How to maximize the following function of two variables?

Question

I have the following function:

\begin{equation} f(q,p) = q \sqrt{p} + (1-q) \sqrt{1 - p} \end{equation}

Here, $q \in [0,1]$ and $p \in [0,1]$.

Now, given some value $q \in [0,1]$ what value should I select for $p$ in order to maximize $f(q,p)$? That is, I need to define some function $g(q)$ such that $f(q, g(q))$ is a local maximum.

I've been thinking about this problem for days and I don't know where to begin. Any help will be greatly appreciated.

Answer 1

By Cauchy-Schwarz, $$ [q \sqrt{p} + (1-q) \sqrt{1 - p}]^2\leq(q^2+(1-q)^2)(p+1-p)=q^2+(1-q)^2. $$ To have equality, we require $$ \frac{\sqrt{p}}{\sqrt{1-p}}=\frac{q}{1-q}\iff \boxed{p=\frac{q^2}{1-2q+2q^2}}\in[0,1]. $$

Answer 2

Hint:

Now we take some constant $q$ so our function becomes a function with only $p$ varying:

$$\begin{equation} f(p) = q \sqrt{p} + (1-q) \sqrt{1 - p} \end{equation}$$

With the constraint $0 \leq p \leq 1$. To maximize one must consider critical points in our interval, and the endpoints. Keep in mind that the constant $q$ is somewhere between $0$ and $1$.

Answer 3

We have that

$$\nabla f=\left(\sqrt{p}-\sqrt{1-p},\frac{q}{2\sqrt p}-\frac{1-q}{2\sqrt{1-p}}\right)$$ doesn't vanish on $(0,1)\times (0,1).$ Thus the maximum is achieved on the boundary. Now we have

\begin{cases}f(0,p)&=1-p\\f(1,p)&=p\\f(q,0)&=\sqrt{1-q}\\ f(q,1)&=\sqrt{q}\end{cases}

So we have $$f(0,0)=f(1,1)=1>f(x,y), \forall (x,y)\in [0,1]\times [0,1]\setminus \{(0,0),(1,1)\}.$$

To maximize $f(p,q)$ for a given $q$ we consider $g(p)=q\sqrt{p} + (1-q) \sqrt{1 - p}$ with $q$ constant. Then $g'(p)\ne 0, \forall p\in (0,1).$ So the maximum value is among

$$f(q,0)=1-q, f(q,1)=q.$$