University:
High School
Course:
High School Mathematics
Academic year:

2022
Views:

433

Pages:

4
Author:

sakurasparrow51xu

5.4 Modeling & Optimization Concept Strategy for Solving Max-Min Problems Understand the Problem Develop a Mathematical Model Graph the Function (if that helps) Identify the Critical Points & Endpoints 5) Solve & Interpret the Solution Examples: Closing off the first Quadrant You are planning to close off a corner of the first quadrant with a line segment 20 units long running (a, 0) to (0, b). Show that the area of the triangle enclosed by the segment is largest when a=b. Prove it. coordinate plane with a line segment connecting (a, 0) and (0, b) in the first quadrant. The area of the triangle formed by this line segment and the x and y axes is shaded. Pythagorean Theorem 400 = a² + b² b² = 400 - a² b = √(400 - a²) Area of Triangle A = 1/2 * a * b A = 1/2 * a * √(400 - a²) Now the derivative A' = 1/2 * a * (1/2) * (400 - a²)^(-1/2) * (-2a) + 1/2 * √(400 - a²) = -a² / (2√(400 - a²)) + √(400 - a²) 0 = 400 - 2a² 400 = 2a² 200 = a² a = √200 400 = 200 + b² 200 = b² b = √200 This proves that a=b Sum x + y = 20 x = 20 - y Product x * y = P P(y) = (20 - y) * y = 20y - y² P'(y) = 20 - 2y 0 = 20 - 2y 2y = 20 y = 10 x + (10) = 20 x = 10 Example: Con-can You have been asked to design a one-liter oil can shaped like a right circular cylinder. What dimensions will use the least materials? 1 right circular cylinder with height h and radius r Volume V = πr²h 1000cm³ = πr²h 1000/πr² = h Surface Area SA = 2πr² + 2πrh SA = 2πr² + 2πr(1000/πr²) = 2πr² + 2000/r SA'(r) = 4πr - 2000/r² 0 = 4πr - 2000/r² 4πr = 2000/r² 4πr³ = 2000 r³ = 2000/4π r³ = 500/π r ≈ 5.419 cm h = ? 1000 = π(5.419)²h 1000 = π(29.365)h 1000/29.365π = h h ≈ 10.84 cm Example: Maximizing Profit Suppose that (r(x) = 9x) and (c(x) = x^3 - 6x^2 + 15x) where x represents thousands of units. Is there a production level that maximizes 1 the profit? (What is it?) Profit P = R - C P = 9x - (x^3 - 6x^2 + 15x) = -x^3 + 6x^2 - 6x P'(x) = -3x^2 + 12x - 6 0 = -3x^2 + 12x - 6 0 = x^2 - 4x + 2 x = (4 ± √(16 - 4(1)(2))) / 2(1) = (4 ± √8) / 2 = (4 ± 2√2) / 2 = 2 ± √2 x = 3.414 and x = 0.586 parabola with a maximum point x_max = ... will be max x_min = ... will be min