Question

legrange multiplier? maximize volume of a cone with surface area of 8. given V=(1/3) pi*h*r^2 and S=8=pi(r)(sqrt(r^2 +h^2))

Answer 1

i know that i have to take the first partial derivatives of the function of volume and the surface are (constraint) but after i have the gradients, and the lambda... its a very confusing system of equations. looks like this i think

\[2/3 \pi rh=\lambda \pi (4r^3 +2\pi h^2r)/(\sqrt(r^4+h^2r^2)\]

\[1/3 \pi r^2=\lambda (2 \pi rh)/(\sqrt(r^2 +h^2)\]

\[8=\pi(r)(\sqrt(r^2 +h^2))\]

ideally i want to find what values of r and h make the statements above true...

Answer 2

\[L f(rh)=g(r,h)\]
\[F(r,h)=g(r,h)-L f(rh)\]
\[F(r,h)=\frac13\pi~r^2h-L(\pi ~r(r^2+h^2)^{1/2}-8)\]
\[F_r=\frac23\pi rh-L\pi(~(r^2+h^2)^{1/2}+\frac{r}{(r^2+h^2)^{1/2}})\]
\[F_r=\frac23\pi rh-L\pi(\frac{r^2+h^2+r}{\sqrt{r^2+h^2}})\]
this is a pain to type :)

Answer 3

http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx

thatll go over the basics of this real good

Answer 4

\[F(r,h)=\frac13\pi~r^2h-L(\pi ~r(r^2+h^2)^{1/2}-8)\]
\[F_h=\frac13\pi~r^2-L\pi ~r\frac{h}{\sqrt{r^2+h^2}}\]
\[F_L=8-\pi ~r\sqrt{r^2+h^2}\]

Answer 5

if i may add,
\[F_r=\frac{2}{3}\pi rh-L\left(\pi\sqrt{r^2+h^2}+\pi\frac{r^2}{\sqrt{r^2+h^2}}\right)\]

Answer 6

lol, i lost track of an r :) good spot

Answer 7

i believe in the end that h and r are framed in terms of L and that reduces the last equation to a single variable. but you are right, these are rather contorted

Answer 8

yeah, its not so much the calculus that bothers me about legrange multipliers... its the fact that there is no standard method to find values that satisfy the equations. the algebra gets me every time.

Answer 9

thanks so much for the attention, though. i wasnt expecting anyone to pick this up.

Answer 10

I used the last equation to say
\[ \sqrt{r^2+h^2} = \frac{8}{\pi r} \]
to get rid of the radicals in the other two equations

Answer 11

\[ \frac13\pi~r^2=L\pi ~r\frac{h}{\sqrt{r^2+h^2}}\]\[ \frac13~r=L ~\frac{h}{\sqrt{r^2+h^2}}\]\[8=\pi ~r\sqrt{r^2+h^2}\]
yeah, phi looks to have been quick on the keybothat oneard with

Answer 12

I used that in Fh to solve for lambda (L)

Answer 13

then all of that into Fr to get (if I did it correctly) h= sqrt(2)*r

Answer 14

laptop has a touchy touch pad, if it get hit while typing it goes to wherever the coursor is

Answer 15

to check the result,

\[8=\pi r\sqrt{r^2+h^2}\\64=\pi^2 r^2(r^2 + h^2)\\\frac{64}{\pi^2r^2}-r^2=h^2\]

now, \[V=\frac{1}{3}\pi r^2h\\9V^2=\pi ^2r^4h^2\]

sub \(h^2\) from the first equation into the second equation. it becomes univariate calculus.

Answer 16

thanks!!!