My Math Forum cylinder inscribed in sphere with given equation

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 October 14th, 2012, 04:12 AM #1 Newbie   Joined: Oct 2012 Posts: 5 Thanks: 0 cylinder inscribed in sphere with given equation Hi guys, i'm in some trouble right now so anyway, this is what the question is. Use the method of Lagrange multipliers to find the cylinder of maximum volume that can be inscribed inside an AFL style mathematical football with the geometric equation, f(x,y,z) = 15x^2 + 15y^2 + 4z^2 = 15. without lose of generality, let the axis of the cylinder be the z axis. a) Using r^2 = x^2 + y^2 , restate f(x,y,z) = 15 as g(r,z) = 15. b) show that the volume of an inscribed cylinder , V(r,z) = 2 pi R^2 z. c) Set-up and solve the Lagrange multiplier equations that maximises V(r,z) subject to g(r,z) = 15. d) state the exact and approximate numberical values of (i) radius, (ii) height, (iii) volume of the inscribed cylinder of maximum volume. SO, does "let the axis of the cylinder be the z axis" mean that z would be the height? if not, I know my function to be maximised is V = pi r^2 h, and my working for part a) is, R^2 + (h/2)^2 - r^2 = 15 (R being the radius of the cylinder and r being the radius of the sphere) i've managed to work it out by setting R^2 as the subject and plugging it back into my volume equation, but since this part i seem to be lost. I also don't fully understand the question, am I right to set my part (a) equation to 15? cause if I am right, i can't seem to prove part (b) , and of course can't move on.
 October 14th, 2012, 10:48 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: cylinder inscribed in sphere with given equation We are given the ellipsoid: $f(x,y,z)=15x^2+15y^2+4z^2=15$ a) Using $r^2=x^2+y^2$ we may then write: $f(r,z)=15r^2+4z^2=15$ b) The base of the cylinder is $\pi r^2$ and the height is $z-(-z)=2z$, so we have: $V(r,z)=2\pi r^2z$ c) We want to maximize: $V(r,z)=2\pi r^2z$ subject to the constraint: $f(r,z)=15r^2+4z^2-15=0$ Using the theorem of Lagrange, we find: $4\pi rz=\lambda 30r$ $2\pi r^2=\lambda 8z$ If we observe that $(r,z)=(0,0)$ represents the minimum for $V$ then we are left with: $z^2=\frac{15}{8}r^2$ and substituting this into the constraint, we find: $15r^2+4$$\frac{15}{8}r^2$$-15=0$ which means: $r^2=\frac{2}{3}$ and $z^2=\frac{5}{4}$ d) Thus, for the maximized cylinder subject to the given constraint, we have: radius: $r=\sqrt{\frac{2}{3}}\approx0.816496580928$ height: $z-(-z)=2\cdot\frac{\sqrt{5}}{2}=\sqrt{5}\approx2.23606 79775$ and thus, the maximum volume is: $V_{\text{max}}=V$$\sqrt{\frac{2}{3}},\frac{\sqrt{5 }}{2}$$=2\pi$$\frac{2}{3}$$$$\frac{\sqrt{5}}{2}$$= \frac{2\pi\sqrt{5}}{3}\approx4.68320982069$
 October 15th, 2012, 06:43 AM #3 Newbie   Joined: Oct 2012 Posts: 5 Thanks: 0 Re: cylinder inscribed in sphere with given equation gosh I feel so stupid right now, thanks!!

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