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 brainly.com
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would like to help but have no access to original question
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.
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!