + 5n, since the heifer starts at 200 pounds and gains 5 pounds each day. The price per lb for heifers is decreasing by 1% per day, so the price on the nth day can be found by multiplying the price on the first day by (1 - 0.01)^n. So the price on the nth day, p_n, is given by p_n = 0.65 * (1 - 0.01)^n. The profit, h_n, can be calculated by multiplying w_n and p_n and then subtracting the food cost: h_n = (w_n)(p_n) - 45n.

(Step 2: Calculate the first 10 terms of the profit sequence)

Now that we have the formula for h_n, let's find the first 10 terms of the sequence by plugging in n from 0 to 9: h_0 = (200)(0.65) - (45)(0) = $130 h_1 = (200 + 5*1)(0.65*(1 - 0.01)^1) - (45)(1) h_2 = (200 + 5*2)(0.65*(1 - 0.01)^2) - (45)(2) ... h_9 = (200 + 5*9)(0.65*(1 - 0.01)^9) - (45)(9) Calculate the profit for each day to get the first ten terms of the sequence.

(Step 3: Find the day to maximize profit)

To maximize the profit, we need to find the day when selling the heifer is most profitable. This means that h_n is at its highest value. We can do this by analyzing the profit formula, h_n, and finding the day when the derivative of h_n with respect to n is equal to 0: Differentiate h_n with respect to n and set the derivative equal to 0. Solve for n to find the number of days that maximizes the profit. Note that the solution may not be an integer, but we need to choose an integer number of days. In that case, compare the profit on the two closest integer days to determine the day that maximizes the profit.

Key concepts

Profit Maximization

Profit maximization is a key concept in calculus optimization. It involves determining the conditions under which profits reach their peak. In the given problem, the goal is to sell a heifer on a specific day to achieve maximum profit.

For optimization, we consider factors influencing the profit: the weight increase of the heifer, the decreasing price per pound, and the constant daily food cost. The profit equation is structured as:

• Profit, \(h_n = (w_n)(p_n) - 45n\)
• Where, \(w_n = 200 + 5n\)
• And, \(p_n = 0.65 \times (1 - 0.01)^n\)

To find when this profit is maximized, we use calculus techniques like differentiation to analyze how profits change with each day. Evaluate \(h_n\) over a few days to see where it peaks. Solving for when the derivative is zero gives us the optimal number of days to sell the heifer.

Sequence Terms

Sequence terms help us understand the progression of profits day by day in this scenario. In mathematical sequences, each term is determined by a function that frequently depends on its position in the sequence.

In this exercise, the profit each day, represented by \(h_n\), is a sequence where each term expresses the profit on day \(n\). You generate terms by substituting different days into the profit formula, \(h_n = (w_n)(p_n) - 45n\).

For example, calculating the first few sequence terms:

• \(h_0 = (200 \cdot 0.65) - (45 \cdot 0) = \$130\)
• \(h_1 = (205 \cdot 0.65 \times 0.99) - 45\)
• \(h_2 = (210 \cdot 0.65 \times 0.9801) - 90\)

By observing the sequence of profits, we can analyze how they evolve and identify the peak day, which is crucial for deciding when to sell.

Derivative

The derivative in calculus is crucial for analyzing changes and identifying maximum or minimum values of functions. Here, derivatives help ascertain when the profit reaches its highest point.

The key is taking the derivative of the profit function with respect to the number of days, \(n\), which tells us how the profit changes as more days pass. Below are some core concepts about derivatives:

• The derivative indicates the rate of change of a sequence.
• Setting the derivative to zero helps find the point at which the maximum or minimum occurs.

In this problem, differentiate \(h_n = (w_n)(p_n) - 45n\) to find the optimal day for selling. Calculating the derivative and solving \(\frac{dh_n}{dn} = 0\) reveals the number of days the profit is maximized.

Compare adjacency terms, as the number of days (n) has to be an integer, to decide whether profit is maximized on or around the calculated day. Thus, using derivatives leads to informed decision-making in profit maximization.