Max/Min (1 Viewer)

[Kutay]

Hey i was wondering if anyone could help me with this question

" A piece of cardboard measuring 60cm by 60cm has 4 squares of side x cm cut from each corner. The remaining cardboard is bent to from a box with a square base. Find the greatest volume this box can enclose'

 

[table for 1]

that Q sounds familiar. except i think i did the rectangle one

anyways, i hate these Q's, but i'll try to help.

we can say that the base of the box has lengths
L = 60 - 2x
so it's area is
A = (60 - 2x)²
so it's volume is
V = x(60 - 2x)²
ie. V = x(360 - 240x +4x²)
V = 4x³ - 240x² + 360x

for max and min problems, you differentiate and let it = to 0
so, differentiating V = 4x³ - 240x² + 360x
gives,
V' = 12x² - 480x + 360 = 0
x² - 40x + 30 = 0
quadratic eqn...
x = { -b + √[b² - 4ac] } / 2a
x = { 40 + √[40² - 4.30] } / 2
x = { 40 + √[1480] } /2
x = { 40 + √[370] } /2
x = 20 + √[370]
x = 0.764..., 39...

if x = 39..., then L = 60 - 2X39... = -18
and since length cannot be negative,
the only value for x is 0.764...
.'. x = 20 + √[370]

sub x into V
V = 4x³ - 240x² + 360
V = 4(20 + √[370]³ - 240(20 + √[370]² + 360(20 + √[370])
V = 136.7368.....
V = 136.74 cm³ (2 d.p)
.'. max volume the box can hold is 136.74 cm³

so yeah...i think it's best you check the answer and see if i got the same thing before you try to understand what i wrote, because i really suck at these questions..... i don't know how the code to make the root sign, so i'll try to do it on paint if you need it.

 

[Xayma]

The code for √ is & radic; without the space between & and radic;

 

[table for 1]

hey ! check my working ! i seriously think i would have done something wrong. i don't have much experience in these type of Qs

does the root sign extend to cover a number of values?

i haven't bothered to start on the paint thing yet. i guess i should start now. sigh...cbb

 

[table for 1]

TESTING

x = { -b + √[b² - 4ac] } /2a

i don't know how to extend the top...it's not working

oh well, it looks better than 'root'. so i'll change it anyway

 

[physician]

that Q sounds familiar. except i think i did the rectangle one

anyways, i hate these Q's, but i'll try to help.

we can say that the base of the box has lengths
L = 60 - 2x
so it's area is
A = (60 - 2x)²
so it's volume is
V = x(60 - 2x)²
ie. V = x(360 - 240x +4x²)

it should be V= x(3600 - 240x + 4x²)

(60 - 2x)² = the above in brackets

 

[physician]

for max and min problems, you differentiate and let it = to 0
so, differentiating V = 4x³ - 240x² + 360x
gives,
V' = 12x² - 480x + 360 = 0
x² - 40x + 30 = 0
quadratic eqn...
x = { -b + √[b² - 4ac] } / 2a
x = { 40 + √[40² - 4.30] } / 2
x = { 40 + √[1480] } /2
x = { 40 + √[370] } /2
x = 20 + √[370]
x = 0.764..., 39...

for maximum don't u find the second derivative?...

then u'll end up with

d2y/dx² = 12x - 480 ...... therefore x = 40

 

[physician]

for maximum don't u find the second derivative?...

then u'll end up with

d2y/dx² = 12x - 480 ...... therefore x = 40

actually i'm not too sure on this.. seeing as though lengh would be negative...

my brain is currenlty out of order.... could u plz post up the answer... kutay

 

[physician]

that Q sounds familiar. except i think i did the rectangle one

anyways, i hate these Q's, but i'll try to help.

we can say that the base of the box has lengths
L = 60 - 2x
so it's area is
A = (60 - 2x)²
so it's volume is
V = x(60 - 2x)²
ie. V = x(360 - 240x +4x²)
V = 4x³ - 240x² + 360x

for max and min problems, you differentiate and let it = to 0
so, differentiating V = 4x³ - 240x² + 360x
gives,
V' = 12x² - 480x + 360 = 0
x² - 40x + 30 = 0
quadratic eqn...
x = { -b + √[b² - 4ac] } / 2a
x = { 40 + √[40² - 4.30] } / 2
x = { 40 + √[1480] } /2
x = { 40 + √[370] } /2
x = 20 + √[370]
x = 0.764..., 39...

if x = 39..., then L = 60 - 2X39... = -18
and since length cannot be negative,
the only value for x is 0.764...
.'. x = 20 + √[370]

sub x into V
V = 4x³ - 240x² + 360
V = 4(20 + √[370]³ - 240(20 + √[370]² + 360(20 + √[370])
V = 136.7368.....
V = 136.74 cm³ (2 d.p)
.'. max volume the box can hold is 136.74 cm³

so yeah...i think it's best you check the answer and see if i got the same thing before you try to understand what i wrote, because i really suck at these questions..... i don't know how the code to make the root sign, so i'll try to do it on paint if you need it.

lol.... the way i worked this question out.. i ended up with 16000 cm³ as a maximum volume

i think ur's sounds a little more realistic...

 

[physician]

that Q sounds familiar. except i think i did the rectangle one

yeh rectangle one... page 239 question 8 '3 unit mathematics book 1.. Jones/Couchman'...

 

[Kutay]

lol.... the way i worked this question out.. i ended up with 16000 cm³ as a maximum volume

i think ur's sounds a little more realistic...

Yeh i redid it jsut then i also got 16000cm^3

Thankz for all the help i will make suer i will get the answer for you guys but i am pretty sure that the answer is 16000cm^3

 

[physician]

Yeh i redid it jsut then i also got 16000cm^3

Thankz for all the help i will make suer i will get the answer for you guys but i am pretty sure that the answer is 16000cm^3

ok... cool... yeh check it up... and if 16000 is incorrect post up the answer and i'll try help out from there...

 

[Kutay]

Yes the correct answer is 16000cm^3 good work

 

[physician]

Yes the correct answer is 16000cm^3 good work

so i'd take it that u now understand how to do the question....
by the way.. did u use the quadratic equation to find the values for 'x'..because it is Unnecessary ... just incase u may have missed this during ur solving.. although u could use the quadratic equation.. but it'll save u time not to...

 

[table for 1]

it should be V= x(3600 - 240x + 4x²)

(60 - 2x)² = the above in brackets

oh nuts. thanks for spotting that mistake.

anyways, i redid the question from that point the same way, and cool, i got the same answer as you guys ! so i guess that's good, except i'm so screwed for the upcoming test.

yeh rectangle one... page 239 question 8 '3 unit mathematics book 1.. Jones/Couchman'...

nah, i got my Q on some random exercise on a sheet of paper. but i'm pretty sure all the textbooks would have a rectangle variation of those Q's.

so i'd take it that u now understand how to do the question....
by the way.. did u use the quadratic equation to find the values for 'x'..because it is Unnecessary ... just incase u may have missed this during ur solving.. although u could use the quadratic equation.. but it'll save u time not to...

yeah...i feel pretty stupid now. so much simplier to solve for x when
V= x(3600 - 240x + 4x²)rather than
V= x(360 - 240x + 4x²)
since x works out to be 10, instead of some long thing with the root

erm...yeah. i'm sorry ! i think i probably confused Kutay even more with that mistake which caused such a long working out! but yes, good that you get it now. if not, you could always ask again, and hope that someone who does decide to help doesn't stuff it up...!