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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?

 Jan 8, 2020
 #1
avatar+106519 
+1

Look at the graph here :  https://www.desmos.com/calculator/vsoqo96xtr

 

The  least possible value  of  8x + 13y  will occur at  the  intersection of  these  lines

 

So  we have that

 

2x + 5y  =10

3x + 4y  =  12          multiply the first eauation through by 4  and the second  by -5

 

8x + 20y  = 40

-15x - 20y = -60      add these

 

-7x  = -20

x = 20/7 =  the x coordinate of the  intersection of these lines

 

And  using 2x + 5y = 10    we can find the y  coordinate  as

2(20/7) + 5y  =10

40/7 + 5y = 70/7

5y=   70/7 - 40/7

5y = 30/7       divide both sides by 5

y = 6/7

 

 

So.....the minimum value of  8x + 13y  =

 

8(20/7) + 13(6/7)  =

 

[160  + 78 ]  / 7  =

 

238 / 7

 

34

 

cool cool cool

 Jan 8, 2020
 #2
avatar+23812 
+2

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 \(8x + 13y\)?

 

\(\begin{array}{|lrcll|} \hline & 2x + 5y &=& 10 \\ & 3x + 4y &=& 12 \\ & 3x + 4y &=& 12 \\ \hline \text{sum} & 8x + 13y &=& 10+12+12 \\ & \mathbf{8x + 13y} &=& \mathbf{34} \\ \hline \end{array}\)

 

laugh

 Jan 8, 2020

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