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Math Help - Problem on Finding Minimum Value - Multivariable Calculus

  1. #1
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    Problem on Finding Minimum Value - Multivariable Calculus

    Hello everyone,

    I am having trouble understanding a solution to the following problem in multivariable calculus. Could someone please answer my question posted beneath the actual problem and solution below?

    Thank you very much!

    --

    Problem: Find an equation of the plane that passes through the point (1, 2, 3) and cuts off the smallest volume in the first octant.

    Solution:Please see below.

    Question:
    Could someone please explain the step in red? I don't understand why the equation of the tetrahedron (which seems to be something fixed) is differentiated with respect to a and b, the x- and y-intercepts of the tetrahedron?

    I know that doing so gives you two constraints for this minimization problem, but why else for these differentiations?
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  2. #2
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    Yes, the equation of the tetrahedron is fixed.
    However, for a small change in "a" (holding "b" fixed) there must be a corresponding change in "c" so that the tetrahedron still passes through (1,2,3).
    This is what that step is trying to find (using the chain rule).
    If you solved for "c" in terms of "a" and "b", and took the partials, it would give the same answer.
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  3. #3
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    Thanks for your reply, snowtea.

    However, I still don't understand the meaning behind this differentiation. If this equation of the tetrahedron is fixed (including its coordinate intercepts), then why are partial derivatives of it taken with respect to a and b?

    Also, you explain that "For a small change in a (holding b fixed), there must be a corresponding chance in c...| However, if the equation of the tetrahedron is fixed, how can these changes happen?
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  4. #4
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    I think you confusion is coming from the fact that you think a,b,c are fixed.
    This is not the case. The only condition is that the plane must pass through (1,2,3) and cut the first octant.
    Think of the plane as pivoting about (1,2,3). i.e. take a piece of cardboard and balance it on a pen. The cardboard still has freedom to move even though it must touch the pen at a point. If you tilt one side of the cardboard the other side will move correspondingly. The partial derivatives are trying characterize this relation. Does it make a little bit more sense now? Think about it for a while.
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  5. #5
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    The general "form" of the equation is fixed: any plane that intercepts the three axes at (a, 0, 0), (0, b, 0), and (0, 0, c) must have equation [tex]\frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= 1[tex]. You can see that by observing that
    1) That is the equation of some plane since it is linear is x, y, and z.
    2) A plane is determined by three points.
    3) (a, 0, 0), (0, b, 0), and (0, 0, c) all satisfy that equation.

    But the individual values of a, b, and c are NOT "fixed". You can "rotate" the plane about the given point (1, 2, 3) (How original mathematicians are!) so that the plane intersect the axes in different places- different values for a, b,and c.
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