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+4

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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+4

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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?

Guest Oct 16, 2019

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CoolStuffYT · Answer 1 · 2019-10-16T05:55:06

Answer: 34

I graphed the equations $2x+5y>10$ and $3x+4y>12$ , then just found the optimal point.

To find the optimal point, I graphed $8x+13y=k$ and then found the point where that line first touched the region.

It was at (2.857, 0.857) to the nearest thousandth.

Then I did the math and got 34.

You are very welcome!

:P

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+4

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Both xx and yy are positive real numbers, and the point (x,y)(x,y) lies on or above both of the lines having equations 2x+5y=102x+5y = 10 and $3x+4

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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+4

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