Optimization (APPLICATIONS OF CALCULUS)

Question

An individual’s ABG oxygen level, L, at time t ∈ [0, π], is given by L(t) = 90 +

10 sin(2t). Find the values of the maximum and minimum oxygen level. When do these values

occur?

Answer 1

L '(t) = 20cos(2t)

Setting this to 0, cos(2t) = 0 when 2t = π/2 and 3π/4

90 + 10sin(π/2) = 100 <--max

90 + 10sin(3π/2) = 80 <--min

These occur at t = π/4 and t = 3π/4

Answer 2

L'(t) = 20cos(2t) = 0, cos (2t) = 0, 2t = π/2, t = π/4; 2t = 3π/2, t = 3π/4.

We have 2 critical points π/4 and 3π/4 and end points of interval: t = 0 and t = π

L(0) = 90

L(π/4) = 90 + 10sin(π/2) = 100 (max level)

L(3π/4) = 90 + 10sin(3π/2) = 80 (min level)

L(π) = 90 + 10sin(3π) = 90