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D1 Question Jan 2005

Hi everyone,
I've been struggling on the change to the linear programming objective function in Q5iii for the past hour and could really do with some help.

Any help would be much appreciated!!
Thanks.
(edited 12 years ago)
Screen shot 2011-05-23 at 13.03.29.png18.0KB
Think about which of the solutions on the diagram it could be referring to; Q=3 can only occur when (0,1) is the solution, so what would 'a' have to be to ensure that this was the solution point?
Reply 2
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nick1987
OP
Thanks for your help. It helped me reach the solution a<=-6 as the I assume the gradient has to be >=2. But, I still don't understand why... why can Q only equal 3 at (0,1) and why does the gradient have to be >=2.

Sorry if I'm being rather thick!
Original post by nick1987
Thanks for your help. It helped me reach the solution a<=-6 as the I assume the gradient has to be >=2. But, I still don't understand why... why can Q only equal 3 at (0,1) and why does the gradient have to be >=2.

Sorry if I'm being rather thick!


Q = ax + 3y, if Q = 3, then:

y = (-a/3)x + 1

The gradient has to be greater than that of y=2x+1, so greater than 2. Think about what would happen if the gradient of the objective function wasn't greater than 2, for example if the objective function went through the feasible region. It would mean that for a gradient of 1.9 for example, the solution at the top would give your maximum Q.
(-a/3) > 2

-a >6

a<-6.

Q= -6x + 3y. Your point (0,1) gives Q=3. The point (9/10, 28/10) (the one at the top) gives:

Q = -54/10 + 84/10 = 3. Any lower value on the coefficient of x wouldn't affect the value of Q at (0,1) but it would affect (9/10, 28/10). The same can be applied to the point (4,0).
Reply 5
Avatar for nick1987
nick1987
OP
Thanks a lot little_wizard123. I get it now!! :smile:

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