Unexpected result from Reduce[ ] of an inequality?

Questions
,
1

I want to use the Reduce command to solve the following inequality set and obtain the value range of [EmptyUpTriangle]t, but running the following code results cannot obtain the correct results. How can I modify my code?

Reduce[{1/
    2 (14758. E^(-267.929 t) - 14758. E^(-186.617 t) - 
      14758. E^(-267.929 (t + \[EmptyUpTriangle]t)) + 
      14758. E^(-186.617 (t + \[EmptyUpTriangle]t))) \
\[EmptyUpTriangle]t <= 0.0005, 0 <= t <= 0.08}, \[EmptyUpTriangle]t]
2

Reduce gives answers for fixed values of t:

Reduce[1/
    2 (14758 E^(-267929 t/1000) - 14758 E^(-186617 t/1000) - 
      14758 E^(-((267929 (t + \[EmptyUpTriangle]t))/1000)) + 
      14758 E^(-((186617 (t + \[EmptyUpTriangle]t))/
        1000))) \[EmptyUpTriangle]t <= 1/2000 &&
  t == 1/25,
 \[EmptyUpTriangle]t]
3

Hello, sir! Is it necessary to give a certain value for t in this inequality group?

4

Someone from Wolfram may give a more informed answer, but my guess is that a full symbolic solution of a trascendental inequality in two variables may be too much to expect.You can visualize the solution set with RegionPlot, for example:

RegionPlot[
 1/2 (14758 E^(-267929 t/1000) - 14758 E^(-186617 t/1000) - 
     14758 E^(-((267929 (t + \[EmptyUpTriangle]t))/1000)) + 
     14758 E^(-((186617 (t + \[EmptyUpTriangle]t))/
           1000))) \[EmptyUpTriangle]t <= 1/2000,
 {t, 0, 2/25},
 {\[EmptyUpTriangle]t, 0, 1/1000},
 FrameLabel -> Automatic]

For small values of t the set becomes a little complicated:

Reduce[1/2 (14758 E^(-267929 t/1000) - 14758 E^(-186617 t/1000) - 
      14758 E^(-((267929 (t + \[EmptyUpTriangle]t))/1000)) + 
      14758 E^(-((186617 (t + \[EmptyUpTriangle]t))/
            1000))) \[EmptyUpTriangle]t <= 1/2000 && 
  t == 1/1000, {t, \[EmptyUpTriangle]t}]