the math, it burns!

[thubby]

so i have an optimization problem I can't seem to make work.

a box has a combined length and girth (perimeter of a cross-section) of 108 inches (figure of a long rectangular box with the part facing us being a square with a side length of x, and the length of the box as y *y is noticeably larger than x*)

find the dimensions that maximize volume.

i know, courtesy of the back of my book, that the box is 18x18x36, but i can't work it backwards either >,<

please help my poor, abused, brain. :smallfrown:

[absolmorph]

Perimeter of the girth: 18*4 = 72
72 + 36 = 108
18^2 = 324
324 * 36 = 11664

If you increase the perimeter of the girth by 4 (each side by 1), you get 11552. If you decrease the perimeter by 4, you get 11560.

[Manga Shoggoth]

More complex solution (which doesn't involve already knowing the answer...):

Volume [V] = x^2 * y (1)

Other known value is 4x + y = 108 (2)

We wish to maximixe volume.

First, get y in terms of x:

Rearrainging (2) gives y = 108 - 4x (3)


Putting (3) into (1) we have:

V = x^2 * (108 - 4x) = 108x^2 - 4x^3

We want the maximum volume, so differentiate with respect to V and equate to zero (the maxima and/or minima occur when dV/dx = 0):

dV/dx = 216x - 12x^2; and so we have 216x - 12x^2 = 0

From which, 216x = 12x^2
and then, 216 = 12x
and finally, x = 216/12 = 18

Now that we have x at the point of maximum volume, we can get y from (2):

4x + y = 108

Thus; y = 108 - 4x = 108 - 72 = 36

So, x = 18; y = 36 and the volume you want is 18 * 18 * 36 = 16644

(What's stunning is that I can still differentiate ay my age...)

[thubby]

thank you Manga Shoggoth!
I was horribly over-thinking it >,<

in an unrelated note, is there any way to make a ti-84 use an answer it got previously other than the most recent one?

[Manga Shoggoth]

thank you Manga Shoggoth!
I was horribly over-thinking it >,<

You're welcome. You should have seen my first attempt at the answer...

[Player_Zero]

in an unrelated note, is there any way to make a ti-84 use an answer it got previously other than the most recent one?

Yes. Read your instructions manual.

[thubby]

Yes. Read your instructions manual.

i would if i hadn't lost it 4 years ago(?):smallfrown:

[drakir_nosslin]

^This (http://education.ti.com/educationportal/sites/US/productDetail/us_ti84pse.html?bid=6) should fix your problem then.

[Castaras]

i would if i hadn't lost it 4 years ago(?):smallfrown:

Shift-Enter shift-enter until you get to it.

[Deathslayer7]

More complex solution (which doesn't involve already knowing the answer...):

Volume [V] = x^2 * y (1)

Other known value is 4x + y = 108 (2)

We wish to maximixe volume.

First, get y in terms of x:

Rearrainging (2) gives y = 108 - 4x (3)


Putting (3) into (1) we have:

V = x^2 * (108 - 4x) = 108x^2 - 4x^3

We want the maximum volume, so differentiate with respect to V and equate to zero (the maxima and/or minima occur when dV/dx = 0):

dV/dx = 216x - 12x^2; and so we have 216x - 12x^2 = 0

From which, 216x = 12x^2
and then, 216 = 12x
and finally, x = 216/12 = 18

Now that we have x at the point of maximum volume, we can get y from (2):

4x + y = 108

Thus; y = 108 - 4x = 108 - 72 = 36

So, x = 18; y = 36 and the volume you want is 18 * 18 * 36 = 16644

(What's stunning is that I can still differentiate ay my age...)

exactly as he did it. Remember that max and mins occur when the derivative of a function =0 or undefined.