Question:
x and y are real numbers s.t 2 < x < 3 and -8<y:
A) x^2 y
B) xy^2
C) 5xy
Solution:
xY^2 is postive. So out of scope
X^2 Y has min value of 4 * -8 = - 32
and 5xy can have min value of 5*2* -8 = - 80
(I know they arent equal. But by approaximation it can be inferred)
Also y has to be of odd degree and thn maximise x.