Optimization:

1) In Physics, an object thrown vertically upward will always start to descend once it reaches its maximum height. Determine the time of ascent if the initial speed is V_o and the acceleration due to gravity is 1/2 of that of Earth's

2) The total surface area of an industrial cylindrical autoclave is to be limited to 20m². If it is constructed to maximize the volume, what should be the ratio of the height to that of the radius?

3) Based on the concept of the previous number. If someone asks you for help in constructing a cylindrical tank with a top cover area of 100 cm², accounting the minimum surface area and maximum volume, what would you recommend for its lateral area?

Can you include the whole process and what formula did you use, so that I can understand it more? Thank you so much

Q1)

?  t  = g2Vo??? is the time of ascent

Q2)

?r∗h∗?   =  2? is the ratio for max volume with surface area equal to 20 m^2

Q3)

lateral area should be ?A=400 cm2?

Step-by-step explanation

Q1)

 The differential equation for this case is

?dtdV?  = a?

where

V : velocity of the object at time t

a : acceleration of the object

Let upward be the positive direction so in our case  ?a= −21?g?  where g is acceleration due to gravity on earth

→  ?dtdV? = −21?g?

Integrating

?∫Vo?V? dV = ∫0t?−21?g dt?

→ ?V−Vo? = −21?g(t−0)?

At maximum height the velocity must go to zero so ?V=0?

→ ?−Vo? = −21?gt  →  t  = g2Vo???   is the time of ascent ANS

Q2)

Volume of a cylinder is given by  ?V = πr2h?

and surface area of cylinder is given by ?S = 2πr2 +2πrh?

now we are given that ?S = 20?

→ ?20 = 2πr2+2πrh?

→  ?h = 2πr20−2πr2? = πr10−πr2??       A

→ ?V =   πr2 (πr10−πr2?)?

=  10r − πr3?

let ?r=r∗?  be radius that maximizes the volume 

so using first order and second order test we have ?drdV?(r∗)=0?  and ?dr2d2V?(r∗)<0?

drdV? = drd?(10r−πr3) = 10−3πr2  ?

→ ?10−3π(r∗)2  =0  → r∗ = 3π10???

and  ?dr2d2V? = drd?(10−3πr2) = −6πr <0  ∀ r>0?

hence ?r∗ = 3π10???   maximizes the volume

now from relation A,  height  ?h∗ =  π 3π10??10−π (3π10??)2?  = π 3π10??320?? = π 3π10??2 ×310?? = π?2?310???

Therefore ratio of height to radius is  ?= r∗h∗? =  3π10??π?2?310???  =  2? ANS

Q3)

 now we have top cover area ,  ?πr2 = 100  →  r=  π100?? = π?10??

using the result of Q2 we have ?rh? =2  →  h = 2r ?

→ ?h=  2  π?10? = π?20?  cm?

so lateral area is  ?A = 2πrh  = 2π  π?10? .π?20? = 400   cm2  ? ANS