what is the area of this rectangle?

Question

the area of a rectangle is given by the relation A= 20x^2 +11x -3. determine the expressions for the possible dimensions of this rectangle.

Answer 1

20x^2 + 11x -3 = Area = LW

= (4x+3)(5x-1) = LW

L= 4x+3

W = 5x-1

20x^2 + 11x -3

= 20(x^2 + 11x/20 + 121/1600) - 3 -121/80

=20(x+11/40)^2 - 361/80 is a parabola with vertex = minimum point = (11/40, -361/80) = (27.5, -45.125) where y > or = -45 1/8

but 20x^2 + 11x -3 >0 to have a positive area

(5x+3)(4x-1)>0 means

either 5x+3 and 4x-1 are both positive

or both are negative

if both are positive, then 5x+3>0 and 4x-1>0

and x >1/4

if x>1/4 then f(1/4) = A = 20x^2 +11x -3 = 20/16 + 44/16 - 48/16= 1

the area>1 and 5(1/4)+3 > 5/4+12/4 = 17/4 = 4.25

length > 4.25 and width>1/4.25 = 4/17. width>4/17

L> 17/4, W > 4/17

or vice versa

If both are negative then 5x+3<0 and 4x-1<0

and x<-3/5 and x>1/4 which is impossible

then

20(-3/5)^2 + 11(-3/5)-3 = 180/25 - 165/25 - 75/5 which is <0 making the area <0 which is impossible

both factors can't be negative.