# Thread: Limiting a parameter in optimization problem

1. ## Limiting a parameter in optimization problem

In the following optimization problem, consider equation (1)

\begin{align}
\max_{0 < t \leq 1} f(t) = t\left(1-\dfrac{t}{m}\right)\left(1-\dfrac{t}{mn}\right)^{d-2} \ \ \ \ \ \ (1)
\end{align}

How can I maximize such that $\max_{f(t)}$ given $0 \leq f(t) \leq 1$.

I have to solve equation (1) such that $f(t)$ always remains $0 \leq f(t) \leq 1$. I know the $t^{**}$ below gives the solution for the optimization problem, but it lies outside feasible solution. But for some of the solutions, $t >1$. (see figure 1 for example). How can I limit it? (note that $m$ and $n$ are variables, while $d$ is a constant)

$50 \leq m\times n\leq 1000$
$10 \leq d\leq 1500$

[![enter image description here][1]][1]

**Solution**:
\begin{align}
f'(t) = -\dfrac{mn^2\left(1-\frac{t}{mn}\right)^d\left(dt^2+\left(\left(1-d\right)m-2mn\right)t+m^2n\right)}{\left(t-mn\right)^3}
\end{align}

letting $f'(t) = 0$, we get

t^* = \dfrac{m\sqrt{4n^2-4n+d^2-2d+1}+2mn+\left(d-1\right)m}{2d}

or
\begin{equation*}
t^{**} = -\dfrac{m\sqrt{4n^2-4n+d^2-2d+1}-2mn+\left(1-d\right)m}{2d}
\end{equation*}

2. ## Re: Limiting a parameter in optimization problem

Note that $f(0)=0$ and $f(t)$ is continuous.
If your $t^{*}$ and $t^{**}$ are out of the feasible set [0,1], just check the sign of $f'(t)$ at any $t\in[0,1]$. If $f'(t)>0$, then $f$ is increasing function and $\min[1,f(1)]$ is the maximum. Otherwise, $f$ is decreasing and $f(0)=0$ is the maximum within the feasible set.
If any of $t^{*}$ and $t^{**}$ is within the feasible set, the maximum will be $\min[1,f(T)]$ where $T=\arg\max_{t\in\{t^{*},t^{**}\}} f(t)$.
In all above cases, there will be a feasible $t$ satisfying above maximum value of $f$ due to the fact that $f$ is continuous.

3. ## Re: Limiting a parameter in optimization problem

The extrema of a smooth function on a closed region $A\subset \mathbb{R}^n$ with boundary $B$ will either be a calculus like local extrema in the interior of $A$ or it will lie on the boundary $B$. So if there are no local maxima of $f(x)$ in $(0,1)$ then the maximum will occur at $x=0\ \rm{ or }\ 1$