# Linear programming (maximizing)

• Feb 4th 2013, 11:21 AM
Yoodle15
Linear programming (maximizing)
I am not sure whether I have done this right or that my answer is correct. I have a strong feeling that tells me it is not right. Could you please check it for me?

Minimize
C = x + y
subject to the constraints
1x +7y ≥ 1
11x + 2y ≥ 1
x, y ≥ 0

The answers I got through solving the system geometrically/graphically is that the minimum of C is 0.244360901 and occurs when x = 0.11842105 and y = 0.125939849.
• Feb 4th 2013, 12:52 PM
Soroban
Re: Linear programming (maximizing)
Hello, Yoodle15!

I have no idea how you got those ugly decimals . . .

Quote:

$\text{Minimize: }\:C \:=\: x + y$

$\text{Subject to the constraints: }\:\begin{Bmatrix}x +7y \:\ge\:1 \\ 11x + 2y \:\ge\:1 \\ x, y\:\ge\: 0\end{Bmatrix}$

$\text{The graph looks like this:}$

Code:

      |::::::       |::::::::     Ro:::::::::       |*:::::::::       | *:::::::::       |  *:::::::::       *  *:::::::::       |  * *:::::::::       |    o::::::::::       |    Q * *::::::::       |      *  *:::::::       |        *    *:::::::     - + - - - - * - - - o - - - -       |                P
$\text{The vertices are: }\:\begin{Bmatrix}P: &(1,0) \\ \\[-4mm] Q: & (\frac{2}{15},\frac{1}{15}) \\ \\[-4mm] R: & (0,\frac{1}{2}) \end{Bmatrix}$

$\text{Test them in the }C\text{-function.}$