Problem 57 - Applied Calculus

Question

The yield, \(Y\), of an apple orchard (measured in bushels of apples per acre) is a function of the amount \(x\) of fertilizer in pounds used per acre. Suppose $$ Y=f(x)=320+140 x-10 x^{2} $$ (a) What is the yield if 5 pounds of fertilizer is used per acre? (b) Find \(f^{\prime}(5)\). Give units with your answer and interpret it in terms of apples and fertilizer. (c) Given your answer to part (b), should more or less fertilizer be used? Explain.

Step-by-Step Solution

Answer

(a) 770 bushels per acre; (b) 40 bushels per pound of fertilizer. Use more fertilizer because yield increases with more fertilizer at 5 pounds.

Step 1: Calculate the Yield with 5 Pounds of Fertilizer

To find the yield when 5 pounds of fertilizer is used, substitute \(x = 5\) into the equation \(Y = f(x) = 320 + 140x - 10x^2\). Calculate the value of \(Y\):\[Y = 320 + 140(5) - 10(5)^2\]\[= 320 + 700 - 250\]\[= 770\]Therefore, the yield is 770 bushels per acre when 5 pounds of fertilizer is used.

Step 2: Find the First Derivative of the Function

To find \(f'(x)\), differentiate the function \(f(x) = 320 + 140x - 10x^2\) with respect to \(x\). Use the power rule of differentiation:\[f'(x) = \frac{d}{dx}(320) + \frac{d}{dx}(140x) - \frac{d}{dx}(10x^2)\]\[= 0 + 140 - 20x\]This gives the first derivative: \(f'(x) = 140 - 20x\).

Step 3: Calculate f'(5) and Interpret

Substitute \(x = 5\) into \(f'(x) = 140 - 20x\) to find \(f'(5)\):\[f'(5) = 140 - 20(5)\]\[= 140 - 100\]\[= 40\]The derivative \(f'(5)\) is 40 bushels per pound of fertilizer. This means that for each additional pound of fertilizer used per acre, the yield increases by 40 bushels at \(x = 5\) pounds.

Step 4: Evaluate Whether to Use More or Less Fertilizer

Since \(f'(5) = 40\) is positive, increasing the amount of fertilizer used will increase the yield. Therefore, it suggests that more fertilizer should be used beyond 5 pounds per acre to improve yield, as the marginal benefit in terms of additional apples per additional pound of fertilizer is still substantial.

Key Concepts

Derivative Calculation

In the context of agricultural yield optimization, understanding derivative calculation is crucial. A derivative helps us understand how a function changes as its input changes. Here, we aim to determine how the yield of an apple orchard responds to different amounts of fertilizer applied per acre. The yield function is given as:\[ Y = 320 + 140x - 10x^2 \]where \(Y\) represents the yield in bushels per acre, and \(x\) represents the fertilizer amount in pounds per acre. To calculate the derivative of this function, we apply the power rule.- The derivative of a constant, like 320, is 0.- The derivative of \(140x\) is 140, since the power of \(x\) drops by 1.- For \(-10x^2\), bringing down the power of 2 gives \(-20x\).This results in the first derivative:\[ f'(x) = 140 - 20x \]The derivative \(f'(x)\) represents the rate of change in yield with respect to the fertilizer amount, providing insights into how each additional pound affects the apple yield.

Fertilizer Application Rate

Optimal fertilizer application is vital for maximizing an orchard's yield without overuse, which can be costly or environmentally harmful. In our function, the fertilizer application rate refers to the \(x\) value in the yield equation \(Y=f(x)=320+140x-10x^2\).To find the yield when using a specific application rate, substitute the desired \(x\) value into the equation. For 5 pounds of fertilizer:\[ Y = 320 + 140(5) - 10(5)^2 = 770 \]Thus, at 5 pounds of fertilizer per acre, the yield is 770 bushels, showing the relationship between fertilizer usage and yield. It's crucial to balance this to achieve an optimal yield without unnecessary fertilizer application.Determining the appropriate amount of fertilizer involves analyzing potential yields against costs. This approach ensures that resources are used efficiently.

Marginal Analysis

Marginal analysis in agriculture involves assessing the impact of small changes in input levels, such as fertilizer, on output, like apple yield. It focuses on understanding the effect of adding one more unit of input.In this case, the derivative \( f'(x) = 140 - 20x \) plays a crucial role in marginal analysis. Specifically, calculating \( f'(5) \):\[ f'(5) = 140 - 20(5) = 40 \]This outcome, 40 bushels per additional pound of fertilizer, indicates a positive marginal benefit. It tells us that using an extra pound of fertilizer at the 5-pound level will increase the yield by 40 bushels per acre. The positive derivative means the yield increases with more fertilizer, suggesting further application beyond 5 pounds could enhance yield. However, considerations such as cost, environmental factors, and diminishing returns should guide the decision on the exact amount of additional fertilizer to apply. Understanding these marginal gains helps in making informed decisions to optimize agricultural practices.