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# i need to understand ._.

0
69
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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
+107107
+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

Jan 8, 2020
#2
+24054
+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 $$8x + 13y$$?

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

Jan 8, 2020