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Maximizing the surface of a cylinder and a box

leprofece

Member
Jan 23, 2014
241
¿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 dont get the book answer R = 25/pi

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

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
re: Max and min 299 y 301

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

leprofece

Member
Jan 23, 2014
241

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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 dont 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

Member
Jan 23, 2014
241
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

Member
Jan 23, 2014
241
Re: Max and min 299 y 301

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