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There is a math problem that you won't ask good people for help!
Let the upper bottom a and the lower bottom b be h, then the area s = (a+b) * h/2-the formula of isosceles trapezoid area.

The base angle is 60, and the auxiliary line cuts a right triangle, so the short right side length of the right triangle is = (b-a)/2.

Long rectangular side, that is, H = (root number 3)/2 times (b-a)/2, that is, H = (root number 3)(b-a)/4- relation 1,

Then the hypotenuse, which is the waist of the isosceles trapezoid, is 2 times (b-a)/2, which is b-a,

Because the circumference is 60, 60 = A+B+2 * (B-A), after sorting, 60 = 3B-A- relation 2 is obtained.

Substituting relation 1 and relation 2 into the isosceles trapezoid area formula, we get S = (3b-60) * (root number 3)*(b+60-3b)/6.

Tidy up,

S=75-(b-25)2

Because in real numbers, the square of a number is not less than zero, when b = 25, S is the largest, which is 75 square centimeters. At this time, the upper bottom edge15cm, the lower bottom edge 25cm, and the waist10cm.

To sum up: when the upper and lower bottom edges are 15cm and 25cm respectively, and the waist circumference is 10cm, the maximum area is 75cm!