# Thread: Linear Programming - Sensitivity Range of One Constraint

1. ## Linear Programming - Sensitivity Range of One Constraint

Hey guys, I've attached the work I have done... I'm not exactly sure why I am wrong, or what appears to be wrong with my math.

The Question: imgur: the simple image sharer
My Answer: imgur: the simple image sharer
The correct answer is C1 = 58.33. C2 = 25.71

The first part of the question is easy, A is obviously optimal. However the second question, as easy as it appears, is tricking me. Finding C1 and C2

The textbook has taught me to rearranged the objective function in order to find its slope, which is -30/50 if I am finding C1 and -50/30 if I am finding C2. I also did the same for constraint line (x1/x2) which is -60/70 for C1 and then -70/60 for C2 (x2/x1) to reflect the same setup as the objective function.

The way they go about doing it is -C1/50 = -70/60 and then -30/C2 = -70/60.

I am totally lost because the textbook has taught me to do what I have done in my answer... Yet it's wrong. Can someone tell me what I am failing to realize? I am having trouble conceptualizing the problem and usually do math based on pattern recognition.

Thanks for your time (apologize for potato quality pictures)

2. ## Re: Linear Programming - Sensitivity Range of One Constraint

To solve for $\displaystyle c_{1}$, we want the tangency condition: $\displaystyle c_{1} * 60 = c_{2} * 70$. Holding $\displaystyle c_{2} = 50$, we get $\displaystyle c_{1} = \dfrac{350}{60} = 58.33$.

To solve for $\displaystyle c_{2}$, we hold $\displaystyle c_{1} = 30$ and solve: $\displaystyle 30 * 60 = 70 * c_{2}$ to get $\displaystyle c_{2} = \dfrac{1800}{70} = 25.71$.