Question

A farmer can get 2 dollars per bushel for their potatoes on July 1. After July 1, the price drops by 2 cents per bushel per extra day. On July 1, a farmer had 80 bushels of potatoes in the field and estimates that the crop is increasing at the rate of 1 bushel per day. When should the farmer harvest the potatoes to maximize his revenue? [Hint: Let x be the number of extra day after July 1 and R be the revenue.] (4 marks) [Total: 25 marks)

Answer 1

To maximize revenue, the farmer should harvest the potatoes 100 days after July 1.

Step-by-step explanation:

To determine when the farmer should harvest the potatoes to maximize revenue, we can use the revenue function. Let x be the number of extra days after July 1. The revenue function is given by R = (80 + x)(200 - 2x).

First, we expand the equation: R = 16000 + 400x - 2x^2.

To find the maximum revenue, we can take the derivative of the revenue function and set it equal to zero. Taking the derivative, we get dR/dx = 400 - 4x.

Setting this equal to zero, 400 - 4x = 0, solving for x gives x = 100.

Therefore, the farmer should harvest the potatoes 100 days after July 1 to maximize revenue.

Answer 2

The farmer should create a revenue function R(x) based on price per bushel and number of bushels over time, take the derivative to find the critical point, and solve for x to determine the optimal day to harvest the potatoes to maximize revenue.

Step-by-step explanation:

To determine when the farmer should harvest the potatoes to maximize revenue, we need to set up a revenue function that incorporates both the change in price and the increase in bushels over time. Let x represent the number of extra days after July 1. The price per bushel p(x) will then be $2.00 - $0.02x. The number of bushels b(x) will be 80 + x, since the farmer is gaining 1 bushel extra per day. The revenue function R(x) is obtained by multiplying the price function by the number of bushels, R(x) = p(x) × b(x) = (2 - 0.02x)(80 + x).

The next step is to find the maximum value of R(x). We do this by taking the first derivative of R and setting it to zero to find the critical points. Upon finding the derivative, set R'(x) = 0 and solve for x. This will give us the number x at which revenue is maximized. In some cases, you might need to check the second derivative to ensure that the critical point represents a maximum.

Example Calculation:

R(x) = (2 - 0.02x)(80 + x) = 160 + 2x - 1.6x - 0.02x₂

R'(x) = 2 - 1.6 - 0.04x

0 = 2 - 1.6 - 0.04x

0.4 = 0.04x

x = 10

Therefore, the farmer should harvest the potatoes 10 days after July 1 to maximize revenue.