Chapter 3, Problem 1EA

Use sensitivity analysis to find the optimal solution to the original 4-H problem, if the profit from raising a goat increases from $12 to $25.

To determine

The new optimal solution for the number of goats and the number of pigs raised by Lisa if the profit increases from $12 to $25 for one goat raised.

Answer to Problem 1EA

Solution: The new optimal solution is to raise 6 goats and 10 pigs.

Explanation of Solution

Given: The maximum number of animals raised at a time is 16. The maximum number of goats raised at a time is 10. The cost to raise one goat is $25, and the cost to raise one pig is $75. The maximum cost available to raise the animals is $900. The profit gained from one pig is $40. The profit increases from $12 to $25 for one goat raised.

Explanation:

The constraints as provided by the original problem are

$x+y\le 16$

$25x+75y\le 900$

$x\le 10$

$x\ge 0$

$y\ge 0$

The original objective function is

$z=12x+40y$

The graph of the linear inequalities as provided by the original solution is

The optimal solution of the original problem is at the corner point $(0,12)$.

Let the profit on one goat increase from $12 to $$\left(12+c\right)$.

The new objective function is

$\begin{array}{l}z=(12+c)x+40y\\ y=-\frac{\left(12+c\right)}{40}x+\frac{z}{40}\end{array}$

The slope of the isoprofit lines for this objective function is $-\frac{\left(12+c\right)}{40}$.

The cost constraint inequality can be written as

$\begin{array}{c}25x+75y\le 900\\ x+3y\le 36\end{array}$

The corresponding boundary line is

$\begin{array}{c}x+3y=36\\ y=-\frac{x}{3}+12\end{array}$

The slope of this boundary line is $-\frac{1}{3}$.

The optimal solution changes from the corner point $(0,12)$ to $\left(6,10\right)$ when the slope of the isoprofit lines becomes less than the slope of the cost constraint boundary line and stays the same until the slope of the next boundary line:

$\begin{array}{c}x+y=16\\ y=-x+16\end{array}$

The slope of this constraint line is −1.

Thus, the optimal solution changes from $(0,12)$ to $\left(6,10\right)$ when:

$\begin{array}{c}-1\le -\frac{\left(12+c\right)}{40}\le -\frac{1}{3}\\ 120\ge 36+3c\ge 40\\ 84\ge 3c\ge 4\\ 28\ge c\ge \frac{4}{3}\end{array}$

The profit on one goat increases from $12 to $25.

$\begin{array}{c}12+c=25\\ c=13\end{array}$

Since c lies in the range $28\ge c\ge \frac{4}{3}$, the optimal solution changes to the corner point $\left(6,10\right)$.

Conclusion: When the profit on one goat raised is increased from $12 to $25, the slope of the isoprofit lines changes such that the optimal solution changes from 0 goats and 12 pigs raised to the next intersection point of the provided constraints, which is 6 goats and 10 pigs raised because the profit margin increased on one goat raised makes it profitable to raise goats also.