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May 7th, 2017, 10:35 AM   #1
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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.
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May 10th, 2017, 04:07 AM   #2
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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 xy-plane 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.
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May 10th, 2017, 09:56 AM   #3
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Quote:
Originally Posted by Country Boy View Post
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 xy-plane 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.
I'm very sorry about that , i'm not from country where people speak english, you would ask me about what i meant, i would explain again and again
So I answered correctly I guess... Thanks a lot!
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