Problem 58 - Applied Calculus

Question

The demand for a product is given, for \(p, q \geq 0\), by $$ p=f(q)=50-0.03 q^{2} $$ (a) Find the \(p\) - and \(q\) -intercepts for this function and interpret them in terms of demand for this product. (b) Find \(f(20)\) and give units with your answer. Explain what it tells you in terms of demand. (c) Find \(f^{\prime}(20)\) and give units with your answer. \(\mathrm{Ex}\) plain what it tells you in terms of demand.

Step-by-Step Solution

Answer

(a) p-intercept is 50, q-intercept is approximately 40.83. (b) f(20) = 38. (c) f'(20) = -1.2.

Step 1: Identify the p-intercept

To find the p-intercept, set the quantity variable \(q\) to 0 in the demand function \(p = f(q)\). Substitute \(q = 0\) into the equation to find \(p\).\[p = 50 - 0.03(0)^2 = 50\]The p-intercept is 50, which means when no product is demanded (q = 0), the price is \(p = 50\).

Step 2: Identify the q-intercept

To find the q-intercept, set the price variable \(p\) to 0 in the demand function. Solve for \(q\) by setting the equation to zero:\[0 = 50 - 0.03q^2\]Rearrange to solve for \(q\):\[0.03q^2 = 50 \q^2 = \frac{50}{0.03} \q^2 = 1666.67 \q = \sqrt{1666.67} \q \approx 40.83\]The q-intercept is approximately \(q = 40.83\). This indicates that at a price of zero, the maximum demand for the product is approximately 40.83 units.

Step 3: Calculate f(20)

Substitute \(q = 20\) into the demand function \(p = f(q)\) to find the price at this quantity:\[p = 50 - 0.03(20)^2 = 50 - 0.03(400) = 50 - 12 = 38\]Thus, \(f(20) = 38\). This indicates that when 20 units are demanded, the price is \(p = 38\).

Step 4: Calculate the derivative f'(q)

Calculate the derivative of the demand function \(p = f(q)\) with respect to \(q\):\[f'(q) = \frac{d}{dq}(50 - 0.03 q^2) = 0 - 0.06 q = -0.06q\]This derivative represents how the price changes with respect to a change in quantity.

Step 5: Evaluate f'(20)

Substitute \(q = 20\) into the derivative expression obtained in Step 4:\[f'(20) = -0.06 \times 20 = -1.2\]This means that at 20 units, the price decreases by 1.2 units per additional unit demanded. It indicates the rate at which price decreases as quantity demanded increases by one unit.

Key Concepts

Understanding the p-intercept

In the world of demand functions, the p-intercept represents the initial price when no products are being demanded. To find it, we set the quantity, denoted as \(q\), to zero in our given demand function, which is \(p = f(q) = 50 - 0.03 q^2\). After substituting \(q = 0\), we get:

• \(p = 50 - 0.03(0)^2 = 50\)

This tells us that when no units of the product are demanded, the price remains at 50. It serves as the maximum possible price that suppliers can charge when demand is nonexistent.

Exploring the q-intercept

The q-intercept provides insight into the highest quantity demanded when the price drops to zero. To find this intercept, we need to set the price, \(p\), to zero and solve the equation for \(q\). Start with

• \(0 = 50 - 0.03q^2\)

Rearrange and solve for \(q\):

• \(.03 q^2 = 50\)
• \(q^2 = \frac{50}{0.03} = 1666.67\)
• \(q = \sqrt{1666.67} \approx 40.83\)

Thus, the q-intercept is approximately 40.83 units and represents the maximum demand achieved at a price of zero. This helps in understanding how the product demand would behave in an ideal scenario of zero pricing.

Calculating the derivative of the demand function

The derivative of a demand function, \(f'(q)\), helps determine how the price is sensitive to changes in the quantity demanded. For the function \(p = f(q) = 50 - 0.03 q^2\), we calculate the derivative with respect to \(q\). The derivative process is as follows:

• \(f'(q) = \frac{d}{dq}(50 - 0.03 q^2) = -0.06q\)

This tells us that the rate of change of price in relation to demand is \(-0.06q\). When the derivative is negative, it indicates an inverse relationship between price and quantity: as the demand increases, the price decreases. This derivative shows how elastic the price is to changes in demand.

Evaluation of demand at specific points

Evaluating the demand function at specific quantities helps to understand the actual price at a given demand. For example, finding \(f(20)\) allows us to pinpoint the price when 20 units are requested.

• \(f(20) = 50 - 0.03(20)^2 = 38\)

Here, a demand of 20 units corresponds to a price of 38, indicating that more units demanded typically mean a lower price.
Similarly, using the derivative at this point, \(f'(20)\), gives insight into the rate of price decline:

• \(f'(20) = -0.06 \times 20 = -1.2\)

This means that each additional unit demanded beyond 20 decreases the price by 1.2 units. Understanding these evaluations assists in making that clear connection between price and demand dynamics.