Problems with geometric application of calculas (1 Viewer)

[bmn]

I have 2 problems, although they double up in places...

1) Soft can drink - Volume is 375ml, looking for minimum.
a) Show surface area is given by S=2nr^2 + 750/r (n= pi) - I done this with ease
b) Find the radius of the can that gives the minimum surface area. - I think I know what to do, just can't do it.

2) A 5m length of timber is used to border a triangular garden bed, with the other sides of the garden against house walls.
a) Show that the area is A = 1/2x(25-x^2)^1/2 - Once again done with ease
b) Find the greatest possible area of the garden bed. - same as above.

 

[gibbo153]

I have 2 problems, although they double up in places...

1) Soft can drink - Volume is 375ml, looking for minimum.
a) Show surface area is given by S=2nr^2 + 750/r (n= pi) - I done this with ease
b) Find the radius of the can that gives the minimum surface area. - I think I know what to do, just can't do it.

2) A 5m length of timber is used to border a triangular garden bed, with the other sides of the garden against house walls.
a) Show that the area is A = 1/2x(25-x)^1/2 - Once again done with ease
b) Find the greatest possible area of the garden bed. - same as above.

to get the max/min areas of things, differentiate S or A. for part 1b) make S' = 0 to find stationary points. use S" to determine their nature. in 1b) you are wanting the minimum area so S" will be >0 (concave up).

in 2b) do the same, obviously you will be looking for the max value (maximum turning point), so A" will be <0

 

[bmn]

to get the max/min areas of things, differentiate S or A. for part 1b) make S' = 0 to find stationary points. use S" to determine their nature. in 1b) you are wanting the minimum area so S" will be >0 (concave up).

in 2b) do the same, obviously you will be looking for the max value (maximum turning point), so A" will be <0

I know. I probably didn't make it clear. My problem is after the differentiation - making it = 0 when I have negative/fraction powers...

 

[Timothy.Siu]

S=2nr^2 + 750/r (n= pi)
S'=4nr-750/r^2
S'=0 when 4nr-750/r^2=0

4nr=750/r^2
4nr^3=750
r^3=750/4n
r=cuberoot(750/4n)

 

[bmn]

S=2nr^2 + 750/r (n= pi)
S'=4nr-750/r^2
S'=0 when 4nr-750/r^2=0

4nr=750/r^2
4nr^3=750
r^3=750/4n
r=cuberoot(750/4n)

My mistake on that one, I was getting that but checking it with answers of the other question... :rofl: Still need help on the other, I was getting stationary points of 0/+-5... which weren't working at all...

 

[Timothy.Siu]

A = 1/2x(25-x^2)^1/2
A'=0.5(-2x^2/((25-x^2)^1/2) + (25-x^2)^1/2 ) [using product and chain rule]

A'=0 when 1/((25-x^2)^1/2) [-2x^2+25-x^2]=0

-3x^2+25=0
x^2=25/3
x=root(25/3)

lol i hope thats right
btw, i dont understand the question!