Calc 3 - Lagrange multipliers

**EliteNewbz** · Feb 25th 2011, 12:51 PM

A plane with the equation

(a,b,c > 0) together with the positive coordinate planes forms a tetrahedron of volume

. Find the plane that minimizes V if the plane is constrained to pass through the point P = (1,1,1).

I'm not sure how to this, trying for

and

I got a=b=c. Which I don't think is right after using Lagrange multipliers. help!

= [(1/6)bc, (1/6)ac,(1/6)ab] and I set it to equal each other :/ help?

**TheEmptySet** · Feb 25th 2011, 01:37 PM

see this previous post
http://www.mathhelpforum.com/math-he...ne-165570.html

**HallsofIvy** · Feb 26th 2011, 03:19 AM

So your target function is $f(a,b,c)= \frac{1}{6}abc$ and the constraint is $g(x,y,z)= \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}= a^{-1}+ b^{-1}+ c^{-1}= 1$ .

Yes, $\nabla f= \frac{bc}{6}\vec{i}+ \frac{ac}{6}\vec{j}+ \frac{ab}{6}\vec{k}=\lambda\nabla g= \lambda\left(-a^{-2}\vec{i}- b^{-2}\vec{j}- c^{-2}\vec{k}\right)$
so we must have $\frac{bc}{6}= -\frac{\lambda}{a^2}$ , $\frac{ac}{6}= -\frac{\lambda}b^{-2}$ , and $\frac{ab}{6}= -\frac{\lambda}{c^2}$ .

Since a specific value of $\lambda$ is not a part of the solution, I find that, for many Lagrange multiplier problems, the best first step is to divide one equation by another, eliminating $\lambda$ .

Here, dividing $\frac{bc}{6}=- \frac{\lambda}{a^2}$ by $\frac{ac}{6}= -\frac{\lambda}{b^2}$ we get $\frac{b}{a}= \frac{a^2}{b^2}$ so that $b^3= a^3$ and either a= b or a= -b. Since these numbers are positive, we must have a= b. Similarly, dividing the first equation by the third gives $c^3= a^3$ so that c= a.

That is, you were correct that a= b= c. And, of course, the condition that $\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}= a^{-1}+ b^{-1}+ c^{-1}= 1$ becomes $\frac{1}{a}+ \frac{1}{a}+ \frac{1}{a}= \frac{3}{a}= 1$ and so a= b= c= 3.

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