# Linear Programming - Sensitivity Range of One Constraint

• Apr 23rd 2014, 02:57 PM
peekay
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)
• Apr 27th 2014, 12:16 PM
macosxnerd101
Re: Linear Programming - Sensitivity Range of One Constraint
To solve for $c_{1}$, we want the tangency condition: $c_{1} * 60 = c_{2} * 70$. Holding $c_{2} = 50$, we get $c_{1} = \dfrac{350}{60} = 58.33$.

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