Im suddenly scared for o levels :(

Hey no, don't be scared! This is from polytechnic math module. This app doesn't have an option for polytechnic math, so I put it under this sec 4 category. (I don't remember having this kind of qn in O lvs when I was sec 4 that time) , so don't worry too much.

Oooooh

My god i was shaken buy the qn

Tks for informing hahahah

No problem. All the best for O lvs!

Thank u

Later I answer

Volume, V = πr²h where r is radius of cylinder and h is height of it.

330 = πr²h

So, h = 330/πr²

Then, surface area of unopened can (call it A)

A = Curved surface area + the 2 circular flat surfaces (one on each end of the can)

A = 2πrh + 2πr²

(Here you do everything in terms of r by changing the h into terms of r)

A = 2πr(330/πr²) + 2πr²

A = 660/r + 2πr²


This needs to be a minimum or minimised.
So find the stationary point where the rate of change of area with respect to radius is 0.
(i.e dA/dr = 0)


(Recall that stationary points can be maximum, minimum points or inflection points)


A = 660/r + 2πr²

A = 660r-¹ + 2πr²

dA/dr = -660r-² + (2)(2πr)

dA/dr = 4πr - 660/r²

When dA/dr = 0 ,

4πr - 660/r² = 0

4πr = 660/r²

4πr³ = 660

r³ = 660/4π = 165/π

r = ³√(165/π)

r ≈ 3.7449

r = 3.75 (3 s.f)


h = 330/πr²

= 330/(π × (³√(165/π))²)

≈ 7.4899

= 7.49 (3s.f)

Show that its a minimum : 2nd derivative test or 1st derivative test (follow Yong Kc's for the former)


By the way, this is pretty standard and similar to some types of question for O level A Math so Jesse, take note.

Zwen didn't take A Math so he/she wouldn't have experienced this in O levels.

Thank you for the clear workings.