Maximizing the surface of a cylinder and a box

[leprofece]

¿The post office has established that the length and the outline of any parcel may exceed the 100 cm. under such restriction find the dimensions for:
299) circular cylinder straight greater possible surface.
Answer R = 50/(2pi-1) H= 100(pi-1)/(2pi-1)

301) rectangular box of square base of greater surface area.
Answer Length 300/7
width Lado 100/7
There are this formula in my book
299) a = 2piRH
H = 100 -2piR
But I don't get the book answer R = 25/pi

301) H = 100 -2piR
and a = 2pirh +2pir²

 

[Prove It]

re: Max and min 299 y 301

Do you mean that the length and outline of the parcel may NOT exceed 100cm?

 

[leprofece]

re: Max and min 299 y 301

do you mean that the length and outline of the parcel may not exceed 100cm?

yeah

 

[HallsofIvy]

re: Max and min 299 y 301

¿The post office has established that the length and the outline of any parcel may exceed the 100 cm. under such restriction find the dimensions for:
299) circular cylinder straight greater possible surface.
Answer R = 50/(2pi-1) H= 100(pi-1)/(2pi-1)

301) rectangular box of square base of greater surface area.
Answer Length 300/7
width Lado 100/7
There are this formula in my book
299) a = 2piRH
H = 100 -2piR
But I don't get the book answer R = 25/pi

That "a" is for the curved surface. The total surface area includes the two circular ends: A= 2pi RH+ 2pi R^2. Replacing H with 100- 2pi R,
A= 2pi R(100- 2piR)+ 2pi R^2= 200pi R- 4pi^2R^2+ 2piR^2= 200pi R- (4pi^2- 2pi)R^2.

You don't say how you are to find the maximum area. Completing the square would work but be tedious. Taking the derivative and setting it equal to 0 give 200pi - (8pi^2- 4pi)R= 0, R= (200pi)/(8pi^2- 4p)= 25/(pi- 1)

301) H = 100 -2piR
and a = 2pirh +2pir²

? 301 was about a rectangular box. Where did "pi" come from? If a box has length L, width W, and height H, then its surface area is A= 2LW+ 2LH+ 2WH. The condition that "total outline plus length be no more than 100" gives 2W+ 2H+ L= 100 at most.

 

[leprofece]

re: Max and min 299 y 301

That "a" is for the curved surface. The total surface area includes the two circular ends: A= 2pi RH+ 2pi R^2. Replacing H with 100- 2pi R,
A= 2pi R(100- 2piR)+ 2pi R^2= 200pi R- 4pi^2R^2+ 2piR^2= 200pi R- (4pi^2- 2pi)R^2.

You don't say how you are to find the maximum area. Completing the square would work but be tedious. Taking the derivative and setting it equal to 0 give 200pi - (8pi^2- 4pi)R= 0, R= (200pi)/(8pi^2- 4p)= 25/(pi- 1)? 301 was about a rectangular box. Where did "pi" come from? If a box has length L, width W, and height H, then its surface area is A= 2LW+ 2LH+ 2WH. The condition that "total outline plus length be no more than 100" gives 2W+ 2H+ L= 100 at most.

Ok I solved 299
but in 301 it is a rectangular box whose bass is a square so L = w = y h
the other equation must be 2L²+4LH
let see if i got the answer 2w+2h+L = 100
si l = w I got 3l + 2h = 100
100- 3L /2 = H
No I got 5

 

[leprofece]

Re: Max and min 299 y 301

Am I right or not?
Who has the reason? me orb the book?