more help maxima and minima (1 Viewer)

me dumb · Jul 14, 2009

i got a ridiculus answer of 123.6
could some1 plz explain it. thnx
part (iii)
this is the diagram, dnt worry bout the lil bit of scribble
View attachment 18852

questions here. also dnt worry bout the little scribble on this too, i was trying to find how to do the last part on (ii) and realised it was sum and difference of 2 cubes
View attachment 18853

kurt.physics · Jul 14, 2009

me dumb said:
i got a ridiculus answer of 123.6
could some1 plz explain it. thnx
part (iii)
this is the diagram, dnt worry bout the lil bit of scribble
View attachment 18852

questions here. also dnt worry bout the little scribble on this too, i was trying to find how to do the last part on (ii) and realised it was sum and difference of 2 cubes
View attachment 18853

$2t + 2 + h = 12$

$\implies h = 10 - 2t$

Also

$V = \left(\frac{4\pi h}{3}\right)\left(t^2 + t + 1\right)$

Substituting h into the equation above

$\therefore V = \left(\frac{4\pi \left(10 - 2t\right)}{3}\right)\left(t^2 + t + 1\right)$

Let $u = \frac{4\pi \left(10 - 2t\right)}{3}$ and $v = t^2 + t + 1$

$u' = \frac{d}{dx}\left(- \frac{8t\pi}{3}\right) = \frac{-8\pi}{3}$

$v' = 2t + 1$

$y' = uv' + vu'$

$= \frac{4\pi \left(10 - 2t\right)}{3} \cdot \left(2t + 1\right) + \left(t^2 + t + 1\right) \cdot \frac{-8\pi}{3}$

For max/min y' = 0

ie

$\frac{4\pi \left(10 - 2t\right)}{3} \cdot \left(2t + 1\right) + \left(t^2 + t + 1\right) \cdot \frac{-8\pi}{3} = 0$

$4 \pi \left(10 - 2t\right)\left(2t + 1\right) -8 \pi \left(t^2 + t + 1\right) = 0$

$8 \pi \left(5 - t\right)\left(2t + 1\right) -8 \pi \left(t^2 + t + 1\right) = 0$

$\left(5 - t\right)\left(2t + 1\right) - \left(t^2 + t + 1\right) = 0$

$- 2t^2 + 9t + 5 - t^2 - t - 1 = 0$

$-3t^2 + 8t + 4 = 0$

$t = \frac{-8 \pm \sqrt{112}}{-6}$

$\approx -0.43, 3.097$

As t>0,

then $t = 3.097$

By checking both sides we can see that t = 3.097 is a max

When t = 3.097,

$V = 218.23\,\,m^3$

Aquawhite · Jul 14, 2009

Nice one kurt ^_^

me dumb · Jul 14, 2009

kurt.physics said:
$2t + 2 + h = 12$

$\implies h = 10 - 2t$

Also

$V = \left(\frac{4\pi h}{3}\right)\left(t^2 + t + 1\right)$

Substituting h into the equation above

$\therefore V = \left(\frac{4\pi \left(10 - 2t\right)}{3}\right)\left(t^2 + t + 1\right)$

Let $u = \frac{4\pi \left(10 - 2t\right)}{3}$ and $v = t^2 + t + 1$

$u' = \frac{d}{dx}\left(- \frac{8t\pi}{3}\right) = \frac{-8\pi}{3}$

$v' = 2t + 1$

$y' = uv' + vu'$

$= \frac{4\pi \left(10 - 2t\right)}{3} \cdot \left(2t + 1\right) + \left(t^2 + t + 1\right) \cdot \frac{-8\pi}{3}$

For max/min y' = 0

ie

$\frac{4\pi \left(10 - 2t\right)}{3} \cdot \left(2t + 1\right) + \left(t^2 + t + 1\right) \cdot \frac{-8\pi}{3} = 0$

$4 \pi \left(10 - 2t\right)\left(2t + 1\right) -8 \pi \left(t^2 + t + 1\right) = 0$

$8 \pi \left(5 - t\right)\left(2t + 1\right) -8 \pi \left(t^2 + t + 1\right) = 0$

$\left(5 - t\right)\left(2t + 1\right) - \left(t^2 + t + 1\right) = 0$

$- 2t^2 + 9t + 5 - t^2 - t - 1 = 0$

$-3t^2 + 8t + 4 = 0$

$t = \frac{-8 \pm \sqrt{112}}{-6}$

$\approx -0.43, 3.097$

As t>0,

then $t = 3.097$

By checking both sides we can see that t = 3.097 is a max

When t = 3.097,

$V = 218.23\,\,m^3$

ahh damn just a simple error. thats what i did but somewhere i made a mistake expanding hte bracket

me dumb · Jul 14, 2009

woops i didnt say thnx.
Thankyou

Sleight_of_Hand · Jul 27, 2009

how did u solve part 2? i cant seem to get (t^2 + t + 1)... i keep gtting (t^2/t-1 + t^2 - 1/t-1)

Sleight_of_Hand · Jul 27, 2009

nvm solved it

Lukybear · Jul 28, 2009

How did you get i)? I kept getting x = h/t

Lukybear · Jul 28, 2009

more help maxima and minima (1 Viewer)

me dumb

Member

kurt.physics

Member

Aquawhite

Retiring

me dumb

Member

me dumb

Member

Sleight_of_Hand

Member

Sleight_of_Hand

Member

Lukybear

Active Member

Lukybear

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)