
Re: Maximize a single variable and solve for the rest
Posted:
May 21, 2011 5:54 AM


thank you Daniel and Bob, this solved it, please excuse my confusion.
On 5/20/11 6:49 AM, Bob Hanlon wrote: > It makes no sense to Maximixe the LHS of an equation since it is fixed to be equal to the RHS. Presumably you are trying to maximize the variable c. > > r1 = 3/4; > r2 = 1; > > Maximize[{c, b == c*r1, h + s + b + d == 1, d == c*r2, h == s, > b + d<= 9/10}, {h, s, b, d, c}] > > {18/35, {h > 1/20, s > 1/20, b > 27/70, d > 18/35, c > 18/35}} > > Solve[{b == c*r1, h + s + b + d == 1, d == c*r2, h == s, b + d<= 9/10}, {h, > s, b, d, c}, Reals] // Quiet > > {{h > ConditionalExpression[1  (7*c)/4 + (1/2)*(1 + (7*c)/4), > c<= 18/35], > s > ConditionalExpression[(1/2)*(1  (7*c)/4), c<= 18/35], > b > ConditionalExpression[(3*c)/4, c<= 18/35], > d > ConditionalExpression[c, c<= 18/35]}} > > Since you want c maximized, > > %[[1]] /. c > 18/35 > > {h > 1/20, s > 1/20, b > 27/70, d > 18/35} > > > Bob Hanlon > >  Ramiro<ramiro.barrantes@gmail.com> wrote: > > ============= > Hello, > > I have a problem where I would like to solve an equation (namely (h+s+b > +d==1) with some constraints, while maximizing for a related variable > "c" (c<=9). Please see below, any suggestions? > > r1 = 3/4; > r2 = 1; > Block[{h, s, b, d, c}, > NMaximize[{h + s + b + d, > b == c*r1&& h + s + b + d == 1&& d == c*r2&& h == s&& > b + d<= 0.9}, {h, s, b, d, c}]] > > {1., {h > 0.5, s > 0.5, (3 c)/4 > 0., c > 0., c > 0.}} > > Should I be using NSolve? > > Thanks in advance, > Ramiro

