- #1
Lennart Stern said:The mathematical problem:
$\theta$ is a constant that equals 0.8.
We consider the set
'$ S=\left\{(F,h):F is a decreasing function from R^{+} to R^{+}, h\in R, 0=1- \frac{\theta+1)}{\theta} \frac {(\int^{h}_{y=0} F(y) dy)}{/F(0)} \frac{F(0)-\frac{1}{2}}{F(0)-F(h)} \right\}$'
The function L is defined on S by
$ L(F,h)= \frac{\int^{h}_{x=0} \int^{h}_{y=x} F(y) dy dx} {\int^{h}_{x=0} \int^{h}_{y=0} F(y) dy dx} h$
We want to find the maximal value of L.
A further problem:
Denote by $L(\theta)$ the maximal value of L. What is $max_{\theta \in (0,1)} \frac {L(\theta)}{\theta}$ ?
A variational problem with the constraint of decreasing function is a type of optimization problem where the goal is to find the function that minimizes a certain quantity, subject to the condition that the function must be decreasing. In other words, the function must have a negative slope.
A variational problem with the constraint of decreasing function is different from a standard optimization problem in that it adds an additional condition or constraint for the function to meet. This constraint makes the problem more challenging and may require different techniques to solve.
A variational problem with the constraint of decreasing function has many applications in fields such as economics, physics, and engineering. For example, in economics, it can be used to model demand curves, while in physics, it can be used to describe the motion of a falling object.
The solution to a variational problem with the constraint of decreasing function is typically determined using the Euler-Lagrange equation, which is a mathematical formula that finds the function that minimizes the given quantity while satisfying the constraint of being decreasing.
Yes, a variational problem with the constraint of decreasing function can have multiple solutions. However, in most cases, there is a unique solution that is the global minimum. In some cases, there may be multiple local minima, but only one global minimum that meets the constraint.