In calculus, finding local and global extrema is essential for optimization problems. Local extrema refer to points where a function reaches a maximum or minimum within a small neighborhood, while global extrema represent the highest or lowest points over the entire domain of the function. This is crucial in various fields, including economics, engineering, and physics. Below are three diverse, practical examples that illustrate how to find these extrema.
A company wants to determine the optimal price for a product to maximize revenue. The relationship between the price per unit and the number of units sold can often be modeled by a quadratic function.
The revenue function is given by:
R(p) = p(100 - 2p)
where R is the revenue and p is the price per unit.
To find the local and global extrema, we need to analyze the revenue function.
To find the critical points, we take the derivative of the revenue function and set it to zero:
R’(p) = 100 - 4p = 0
Solving for p gives:
4p = 100
p = 25
Next, we check the second derivative to confirm whether this point is a maximum or minimum:
R’’(p) = -4
Since R’’(25) is negative, this indicates that we have a local maximum at p = 25.
To find the maximum revenue, we substitute p back into the revenue function:
R(25) = 25(100 - 2(25)) = 25(50) = 1250
This example shows how businesses can use calculus to optimize pricing strategies, ensuring maximum revenue for a given product.
Consider a scenario where a company needs to determine the shortest distance from a specific point to a delivery route represented by a linear equation. This problem can be solved using calculus to find the minimum distance.
Let’s say the delivery route is represented by the line:
y = 2x + 3
and the point is (1, -1). We need to find the point on the line that minimizes the distance to (1, -1).
The distance D between the point (x, y) on the line and (1, -1) is given by:
D = √[(x - 1)² + (y + 1)²]
Substituting the line equation into the distance formula:
D = √[(x - 1)² + (2x + 4)²]
To minimize D, we can minimize D²:
D² = (x - 1)² + (2x + 4)²
Expanding and combining terms:
D² = (x - 1)² + (4x² + 16x + 16)
Setting the derivative of D² equal to zero, we find critical points and determine which gives us a minimum distance.
This example demonstrates how calculus can be applied to real-world scenarios involving distance minimization, which is useful in logistics and transportation.
A farmer wants to maximize the area of a rectangular fenced region with a fixed perimeter. This is a classic optimization problem that involves finding local and global extrema.
Let’s say the perimeter of the fenced area is 100 meters. The relationship between the length (L) and width (W) of the rectangle can be defined by the equation:
P = 2L + 2W = 100
From this, we can express W in terms of L:
W = 50 - L
The area A of the rectangle is given by:
A = L * W = L(50 - L) = 50L - L²
To find the maximum area, we take the derivative of A and set it to zero:
A’(L) = 50 - 2L = 0
Solving for L gives:
2L = 50
L = 25
Next, we check the second derivative to confirm it’s a maximum:
A’’(L) = -2
Since A’’(25) is negative, we have a local maximum. The corresponding width is:
W = 50 - 25 = 25
Thus, the maximum area is:
A = 25 * 25 = 625 m²
This example illustrates how calculus can be utilized in agricultural planning, enabling farmers to maximize their land usage effectively.