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?

Question

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?

irishboy123 · Answer

would like to help but have no access to original question

phi · Answer

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 · Answer

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!