6.2 – Optimization using Trigonometry

6.2 – Optimization using Trigonometry IB Math SL/HL, MCB4U - Santowski

(A) Review • When a function has a local max/min at x = c, then the value of the derivative is 0 at c • Alternatively, some functions will have local max/min points, even though the derivative is not zero  i.e. the absolute value function (which will have an undefined derivative at the point of the max/min) • However, the converse  when a function has a zero first derivative, the critical point may or may not be an extreme value (i.e. max/min) • So we need to “test” whether the critical point is a max or a min or neither  thus we have either our first derivative test or our second derivative test • In the first derivative test, we look for sign changes in the derivative values before and after the critical point  if the derivative changes from +ve to –ve, then the fcn has a max and vice versa for a min • In the second derivative test, we test for concavity  a fcn will have a max only if the curve is concave down, so if the second derivative is negative, then the critical point is a maximum and vice versa for a min • The maximum and minimum points and values have important roles to play in mathematical modelling. • In optimizing problems, we try to solve problems involving maximizing areas, volumes, profits and minimizing distances, times, and costs

(B) Examples • Find the maximum perimeter of a right triangle with a hypotenuse of 20 cm • We can set up a Pythagorean relationship (P = x + 20 + (400 – x2) ) • Or we can draw a right triangle, set  as the base angle  then the adjacent side measures 20cos and the opposite side measures 20sin and of course the hypotenuse is 20 cm • p() = 20cos + 20sin + 20 • Then p`() = -20sin() + 20cos() = 0 • So sin() = cos() or 1 = tan()    = /4 or 45° • Then p(/4) = 20(1 + cos(/4) + sin(/4)) = 20 + 202 = 48.3 cm • To verify a maximum, we can take the second derivative  • p``() = -20cos() – 20 sin() • p``(/4) = -20cos(/4) – 20 sin(/4) = -40/2 • So our second derivative is –ve meaning the function curve is concave down, meaning we have a maximum point at /4 giving 48.3 as the max perimeter

(C) Examples • Ex 2. Find the area of the largest rectangle that can be inscribed in a semicircle of radius r • Show how to develop the formula: • A() = 2rcos()  rsin()

(D) Internet Links • Optimization from Calculus 1 - Problems and Solutions from Pheng Kim Ving • Maximum/Minimum Problems and Solutions from UC Davis • Calculus I (Math 2413) - Applications of Derivatives - Optimization from Paul Dawkins • Calculus@UTK 4.6 - Optimization Problems and Applets from Visual Calculus

(E) Homework • Stewart, 1989, Chap 7.4, p325, Q3-8 • Photocopy from Stewart, 1998, Chap 4.6, p318, Q22-33

6.2 – Optimization using Trigonometry