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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?

Mar 19, 2020

#1
+1955
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We can write the inequalities 2x + 5y ≥ 10 and 3x + 4y ≥ 12. To get the value of 8x + 13y, add the first inequality to twice the second inequality. This gives you 8x + 13y = (2x + 5y) + 2(3x + 4y) ≥ 10 + 2 · 12 = 34 . Hope this helped!

Mar 19, 2020
#2
+389
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Thank you!! This really helped!!

qwertyzz  Mar 19, 2020
#3
+111360
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Thanks, Cal  !!!

Here's another way to solve this with linear programming

Set the equations up as equalities

2x + 5y  =10 ⇒    6x  + 15y  = 30

3x + 4y  = 12 ⇒   -6x  -  8y   = -24          add these

7y = 6

y = 6/7

And

2x  + 5(6/7)  = 10

2x  + 30/7 = 70/7

2x  = 40/7

x = 20/7

These  are  the  (x, y)  values  that will  maximize   8x + 13y

And  this  max  is    8(20/7) + 13(6/7)  =   [160 + 78 ] / 7  = 238/7   = 34

This can be  confirmed with this graph :    https://www.desmos.com/calculator/nwl5zh1adg

The max  will occur at  the "corner" point of  the intersection of the above inequalities ⇒  (20/7, 6/7)

Mar 19, 2020