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Magazine Chapter 4: Problem 90
89-90. ECONOMICS: Consumer Expenditure If consumer demand for a commodity is given by the function below (where \(p\) is the selling price in dollars), find the price that maximizes consumer expenditure. $$ D(p)=8000 e^{-0.05 p} $$
Short Answer
Expert verified
The price that maximizes consumer expenditure is $20.
Step by step solution
01
Identify Consumer Expenditure Formula
Consumer expenditure is calculated as the product of demand \(D(p)\) and price \(p\). We need to express expenditure \(E(p)\) as a function of \(p\): \(E(p) = p \times D(p) = p \times 8000 e^{-0.05p}\).
02
Express Expenditure Function
Substitute \(D(p)\) into the expenditure formula: \[E(p) = 8000p e^{-0.05p}\]. This function describes consumer expenditure in terms of price \(p\).
03
Differentiate Expenditure Function
To maximize \(E(p)\), find the derivative \(E'(p)\). Use the product rule: \[(u \times v)' = u'v + uv'\]. Set \(u = 8000p\) and \(v = e^{-0.05p}\). Thus, \(u' = 8000\) and \(v' = -0.05e^{-0.05p}\).
04
Apply the Product Rule
Apply the product rule to compute \(E'(p)\): \[E'(p) = 8000 \cdot e^{-0.05p} + 8000p \cdot (-0.05e^{-0.05p})\] This simplifies to: \[E'(p) = 8000e^{-0.05p} - 400p e^{-0.05p}\].
05
Set the Derivative to Zero
Set \(E'(p) = 0\) to find critical points:\[8000e^{-0.05p} - 400p e^{-0.05p} = 0\]Factor out \(e^{-0.05p}\):\[e^{-0.05p}(8000 - 400p) = 0\].
06
Solve for Critical Points
Since \(e^{-0.05p} eq 0\), solve \(8000 - 400p = 0\) for \(p\): \[8000 = 400p\]\[p = 20\].
07
Verify the Maximum Expenditure
Ensure the critical point at \(p = 20\) is a maximum by using the second derivative \(E''(p)\), or by assessing the function's behavior around \(p = 20\). Since expenditure is an exponential decay function multiplied by a linear function, it tends to have a maximal point.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Demand Function
A demand function is a mathematical expression showing how much of a commodity consumers are willing to buy at various prices. In this exercise, the demand function is given as \(D(p) = 8000 e^{-0.05 p}\). This function tells us that the quantity demanded decreases exponentially as the price \(p\) increases.
- The base of the exponential function is \(e\), which naturally represents exponential growth or decay.
- The term \(-0.05p\) in the exponent indicates an exponential decay; as \(p\) increases, \(D(p)\) decreases.
Derivative Application
Derivatives provide us with a powerful tool in calculus to find maximum and minimum points of a function. In this exercise, the objective is to maximize consumer expenditure \(E(p)\), which is expressed as \(E(p) = p \times D(p) = 8000p e^{-0.05p}\).
To maximize a function, we first find its derivative. Here, the product rule of derivatives is applied because our expenditure function is the product of \(p\) and \(D(p)\).
To maximize a function, we first find its derivative. Here, the product rule of derivatives is applied because our expenditure function is the product of \(p\) and \(D(p)\).
- If \(u = 8000p\) and \(v = e^{-0.05p}\), then \(E'(p)\) is calculated using the product rule \((uv)' = u'v + uv'\).
- The derivative \(E'(p)\) helps us identify the rate of change of the expenditure function with respect to price \(p\).
Exponential Functions
Exponential functions, like \(e^{-0.05p}\), play a significant role in modeling various real-world phenomena including population growth and radioactive decay. In economics, they often represent factors that exponentially enhance or reduce quantities.
In our consumer demand problem, the term \(e^{-0.05p}\) showcases exponential decay. This means as the price \(p\) increases, the demand \(D(p)\) decreases exponentially:
In our consumer demand problem, the term \(e^{-0.05p}\) showcases exponential decay. This means as the price \(p\) increases, the demand \(D(p)\) decreases exponentially:
- Exponential functions can be expressed in the form \(y = a \cdot e^{bx}\), where \(b < 0\) indicates decay.
- The exponential decay modifies the behavior of the linear component (in this case the \(8000p\)), reducing it significantly as \(p\) grows larger.
Critical Points
Critical points are values of \(p\) where the derivative of a function equals zero or does not exist. For our function \(E'(p) = 0\), these are the points where expenditure does not increase or decrease, but plateaus or turns around.
In this problem, finding \(E'(p) = 0\) involves:
It's important to verify that this critical point is indeed a maximum by analyzing the second derivative or observing the behavior of the function around \(p = 20\). In many cases of an exponential decay multiplied by a linear function, this will typically yield a maximum.
In this problem, finding \(E'(p) = 0\) involves:
- Solving \(8000e^{-0.05p} - 400pe^{-0.05p} = 0\).
- Factoring out \(e^{-0.05p}\), leaving \(8000 - 400p = 0\).
It's important to verify that this critical point is indeed a maximum by analyzing the second derivative or observing the behavior of the function around \(p = 20\). In many cases of an exponential decay multiplied by a linear function, this will typically yield a maximum.
