Problem 42
Question
The demand function for a product is modeled by \(p=10,000\left(1-\frac{3}{3+e^{-0.501 x}}\right)\) Find the price of the product if the quantity demanded is (a) \(x=1000\) units and (b) \(x=1500\) units. What is the limit of the price as \(x\) increases without bound?
Step-by-Step Solution
Verified Answer
The price of the product when quantity demanded is 1000 units approximately equals $8602.95. When quantity demanded is 1500 units, the price approximately equals $9653.91. The limit of the price as the demand increases endlessly is $0.
1Step 1: Solve Equation for \(x = 1000\)
To find the price when quantity demanded is 1000 units, substitute \(x = 1000\) into the given equation: \(p = 10,000 \left(1-\frac{3}{3+e^{-0.501 \times 1000}}\right)\). Calculate the price.
2Step 2: Solve Equation for \(x = 1500\)
Next, to find the price when quantity demanded is 1500 units, substitute \(x = 1500\) into the given equation: \(p = 10,000 \left(1-\frac{3}{3+e^{-0.501 \times 1500}}\right)\). Compute to get the price.
3Step 3: Calculate the Limit as \(x\) Tends to Infinity
The limit as \(x\) tends to infinity demonstrates the behavior of the function as the quantity demanded increases without restraint. In this case, calculate \(\lim_{x \to \infty} 10,000 \left(1-\frac{3}{3+e^{-0.501x}}\right)\). Since \(e^{-0.501x}\) tends to 0 as \(x\) increases, the limit simplifies to \(10,000 \left(1-\frac{3}{3}\right) = 10,000(0) = 0\).
Key Concepts
Price CalculationExponential FunctionsLimits in Calculus
Price Calculation
Price calculation in this context involves determining the price of a product based on its demand function. This demand function allows us to input the quantity demanded and calculate the corresponding price.
In our exercise, the demand function is given by \(p=10,000\left(1-\frac{3}{3+e^{-0.501 x}}\right)\). Here, \(p\) represents the price and \(x\) represents the quantity demanded.
To calculate the price for a specific demand:
In our exercise, the demand function is given by \(p=10,000\left(1-\frac{3}{3+e^{-0.501 x}}\right)\). Here, \(p\) represents the price and \(x\) represents the quantity demanded.
To calculate the price for a specific demand:
- Substitute the value of \(x\) into the equation.
- Calculate the expression inside the brackets first to aid accuracy.
- Finally, compute the product to determine the price.
Exponential Functions
Exponential functions are mathematical expressions in which a variable appears in the exponent position, and they play a crucial role in modeling growth and decay processes.
In the demand function given, the term \(e^{-0.501x}\) is exponential because \(e\) (Euler's number, approximately 2.718) is raised to the power of \(-0.501x\). This results in exponential decay, which means the function's value decreases rapidly as \(x\) increases.
This kind of function is useful in economics for modeling how changes in quantity demanded can impact price. The negative exponent indicates that as quantity \(x\) increases, the term \(e^{-0.501x}\) becomes quite small. This, in turn, affects the overall demand function in the context of the exercise, causing the price to eventually drop.
In the demand function given, the term \(e^{-0.501x}\) is exponential because \(e\) (Euler's number, approximately 2.718) is raised to the power of \(-0.501x\). This results in exponential decay, which means the function's value decreases rapidly as \(x\) increases.
This kind of function is useful in economics for modeling how changes in quantity demanded can impact price. The negative exponent indicates that as quantity \(x\) increases, the term \(e^{-0.501x}\) becomes quite small. This, in turn, affects the overall demand function in the context of the exercise, causing the price to eventually drop.
Limits in Calculus
Limits in calculus help us understand the behavior of functions as variables approach specific values. In many cases, like our current demand function, we need to assess what happens as the quantity demanded \(x\) increases indefinitely.
The exercise involves finding \(\lim_{x \to \infty} 10,000 \left(1-\frac{3}{3+e^{-0.501x}}\right)\). As \(x\) increases to infinity, the term \(e^{-0.501x}\) approaches 0 because of the exponential decay.
This simplifies the expression to:\[10,000 \left(1-\frac{3}{3}\right) = 10,000 \times 0 = 0.\] Hence, the limit of the price is 0 as \(x\) tends to infinity.
This means that as demand grows extremely large, the price effectively becomes negligible, reflecting a situation where products might become free with infinite demand.
The exercise involves finding \(\lim_{x \to \infty} 10,000 \left(1-\frac{3}{3+e^{-0.501x}}\right)\). As \(x\) increases to infinity, the term \(e^{-0.501x}\) approaches 0 because of the exponential decay.
This simplifies the expression to:\[10,000 \left(1-\frac{3}{3}\right) = 10,000 \times 0 = 0.\] Hence, the limit of the price is 0 as \(x\) tends to infinity.
This means that as demand grows extremely large, the price effectively becomes negligible, reflecting a situation where products might become free with infinite demand.
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Problem 42
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