Arain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of the sheet on each side through an angle θ. how should θ be chosen so that the gutter will carry the maximum amount of water?

  θ = 60°

Step-by-step explanation:

The cross sectional area of the trapezoid shape will be that of a trapezoid with bases of 10 cm and (10 cm + 2·(10 cm)·cos(θ)) and height (10 cm)·sin(θ).

That area in cm² is ...

  A = (1/2)(b1 +b2)h = (1/2)(10 + (10 +20cos(θ))(10sin(θ)

  A = 100sin(θ)(1 +cos(θ))

A graphing calculator shows this area to be maximized when ...

  θ = π/3 radians = 60°

_____

A will be maximized when its derivative with respect to θ is zero. That derivative can be found to be 2cos(θ)² +cos(θ) -1, so the solution reduces to ...

  cos(θ) = 1/2

  θ = arccos(1/2) = π/3

answer: a and c are correct

step-by-step explanation:

the +5 is the vertical translation

the added negative symbol in front is the reflection over the y-axis

can you put the picture up?

step-by-step explanation: