Thread: Calc 3 - Lagrange multipliers

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?

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

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

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

Since a specific value of $\displaystyle \lambda$ 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 $\displaystyle \lambda$.

Here, dividing $\displaystyle \frac{bc}{6}=- \frac{\lambda}{a^2}$ by $\displaystyle \frac{ac}{6}= -\frac{\lambda}{b^2}$ we get $\displaystyle \frac{b}{a}= \frac{a^2}{b^2}$ so that $\displaystyle b^3= a^3$ 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 $\displaystyle c^3= a^3$ so that c= a.

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