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 $(x<em>1, y</em>1)$ and $(x<em>2, y</em>2)$ is given by:
  • $d = \sqrt{(x<em>1 - x</em>2)^2 + (y<em>1 - y</em>2)^2}$

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.