Hello, Bryankek!

Your work on #1 is correct . . . excellent work!

. . I didn't approximate like you did.

The Muhibbah Company is a manufacturer of cylindrical aluminium tins.

The manager plans to reduce the cost of production.

The production cost is proportional to the area of the aluminium sheet used.

The volume that each tin can hold is 1000 cm³(1 Litre)

1. Determine the value ofh,rand hence calculate the ratio ofh/r

when the total surface area of each tin is minimum.

Here,hdenotes the height andrthe radius of the tin.

My progress so far:

I assumed that Q1 can be solved by simultaneous equation so i got these

A = Area, h = height, r = radius

1000 = πr²h . → . h = 1000/πr² . (1)

sub (1) into: .A .= .2πrh + 2πr²

. . A .= .2πr(1000/πr²) + 2πr²

. . A .= .2000πr/πr² + 2πr²

. . A .= .2000r^{-1} + 2πr²

dA/dr .= .-2000/r² + 4πr .= .0

Multiply by r²: . -2000 + 4πr³ .= .0

. . 4πr³ .= .2000

. . . .r³ .= .1750/11 . . . ok

We have: .4πr³ .= .2000

. . . . . . . . . r³ .= .500/π

. . . . . . . . . .r .= .cbrt(500/π) .≈ .5.419

Then: .h .= .1000/πr² .= .1000/π[cbrt(500/π)]² .≈ .10.838

. . . . . . . . . . . . . . .h . . . 1000/πr² . . .1000

We have the ratio: . -- .= .---------- .= .------

. . . . . . . . . . . . . . .r . . . . . . r . . . . . . πr³

. . . . . . . . . . . . . . . . . . . . . . . .h . . . . 1000

Since r = cbrt(500/π), we have: . -- .= .----------- .= .2

. . . . . . . . . . . . . . . . . . . . . . . .r . . . π(500/π)

The height is always twice the radius.

The side view of the tin is asquare.