• anonymous
I have problems with question in supplemental problem set 2. In, F-3 the product of the intercepts are D = c^3/(ab) where c = 1-2a-b and where z = ax + by + c. The partial derivative of D with respect to a I found to be b(b-4a-1)(1-2a-b)^2 =0 and the derivative of D with respect to b I found to be a(2a-2b-1)(1-2a-b)^2 = 0. How do I proceed to find out the values of a, b and c that give the minimum product of the intercepts?
OCW Scholar - Multivariable Calculus
  • Stacey Warren - Expert
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  • jamiebookeater
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  • IrishBoy123
would like to help but have no access to original question
  • phi
In the form z= ax + by+ c the (stipulated positive) intercepts on the 3 axes occur at c, -c/a, -c/b thus c>0, a<0, b<0 we have two (ugly) equations and 2 unknowns. \[ b(b-4a-1)(1-2a-b)^2 =0 \\ a(2a-2b-1)(1-2a-b)^2 = 0\] a,b,c are not zero. As the (1-2a-b) term is in both equations, having it be 0 adds no information, thus I would solve using \[ (b-4a-1)=0 \\(2a-2b-1) = 0\] and along with \[ c = 1-2a-b \] we can solve for a, b, and c in z= ax + by+ c that minimize the the product of the intercepts.
  • anonymous
I did that as well, and my answer is a = -1/2 b = -1 and c = 3. Using Lagrange multipliers, I arrive at the same answer. Thanks!

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