Maximum and Minimum in Applications

Optimization Problems

• Optimization involves finding the highest or lowest value of something, such as:
  • Highest value
  • Lowest cost
  • Shortest distance
  • Maximum speed
  • Maximum temperature
  • Maximum force
• We will translate problem questions into a format suitable for applying calculus methods.

Types of Problems

• Maximizing or minimizing a continuous function in two main categories:
  • Finite Interval: Function considered between two endpoints.
    • The extreme value theorem guarantees a maximum and minimum within the interval.
  • Infinite Interval: Function considered over an unbounded domain.
    • A maximum may not always exist if the function increases indefinitely.

Procedure for Solving Optimization Problems

• Draw a Graph: Visualize the problem and the function to check results.
• Find an Equation: Determine the function of a single variable to optimize.
• Consider the Domain: Identify the appropriate domain for the function, as it affects global maxima and minima.
• Differentiate: Find critical points by differentiating the function.
• Check Endpoints: Evaluate the function at the interval's endpoints to find the maximum and minimum values.

Exercise 19: Maximizing a Constrained Area

• Task: Fence in a rectangular play area within a triangular plot to maximize its area.
• Triangle dimensions: 4 meters (height) and 12 meters (base).

Problem Setup

• Define Variables:
  • Let x be the width of the rectangular play area.
  • Let y be the length of the rectangular play area.
• Objective Function:
  • Area A = x \, y. We want to maximize A.

Geometric Constraint

• The rectangle must fit within the triangular area.
• Similar Triangles:
  • The large triangle and the triangle above the rectangle are similar.
  • \frac{4 - y}{4} = \frac{x}{12}
  • x = 3(4 - y)

Function of a Single Variable

• Substitute the constraint into the area equation:
  • A = 3(4 - y)y
  • A = 12y - 3y^2

Calculus

• Domain: 0 \le y \le 4
• Differentiate:
  • \frac{dA}{dy} = 12 - 6y
• Critical Points:
  • Set \frac{dA}{dy} = 0
  • 12 - 6y = 0
  • y = 2
  • When y = 2, x = 3(4 - 2) = 6
  • A = 6 \times 2 = 12 \text{ m}^2

Checking Endpoints

• At y = 0, A = 0
• At y = 4, A = 0

Conclusion

• The maximum area is 12 square meters, occurring when the rectangle has a height of 2 meters and a width of 6 meters.

Exercise 20

Part A

• Show that the distance between a point on the parabola y = \frac{1}{4}x^2 and the point (0, 4) is given by a specific equation.

Distance Formula

• The distance between two points (x1, y1) and (x2, y2) is given by:
  • d = \sqrt{(x1 - x2)^2 + (y1 - y2)^2}

Applying the Formula

• Let (x, y) be a point on the parabola, and (0, 4) be the fixed point.
  • d = \sqrt{(x - 0)^2 + (y - 4)^2}
  • d = \sqrt{x^2 + (y - 4)^2}

Incorporating the Parabola Equation

• Since y = \frac{1}{4}x^2, then x^2 = 4y
  • d = \sqrt{4y + (y - 4)^2}

Eliminating Square Root for Simplicity

• Working with d^2 = s^2 to avoid square roots:
  • s^2 = 4y + (y - 4)^2

Differentiation

• Differentiate both sides with respect to y:
  • 2s \frac{ds}{dy} = 4 + 2(y - 4)

Critical Points

• Setting \frac{ds}{dy} = 0:
  • 4 + 2(y - 4) = 0
  • 4 + 2y - 8 = 0
  • 2y = 4
  • y = 2

Finding Corresponding x Values

• Using x^2 = 4y:
  • x^2 = 4(2) = 8
  • x = \pm \sqrt{8}

Calculating the Distance

• s = \sqrt{4y + (y - 4)^2}
  • s = \sqrt{4(2) + (2 - 4)^2} = \sqrt{8 + 4} = \sqrt{12}

Domain Considerations

• The domain of y is [0, \infty).

Endpoint Analysis

• At y = 0, s = \sqrt{4(0) + (0 - 4)^2} = \sqrt{16} = 4
• As y \to \infty, s \to \infty

Conclusion

• The minimum distance occurs at y = 2, where s = \sqrt{12}.
• The points are (\sqrt{8}, 2) and (-\sqrt{8}, 2).

Part B

• Find the closest point on the parabola y = \frac{1}{4}x^2 to the point (0, 1).

Applying the Distance Formula

• s^2 = x^2 + (y - 1)^2

Incorporating the Parabola Equation

• Since x^2 = 4y:
  • s^2 = 4y + (y - 1)^2
  • s^2 = 4y + y^2 - 2y + 1
  • s^2 = y^2 + 2y + 1

Differentiation

• Differentiate both sides with respect to y:
  • 2s \frac{ds}{dy} = 2y + 2

Critical Points

• Setting \frac{ds}{dy} = 0:
  • 2y + 2 = 0
  • y = -1

Domain Check

• Since the domain is [0, \infty), y = -1 is not a valid solution.

Endpoint Analysis

• At y = 0:
  • s = \sqrt{0^2 + 2(0) + 1} = 1
• As y \to \infty, s \to \infty

Derivative Analysis

• \frac{ds}{dy} is positive for all y in the domain.

Conclusion

• The minimum distance occurs at the endpoint, y = 0.
• The closest point on the parabola to (0, 1) is (0, 0), and the minimum distance is 1.