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Let $x_a$ denote amount of feed A and $x_b$ denote amount of feed B.

Minimize $10x_a+12x_b$
subject to $4x_a+2x_b \ge 12$
$4x_a+8x_b \ge 24$
$8x_b \ge 8$

Have I got the numbers right?

Also to solve it graphically I would get 3 lines that intersect at most 3 times and just test each vertex point, yes?

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You are on the right track. You can test each vertex by calculation. But you can evaluate graphically the optimal point as well. I´ve solved the constraints for $x_b$. Therefore the y-axis is denoted as $x_b$. If I understand you right you are able to draw the constraints into the coordinate system. It follows that the feasible region is the green one.

Let´s denote $z$ as the value of the objective function. Then we have the equality

$z=10x_a+12x_b$

This equality can be solved for $x_b$ as well.

$x_b=\frac{z}{12}-\frac{10}{12}x_a$. Now you set $z$ equal to $0$.

$x_b=-\frac{10}{12}x_a$. That means your objective function has the value $0$. This happens if $x_a=x_b=0$. Thus the line go through the origin with the slope $-\frac{10}{12}$. You need one more point to draw the line, for instance $P_2(3,-2.5)$. Now you have the objective function as a line with the initial level 0. This is the red line.

Then you push the red line right upwards parallel until the line touches the feasible region (green) the $ \texttt{first time}$. The blue lines illustrates the process of moving the line upward.

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