We want to blend corn and soybean meal to create at least 800 lbs. of livestock feed per day. The feed must be at least 30% protein and at most 5% fiber. Each pound of corn contains .09 lbs. of protein and .02 pounds of fiber. Each pound of soybean meal contains .60 lbs. of protein and .06 lbs. of fiber. If corn costs $0.30 per pound and soybean meal costs $0.90 per pound, what blend will minimize daily cost?
I'm stuck in the process of expressing the percentage of each product in regards to 800 lbs. The important thing here is where it says at least 800 lbs. With that being said, we don't know exactly how much they will be produced. Any help would be greatly appreciated!
Let's say we use $x$ pounds of corn and $y$ pounds of soybean meal. The 800 pound requirement says that \begin{equation} x+y\geq 800. \end{equation} Next, let's calculate the amount of protein that this mix would use. Since each pound of corn and soybean meal provides $.09$ pounds and $.60$ pounds of protein, respectively, the total protein provided is \begin{equation} .09x + .60y. \end{equation} To determine the percentage of the feed that is protein, we divide this protein weight by the total weight: \begin{equation} \frac{.09x + .60y}{x+y}. \end{equation} The protein requirement can then be written as \begin{equation} \frac{.09x + .60y}{x+y} \geq 0.3. \end{equation} This can be rearranged as \begin{equation} 0.21x - 0.30y \leq 0, \end{equation} but I'll leave that to you. Can you now produce an inequality for the fiber? Do you then know how to minimize cost? (Finding a feasible region, checking the cost at boundary points, etc.)