Optimising function of two variables with Lagrange multipliers I am trying to find formulae for the $x$ and $y$ that maximise the function $f(x,y)=a(x + p)^bc(y + q)^d$, subject to the constraints:
$$x \geq 0$$
$$y \geq 0$$
$$x + y + p + q \leq M$$
Where $a$, $b$, $c$, $d$, $p$, $q$ and $M$ are positive constants, $b < 1$ and $d < 1$. All the constants can be determined experimentally beforehand, except $M$, which is set by a 'user'. I want to work out general formulae for $x$ and $y$ so that I can easily find the best values given a combination of the constants.
Ignoring the constraints $x \geq 0$ and $y \geq 0$ for the moment, I have used the method of Lagrange multipliers to derive that, for the maximum, $x=(bM)/(b+d) - p$ and $y=(dM)/(b+d) - q$. However, these formulae sometimes give values of $x$ and $y$ that are zero or negative.
How can I take account of the other two constraints? When I try to add them to the Lagrange multiplier method, I can't seem to eliminate the extra Lagrange multiplier variables - and I'm not even sure I can use those constraints with Lagrange. What is the right way to do this?
 A: $a,c$ are useless. Put $x+p=X,y+q=Y$. Then $f(X,Y)=X^bY^d$ and $X\geq p,Y\geq q,X+Y\leq M$. There is no free maximum because $X>0,Y>0$. Then the max (that exists because we are in a compact set) is reached on the edge.
Case 1. $X+Y=M$ and $f(X,Y)=g(X)=X^b(M-X)^d$. Then $g'$ is $0$ for $X_1=\dfrac{b}{b+d}M$; if $p\leq X_1\leq M-q$, then the first candidate is $C_1=(X_1,M-X_1)$, else we do not consider this possibility.
Case 2. $X=p$ and $f(X,Y)=g(Y)=p^bY^d$. The second candidate is $C_2=(p,M-p)$.
Case 3. In the same way , the third candidate is $C_3=(M-q,q)$.
It remains to compare $f(C_1),f(C_2),f(C_3)$.
A: The Lagrange-multiplier method doesn't always give much insight into an optimization problem.  Here, regular old homely single-variable extremization is probably more helpful.  I will largely be following what loup blanc has already posted, but I think a little more can be said.

It does make matters a bit clearer to use the variable $ \ X \ = \ x \ + \ p \ $ and $ \ Y \ = \ y \ + \ q \ $ .  The function to be maximized becomes $ \ f(X, \ Y) \ = \ ac \ X^b \ Y^d \ $ , under the constraints $ \ X \ \ge \ p \ $ , $ \ Y \ \ge \ q \ $ , and $ \ X + Y \ \le \ M \ $ .  In the graph above, the region to be considered is the white triangle, including its "edges".
Since all of the values $ \ a \ , \ b \ , \ c \ , \ d \ , \ p \ , \ q \ $ and $ \ M \ $ are positive, and presumably $ \ M \ > \ p \ + \ q \ $ (or we have no region to consider at all), then  $ \ f(X, \ Y) \ $ is an increasing function over this region.  This will "drive" the maximization "all the way up" to the line $ \ X \ + \ Y \ = \ M \ $ (the minimum of the function will lie at $ \ (X , \ Y) \ = \ ( p, \ q) \ $ or $ \ (x, \ y) \ = \ (0, \ 0) \ $ ).
With any possible solutions for the maximum falling on that line, we can "eliminate" one of the variables to simply work with, say, $ \ Y \ = \ M \ - \ X \ $ and the function $ \ g(X) \ = \ ac \ X^b \ (M - X)^d \ $ on the interval $ \ p \ \le \ X \ \le \ M-q \ $ .  Extremization of the function produces
$$ \ \frac{dg}{dX} \ = \ ac \ [ \ b X^{b-1} \ (M-X)^d \ + \ X^b \ d \ (M-X)^{d-1} \ (-1) \ ] \ = \ 0 \ \ $$
$$ \Rightarrow  \ \ ac \ X^{b-1} \ (M-X)^{d-1} \  [ \ b  \ (M-X) \ - \ dX  \ ] \ = \ 0  \ \ ; $$
since $ \ X \ = \ 0 \ $ and $ \ X \ = \ M \ $ are excluded by the constraints, there is a single critical point given by $ \ b \ (M-X) \ = \ dX \ \ \Rightarrow \ \ X_c \ = \ \frac{b}{b+d} M \ $ , all quite as already described by loup blanc.
This leads us to $ \ Y_c \ = \ M \ - \ \frac{b}{b+d} M \ = \ \frac{(b+d) \ - \ b}{b+d} M  \ = \ \frac{d}{b+d} M \ $ , marked by point 3 on the graph above; in a sense, the location of the critical point is a "weighted average" of the two exponents in $ \ f(x, \ y) \ $ .  The value of our function there is
$$ f(X_c, \ Y_c ) \ = \ f(x_c, \ y_c ) \ = \ ac \ \left( \frac{b}{b+d} M \right)^b \ \left( \frac{d}{b+d} M \right)^d \ = \ ac \ b^b \ d^d \ \left( \frac{M}{b+d} \right)^{b+d}  \ \ . $$
[In terms of the original variables, this critical point lies at 
$$ \ x_c \ =  \ \frac{b}{b+d} M \ - \ p \  , \ y_c \ =  \  \frac{d}{b+d} M \ - \ q \ \ . ] $$  
We should, of course, also consider the values of the function at the endpoints of the relevant interval:
point 1 -- $ \ f(X=M-q, \ Y=q) \ = \ ac \ (M-q)^b \ q^d \ $ ,
point 2 -- $ \ f(X=p, \ Y=M-p) \ = \ ac \ p^b \ (M-p)^d \ $ .
However, unless $ \ M \ $ is only slightly larger than $ \ p + q \ $ , the maximum value of the function will clearly be attained at $ \ (x_c, \ y_c)  \ $ .
$$ \  \ $$
If we use the values of the constants set in your comment on Marty's answer 
( $ \ a = 2 \ , \ c = 3 \ , $ $ b = 0.5 \ , \ d = 0.3 \ , \ p = 10 \ , \ q = 20 \ , \ $ and  $ \ M = 200 \ $ ) , we can compute the values of the function as
point 1 -- $ \ f(X = 180, \ Y = 20) \ = \ 2 \cdot 3 \cdot 180^{0.5} \cdot 20^{0.3} \ \approx \ 197.7 \ , $
point 2 -- $ \ f(X = 10, \ Y = 190) \ = \ 2 \cdot 3 \cdot 10^{0.5} \cdot 190^{0.3} \ \approx \ 91.6 \ , $
point 3 -- $ \ f(X_c = \frac{0.5}{0.5 + 0.3} \cdot 200 = 125, \ Y_c = 75) $
$$ = \ f(x_c =  125 - 10 = 115 \ , \ y_c = 75 - 20 = 55) \ = \ 2 \cdot 3 \cdot 125^{0.5} \cdot 75^{0.3} $$
$$ = \ 2 \cdot 3 \cdot 0.5^{0.5} \cdot 0.3^{0.3} \ \left( \frac{200}{0.8} \right)^{0.8}  \ \approx \ 245.0 \ . $$
The graph below shows the values of the function along the line $ \ X + Y \ = \ 200 \ $ :

the regions shaded in green are excluded by the constraints
A: You could use weighted A.M-G.M inequality, as all the quantities are non-negative.
[b(x+p) + d(y+q)]/(b+d) ≥ [(x+p)^b + (y+q)^d]^{1/(b+d)}
Raising both sides to power (b+d) and multiplying by ac, you get the maximum value, achieved when both sides are equal. This is possible only when
x+p = y+q        (i.e. all terms in the A.M are equal)
Also, x+y+p+q ≤ M
So, for maximum value, x = M/2 - p and y = M/2 -q
(Sorry for the style of writing. Am new to this forum) 
