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2x+3y+4z=100

x+y+z=35

if x,y,z are at least over 1, what  is the maximum and minimum of x?

 Aug 10, 2015

Best Answer 

 #4
avatar+28471 
+10

If x = 33 then 

 

3y + 4z = 34      (1) and

y + z = 2           (2)

 

Multiply (2) by 3

3y + 3z = 6       (3)

 

Subtract (3) from (1)

z = 28

 

Substitute this into (2)

y = -26

 

This violates the condition "x,y,z are at least over 1"

.

 Aug 11, 2015
 #1
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0

Max=33

Min=1

 Aug 10, 2015
 #2
avatar+28471 
+10

2x + 3y + 4z = 100            ...(1)

x + y + z = 35                   ...(2)

 

Multiply (2) by 2 and subtract from (1) to get  y = 30 - 2z    ...(3)

Multiply (2) by 3 and subtract (1) from the result to get x = 5 + z   ...(4)

 

I'll assume you want integer solutions, and that "at least over 1" means 2 or more (you can modify the following appropriately if you mean something else).

Then, using (3), z = 2 means y = 26 (which is greater than 1, so ok),  and y = 2 means z = 14, so z varies from 2 to 14.

From (4), because x is a linear function of z, we have minimum(maximum) of x corresponds to minimum(maximum) of z, so

minimum x = 5+2 = 7

maximum x = 5+14 = 19

 

 Aug 10, 2015
 #3
avatar
+10

I assumed any real solution.

I think mine is right

 

Max 33.     Min 1

 Aug 10, 2015
 #4
avatar+28471 
+10
Best Answer

If x = 33 then 

 

3y + 4z = 34      (1) and

y + z = 2           (2)

 

Multiply (2) by 3

3y + 3z = 6       (3)

 

Subtract (3) from (1)

z = 28

 

Substitute this into (2)

y = -26

 

This violates the condition "x,y,z are at least over 1"

.

Alan Aug 11, 2015

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