The problem states that goats (G) can be fed corn-based feed (C) or soybean-based feed (S). The production function is given by $G = 2C + 5S$. The price of corn feed is $4 and the price of soybean feed is $5. We need to find the cost-minimizing feed combination that produces $G = 200$.

1. Problem Description

The problem states that goats (G) can be fed corn-based feed (C) or soybean-based feed (S). The production function is given by $G = 2C + 5S$. The price of corn feed is $4 and the price of soybean feed is$

5. We need to find the cost-minimizing feed combination that produces $G = 200$.

2. Solution Steps

First, we are given the production function:

$G = 2C + 5S$

We are given that $G = 200$. Therefore, we have:

$200 = 2C + 5S$

We want to minimize the cost, which is given by:

$Cost = 4C + 5S$

We can solve the production function for either C or S. Let's solve for C:

$2C = 200 - 5S$

$C = 100 - \frac{5}{2}S$

Now, substitute this expression for C into the cost function:

$Cost = 4(100 - \frac{5}{2}S) + 5S$

$Cost = 400 - 10S + 5S$

$Cost = 400 - 5S$

To minimize the cost, we want to maximize S. Since $C = 100 - \frac{5}{2}S$, $C$ must be non-negative. Therefore, $100 - \frac{5}{2}S \geq 0$, which implies $\frac{5}{2}S \leq 100$, or $S \leq 40$.

If $S = 40$, then $C = 100 - \frac{5}{2}(40) = 100 - 100 = 0$.

In this case, the cost is $4(0) + 5(40) = 200$.

Now consider option b, $C = 20$, $S = 50$. We have $2(20) + 5(50) = 40 + 250 = 290$, which is not equal to

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0. So, option b is not feasible.

Also the cost would be $4(20)+5(50)=80+250=330$.

Next consider option d, $C = 50$, $S = 20$. We have $2(50) + 5(20) = 100 + 100 = 200$. The cost is $4(50) + 5(20) = 200 + 100 = 300$.

Now consider option a, $C = 100$. In this case $S=0$, so $G = 2(100) + 5(0) = 200$, and the cost is $4(100) + 5(0) = 400$.

Comparing the costs of the feasible options, $S=40, C=0$ yields the lowest cost of

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0. The options do not include $C=0$. Of the remaining options, we want to minimize $4C + 5S$ subject to the constraint $2C + 5S = 200$.

In option c, $S=40$ implies $2C+5(40)=200$ or $2C+200=200$, so $C=0$. But this is not an option.

Option b is not feasible since $2C+5S = 2(20) + 5(50) = 40+250 = 290\ne 200$.

For option d, the cost is $4(50)+5(20)=300$.

For option a, if $C=100$, then $2C+5S = 2(100)+5S=200+5S=200$ so $S=0$ and $Cost = 400$.

In option c $S=40$ which means $C=0$, so G=

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0. Cost $= 200$. However, the question asks what is the cost-minimizing *feed combination*, and option c only gives $S$.

From $C=100-\frac{5}{2}S$ and the cost equation, we know that we want to increase $S$ to minimize cost. Thus we try $S=20$, which implies $C=100-50=50$, which matches option d. The cost is

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3. Final Answer

d. C = 50, S = 20