Question-88218

Question Number 88218 by mathocean1 last updated on 09/Apr/20

Commented by mathocean1 last updated on 09/Apr/20

This is face view of an evacuation

canal . It has a trapeze form . 4 m

represents its small base .

1) Determinate θ \in [60; 90] such as the

capacity of canal is maximal .

Commented by mathocean1 last updated on 09/Apr/20

Answered by ajfour last updated on 12/Apr/20

A = \frac{1}{2} (2 \sin θ) (4 + 4 + 4 \cos θ)

\Rightarrow A = 8 \sin θ + 4 \sin θ \cos θ

\frac{d A}{d θ} = 8 \cos θ + 4 \frac{d}{d θ} (\sin θ \cos θ)

= 8 \cos θ + 4 {\cos θ \frac{d}{d θ} (\sin θ) + \sin θ \frac{d}{d θ} (\cos θ)}

= 8 \cos θ + 4 (\cos^{2} θ - \sin^{2} θ)

\frac{d A}{d θ} = 8 \cos θ + 4 \cos^{2} θ - 4 \sin^{2} θ = 0

l e t \cos θ = t

8 t + 4 t^{2} + 4 t^{2} - 4 = 0

2 t^{2} + 2 t - 1 = 0

t = - \frac{1}{2} + \frac{\sqrt{3}}{2}

θ = \cos^{- 1} (\frac{\sqrt{3} - 1}{2}) \approx 68.5293 ° .

Commented by mathocean1 last updated on 12/Apr/20

please sir can you explain the 2^{nd} line .

Commented by mathocean1 last updated on 12/Apr/20

Thanks sir i have understood \dots

{it}^{'} s great .