Finding a constraint on one variable of a multivariable function to constrain the entire function

Question

I have a function. Now I want to let my variables only take values between 0, and 1. The problem is as follows. For what values of Y, is L(x,y) < 0. That is, without putting a further constraint on x(other than its between 0, 1), how can you bind Y, such that you bind L(x,y) below 0. I graphed it out, and there is visually a range of y such that this is true. I was unable to proceed. I thought of trying the mins and max for x, but then I noticed that really doesn't give me any useful information. I understand that the last inequality must be true for whenever L<0, including the range im looking for, but there must be some other point I am not taking into account that will give me a range independent of x.

Answer 1

Consider the four optimization problems, which cover the combinations of minimizing or maximizing y, when x is fixed at 0 or 1.

a) min w.r.t. y subject to $0 \le y \le 1, L(x,y) < 0, x = 0$

b) max w.r.t. y subject to $0 \le y \le 1, L(x,y) < 0, x = 0$

c) min w.r.t. y subject to $0 \le y \le 1, L(x,y) < 0, x = 1$

d) max w.r.t. y subject to $0 \le y \le 1, L(x,y) < 0, x = 1$

Solving these as having non-strict inequality $L(x,y) \le 0$, the following solutions are obtained:

a) x = 0, y = 1/3

b) x = 0, y = 1

c) x = 1, y = 0

d) x = 1, y = 2/5.5

Given the linearity of L(x,y) for a fixed x, we can conclude that

when x = 0, then y in the interval (1/3,1] satisfies L(x,y) < 0

when x = 1, then y in the interval [0,2.5/5) satisfies L(x,y) < 0.

Due to the bilinearity of L(x,y), x = 0 and 1 are the extreme cases for x in [0,1] with regard to minimum and maximum possible values of y satisfying L(x,y) < 0.

Noting that 2.5/5 = 0.3626... we can therefore conclude that every value of y in the interval (1/3,2.5/5), and only values of y in that interval, will satisfy L(x,y) < 0 no matter which (i.e., "worst case") value of x in [0,1] is chosen.