Distance to Sun at Perihilion and Aphelion at max and min Eccentricity Title says it all. I'm putting together a presentation on Milankovitch Cycles, and I would like to bring up the difference in distance to the sun (km) at aphelion and perihelion when the earths orbit is at maximum and minimum eccentricity. The numbers I've found are that the maximum eccentricity of Earth's orbit is 0.0679, and the minimum is 0.000055.
I like the math of ellipses cause there's enough parts to keep it interesting, but it's not super complex either.   The answer to your question as you put it, is remarkably simple.  I've bolded the letters that represent distances for easier reading.

Eccentricity or $e = c/a$, where $a$ is the semi major axis and $c$ is the distance between the focal point (sun's location) and the center of the ellipse. $ea$ is the distance between the center point and the focal point.

So, the answer to your question is simply $a$ plus $ea$ and $a$ minus $ea$.   For the Earth and sun, the semi major axis or $a$, is $149,600,000\:\mathrm{km}$, then simply add and subtract $ea$ to both.

When $e = .0679$, $ea = 10.15\:\mathrm{million\:km}$. So $149,600,000 \pm 10,150,000$. $159,750,000$ and $139,450,000\:\mathrm{km}$ at furthest and closest.

When $e = .000055$, $ea = .008\:\mathrm{million\:km}$, so $149,680,000$ and $149,520,000$ at most circular.

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Maths and sources of pictures, here and here

footnote:  I hope this wasn't a homework question.