S = 6hs+(3s^2/2){[sqrt(3)-cos(θ)]/(sinθ)} and h and s are positive constants. Find the angle θ that minimizes the surface area S. π/6 ≤ θ ≤ π/2

Question

ikram002p · Answer

$$S = 6hs+(3s^2/2){[\sqrt(3)-\cos(θ)]/(\sinθ)}   $$

ikram002p · Answer

u need to find   $$S'=\frac{ d \theta }{ dS } $$
and see wich values that make S'(theta) =0.

anonymous · Answer

Okie dokie thank you @ikram002p

ikram002p · Answer

so did u got it ?!

anonymous · Answer

@ikram002p I haven't started it yet. What do you do about the h and s? That's the only part I don't understand.

ikram002p · Answer

will do nothing about them coz they are constants ,treat them like any number ok ?!

ikram002p · Answer

u can show me ur work

anonymous · Answer

@ikram002p Okay! Makes sense. I'll try and do it later. Thank you!

ikram002p · Answer

ok , good luck then.