A person wants to reserve a little square of x*x cm at the upper right corner of a square that measures

(60+x)cm in lenght and 80cm in width.

What's the best value for X so that the difference between the panel's area and the square's area is the greatest possible?

80(x + 60) = 80x + 4800
y = 80x + 4800 - x^2
y = -(x - 40)^2 + 6400 => inverted parabola since a = -1 < 0, hence the vertex is the global max, vertex ia at(40 , 6400) => x = 40 => gives the greatest possible difference between the area's.
x = 40 => answer

By panel, do you mean the whole original square?

If so, its area is 80(60+x), while small square is x²

Difference in area = 80(60+x) − x²
= −x² + 80x + 4800
= −(x² − 80x + 1600) + 4800 + 1600
= −(x − 40)² + 6400

The greatest possible difference is 6400 and occurs when x = 40

We want to maximize:

A(x) = (60 + x)(80) - x^2

A '(x) = 80 - 2x = 0 => x = 40 cm

Note that this is a maximum since A '(39) > 0 and A '(41) < 0.