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Both  x and y are positive real numbers, and the point (x,y) lies on or above both of the lines having equations 2x+5y=10 and 3x+4y=12. What is the least possible value of 8x+13y?

 

What is the x-intercept of the line perpendicular to the line defined by 3x-2y=6 and whose y-intercept is 2?

 Apr 11, 2019
 #1
avatar+106515 
+2

Not sure about the first...but I believe that the least possible value will occur at the intersection of the two given lines...so...

 

2x + 5y = 10    ⇒   6x + 15y  = 30     (1)

3x + 4y = 12    ⇒   -6x - 8y = -24      (2)      add (1) and (2)  and we have that

 

7y = 6

y = 6/7

And we can find x as

2x + 5(6/7) = 10

2x = 10 - 30/7

x = 40/14  = 20/7

 

So ( x , y )  =  ( 20/7, 6/7)

 

So  8(20/7) + 13(6/7)  =  34 = the least possible value

 

See the graph here : https://www.desmos.com/calculator/orm3rbjxft

 

cool cool cool

 Apr 11, 2019
edited by CPhill  Apr 11, 2019
 #2
avatar+106515 
+2

What is the x-intercept of the line perpendicular to the line defined by 3x-2y=6 and whose y-intercept is 2?

 

The slope of the given line  =  (3/2)

The slope of the perpendicualr line  = -2/3

If the y intercept of this line is 2, we have

 

y = (-2/3)x + 2     the x intercept occurs when y = 0

 

0 = (-2/3)x + 2

 

-2 = (-2/3)x

 

x = 3  = the x intercept

 

 

cool cool cool

 Apr 11, 2019

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