+980 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?
+2095 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!
+129852 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)
