let y = (14sinØ + 5cosØ)^2
dy/dx = 2(14sinØ + 5cosØ)(14cosØ - 5sinØ)
= 0 for a max/min of y
so 14sinØ = -5cosØ = 0 or 14cosØ = 5sinØ
first case:
sinØ/cosØ = -5/14
tanØ = -5/14
Ø = 160.346° or Ø = 340.346°
for Ø = 160.346° , y = 0
for Ø = 340.346° , y = 0
second case
14cosØ = 5sinØ
14/5 = sinØ/cosØ
tanØ = 14/5
Ø = 70.346° or Ø = 250.346°
for Ø = 70.346° , y = 221
for Ø = 250.346° , y = 221
the maximum value is 221
As θ ranges over all real values, what is the maximum value of
(14sinθ+5cosθ)^2
1 answer
2 answers