I want to find the optimal solution to the following seemingly simple function:
$$\max_{x \in [0, 1]}~f(x) = x\mathrm{e}^{-a\left(bx^t+1\right)^\frac{1}{1-t}},$$ where $a, b > 0$ and $t \in (0, 1)$.
My attempt: I tried obtaining the optimal solution $x^*$ by exploring its properties. For example, I took the first derivative of $f(x)$ with respect to $x$, which is $$\frac{\mathrm{d} f(x)}{\mathrm{d} x} = \dfrac{\left(\left(t-1\right)\left(bx^t+1\right)^\frac{t}{t-1}+abtx^t\right)\exp\left({-\frac{a}{\left(bx^t+1\right)^{1/(t-1)}}}\right)}{\left(t-1\right)\left(bx^t+1\right)^\frac{t}{t-1}}.$$ The above derivative seems to not yield $x^*$ when the derivative is equated to $0$. How can one get the optimal solution or at least some properties regarding the optimal solution such as uniqueness and reasonable bounds on the optimal solution?
First, note that it is true that:
$$ x^* = \text{argmax}_{x\in[0,1]} f(x) = \text{argmax}_{x\in[0,1]} \log(f(x)) $$ since $\log(\bullet)$ is an increasing function. Note that: $$ \log(f(x)) = \log(x) - a(bx^t+1)^{\frac{1}{1-t}} $$ Hence, $$ \frac{d}{dx}\log(f(x)) = \frac{1}{x} - \frac{abt}{1-t} (bx^t+1)^{\frac{1}{1-t} - 1}\frac{x^{t}}{x} $$ if we set $\frac{d}{dx}\log(f(x))=0$ we obtain: $$ \frac{abt}{1-t} (bx^t+1)^{\frac{1}{1-t} - 1}\frac{x^t}{x} = \frac{1}{x} $$ Assume, $x\neq 0$. Then,$\frac{abt}{1-t} (bx^t+1)^{\frac{1}{1-t} - 1}x^{t} = 1$.
Now, note that $b(\bullet)^t+1$ is an increasing and positive function, and the same for $(\bullet)^{\frac{1}{1-t}-1}$ and $(\bullet)^t$ since $\frac{1}{1-t}-1>0$. Hence, $g(x;a,b,t) = \frac{abt}{1-t} (bx^t+1)^{\frac{1}{1-t} - 1}x^t $ is an increasing function.
Now, to see if there is a solution to $g(x^*;a,b,t)=1$ in $(0,1)$ we use the fact that $g(0;a,b,t)=0$ and $g(1;a,b,t) = \frac{abt}{1-t} (b+1)^{\frac{1}{1-t} - 1}$. Since, $g(x;a,b,t)$ is continuous and increasing, there exists a solution of $g(x^*;a,b,t)=1$ with $0< x^*< 1$ only if $g(1;a,b,t) = \frac{abt}{1-t} (b+1)^{\frac{1}{1-t} - 1}> 1$.
Once you find this unique $x^*$, the only thing left is to compare the cost function $f(x^*)$ with the value at the boundaries of the interval, this is with $f(0)$ and $f(1)$. Note that $f(0)=0$ and $f(1) = \exp(-a(b+1)^\frac{1}{1-t})\geq f(0)=0$. If $x^*$ inside $(0,1)$ doesn't exists (if $\frac{abt}{1-t} (b+1)^{\frac{1}{1-t} - 1}<1$) or $f(x^*)\leq f(1)$ for such $x^*\in(0,1)$, then the optimal is certainly $x=1$.
Hope this helps!