Oh wait, yeah, this is just an example of the general principle “when you’re optimizing for xy, and you have a limited budget with linear costs on x and y, the optimal allocation is to spend equal amounts on both.”
Formally, you can show this via Lagrange-multiplier optimization, using the Lagrangian L(x,y)=xy−λ(ax+by−M). Setting the partials equal to zero gets you λ=y/a=x/b, and you recover the linear constraint function ax+by=M. So ax=by=M/2. (Alternatively, just optimizing xM−axb works, but I like Lagrange multipliers.)
In this case, we want to maximize pq+(1−p)rq0=p(q−rq0)−rq0, which is equivalent to optimizing p∗(q−rq0). Let’s define w=q−rq0, so we’re optimizing p∗w.
Our constraint function is defined by the tradeoff between p and w. p(k)=(.5−p0)k+p0, so k=p−p0.5−p0. w(k)=(r−1)q0k+q0−rq0=(r−1)q0(k−1), so k=−w(1−r)q0+1=p−p0.5−p0 .
Rearranging gives the constraint function .5−p0(1−r)q0w+p=.5. This is indeed linear, with a total ‘budget’ M of .5 and a p-coefficient b of 1. So by the above theorem we should have 1∗p=.5/2=.25.
Oh wait, yeah, this is just an example of the general principle “when you’re optimizing for xy, and you have a limited budget with linear costs on x and y, the optimal allocation is to spend equal amounts on both.”
Formally, you can show this via Lagrange-multiplier optimization, using the Lagrangian L(x,y)=xy−λ(ax+by−M). Setting the partials equal to zero gets you λ=y/a=x/b, and you recover the linear constraint function ax+by=M. So ax=by=M/2. (Alternatively, just optimizing xM−axb works, but I like Lagrange multipliers.)
In this case, we want to maximize pq+(1−p)rq0=p(q−rq0)−rq0, which is equivalent to optimizing p∗(q−rq0). Let’s define w = q−rq0, so we’re optimizing p∗w.
Our constraint function is defined by the tradeoff between p and w. p(k)=(.5−p0)k+p0, so k=p−p0.5−p0. w(k)=(r−1)q0k+q0−rq0=(r−1)q0(k−1), so k=−w(1−r)q0+1=p−p0.5−p0 .
Rearranging gives the constraint function .5−p0(1−r)q0w+p=.5. This is indeed linear, with a total ‘budget’ M of .5 and a p-coefficient b of 1. So by the above theorem we should have 1∗p=.5/2=.25.