I assume you're in precalculus, correct? i'll try to keep the solutions as simple as possible.

let the two numbers be x and y, we have:

x - y = 20

P = xy ..........where P is the product

since x - y = 20

=> x = 20 + y

substitute 20 + y for x in P, we get:

P = (20 + y)y

=> P = 20y + y^2

this is an upward opening parabola, the minimum value is at the vertex. the vertex occurs when:

y = -b/2a , where b is the coefficient of y and a is the coefficient of y^2

so for vertex, y = -20/2 = -10

but x - y = 20

=> x + 10 = 20

=> x = 10

so the two numbers are 10 and -10

let the length be x

2) What are the dimensios if the largest rectangle with a fixed perimetre "P"?

let the width be y

the largest rectangle will have the largest area

then P = 2x + 2y where P is the perimeter

now A = xy

since P = 2x + 2y

=> y = (P - 2x)/2

substitute (P - 2x)/2 for y in A, we get:

A = x(P - 2x)/2

=> A = (1/2)(Px - 2x^2) = (P/2)x - x^2

this is a downward opening parabola, the max value occurs at the vertex. for the vertex, x = -b/2a

=> x = -(P/2)/-2 = P/4

but y = (P - 2x)/2

=> y = (P - 2(P/4))/2

=> y = (P - P/2)/2

=> y = (P/2)/2

=> y = P/4

so the largest rectangle with a fixed perimeter P, is a square with dimentions length = width = P/4