Algebra Two Help

Question

Use the equation Q=5x+3y and the following constraints:

3y+6 ≥ 5x

y ≤ 3

4x  ≥ 8

a. Maximize and minimize the equation 

b. Suppose the equation Q=5x+3ywas changed to Q=-5x-3y, describe how the maximum and minimum would change in the new equation compared to the original equation and explain why you think it is that way.

Answer 1

4x ≥ 8 implies that x ≥ 2, and 3y + 6 ≥ 5x and both together imply that 10 ≤ 5x ≤ 3y + 6,

so we get that 10 + 3y ≤ 5x + 3y ≤ 6y + 6 ≤ 6*3 + 6 = 18 + 6 = 24, since we are given that y ≤ 3, so we also get 10 + 3y ≤ 6y + 6, which implies 4 + 3y ≤ 6y, which then gets us to 4 ≤ 3y, so now we have 14 = 10 + 4 ≤ 10 + 3y ≤ 5x + 3y ≤ 24, and then also the following:

a. The maximum of Q = 5x + 3y is 24, and the minimum of Q is 14.

b. If the equation Q = 5x + 3y was changed to Q = -5x - 3y, the new maximum would be the negative of the original minimum, and the new minimum would be the negative of the original maximum, since from 14 ≤ 5x + 3y ≤ 24, we get 14 ≤ 5x + 3y and 5x + 3y ≤ 24, then we can multiply both sides of both inequalities by -1, but we have to switch the direction of these inequalities, since if we don't then we get non-sense like if 1 ≤ 2, then -1 ≤ -2, which is not true, and that, in turn, gets us -14 ≥ -5x - 3y and -5x - 3y ≥ -24, so:

-14 ≥ -5x - 3y ≥ -24, or put in the correct order, from smallest to largest, we get:

-24 ≤ -5x - 3y ≤ -14.