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Math Help - Calc 3 - Lagrange multipliers

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    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?
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  3. #3
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    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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