May 7th, 2017, 10:35 AM  #1 
Member Joined: Mar 2017 From: Israel Posts: 30 Thanks: 2  Review
Hello Can you write please if my solution is right or not? The exercise: There is an equivalent curve of function f(x, y) = min{2x, 3y} which passes through a point (6, 7). Find the index C of the curve and the point where the curve divides into 2 parts. My solution: f(x, y) = min{2x, 3y} f(6, 7) = min{2*6, 3*7} = min{12, 21} 21 > 12 f(6, 7) = 12 C = 12 f(x, y) = 2x while > x ≤ y 2x = 12 x = 6 f(x, y) = 3y while > y < x 3y = 12 y = 4 The point is (6, 4) Is that right? Thanks a lot! Last edited by IlanSherer; May 7th, 2017 at 10:40 AM. 
May 10th, 2017, 04:07 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,731 Thanks: 707 
It took me a moment to figure out what you meant by "equivalent curve". It is what I would call a "level curve" the curve in the xyplane where the function has a constant value. Yes, since the curve passes through (6, 7) that constant value is f(6, 7)= 12. So the problem is to find all (x, y) such that the smaller of 2x and 3y is 12. If 2x< 3y, then we must have 2x= 12 so x= 6 and 12< 3y so y> 4. One part of the curve is the vertical line x= 6 for all y> 4. If 3y< 2x, then we must have 3y= 12 so y= 4 and 12< 2x so x> 6. The other part of the curve is the horizontal line y= 4 for all x> 6. Those two lines meet at (6, 4), just as you say. 
May 10th, 2017, 09:56 AM  #3  
Member Joined: Mar 2017 From: Israel Posts: 30 Thanks: 2  Quote:
So I answered correctly I guess... Thanks a lot!  

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