图 3 显示了 $x$ 和 $y$ 的线性规划问题的约束条件，其中 $R$ 是可行域。图 3 中已显示了两条直线的方程。

已知 $k$ 是正常数，

(a) 确定定义 $R$ 的不等式，必要时用 $k$ 表示。

目标是最大化 $P = 5x + ky$。

已知 $P$ 在 $R$ 的最优顶点处的值为 38，

(b) 确定 $k$ 的可能值。必须展示代数计算并清楚说明方法。

逐条观察可行域 $R$ 位于边界线的哪一侧。第三条边界经过 $(0,8)$ 与 $(4,0)$，先由这两个截距求出直线方程，再根据阴影区域选择不等号方向。

The region $R$ lies below the line $x+y=8$, so

It lies above the line $5y=x+k$, so

The remaining boundary passes through $(0,8)$ and $(4,0)$. Its equation is

Since $R$ lies above this line,

or equivalently,

Therefore the inequalities defining $R$ are

因为目标函数的两个系数都是正数，最大值只可能出现在可行域东北侧边界 $x+y=8$ 的两个端点：$(0,8)$，或 $x+y=8$ 与 $5y=x+k$ 的交点。分别令这些候选顶点的目标值为 38，再检查它是否真的优于另一候选顶点。

At the vertex $(0,8)$,

If this vertex gives $P=38$, then

and hence

The other possible optimal vertex is the intersection of

and

Using $y=8-x$ gives

so the coordinates of this vertex are

Setting the objective value at this vertex equal to $38$ gives

Multiplying by $6$ and simplifying,

so

Therefore

Since $k$ is positive,

It remains to check which candidate really gives a maximum value of $38$.

If $k=4$, the intersection vertex is $(6,2)$ and

whereas

Thus $k=4$ is valid.

If $k=\frac{19}{4}$, the other vertex is

at which

Therefore $(0,8)$ is not optimal when $k=\frac{19}{4}$, so this value must be rejected.

Hence the only possible value is

在两条边界的交点处，把 $y$ 和 $k$ 都表示成 $x$ 的式子。代入目标值 38 后解关于 $x$ 的二次方程，再回代求 $k$。还要检查另一个候选顶点 $(0,8)$，不能只求出一个交点就结束。

The possible optimal vertices are $(0,8)$ and the intersection of $x+y=8$ with $5y=x+k$.

At the intersection,

Setting $P=38$ gives

Thus $x=\frac{47}{6}$ or $x=6$. The first gives

which is rejected because $k>0$. The second gives

For $k=4$, $P(6,2)=38$ whereas $P(0,8)=32$, so the maximum is $38$.

If $(0,8)$ were optimal, $8k=38$ would instead give $k=\frac{19}{4}$. However, the intersection would then be $\big(\frac{47}{8},\frac{17}{8}\big)$, where

Hence $k=\frac{19}{4}$ is rejected, and the only possible value is

也可以在交点处把 $x$ 和 $k$ 都表示成 $y$ 的式子，得到关于 $y$ 的二次方程。求出根后回代，排除负的 $k$，再比较两个候选顶点的目标值。

The possible optimal vertices are $(0,8)$ and the intersection of $x+y=8$ with $5y=x+k$.

At the intersection,

Setting $P=38$ gives

Thus $y=\frac16$ or $y=2$. The first gives

which is rejected because $k>0$. The second gives

For $k=4$, $P(6,2)=38$ whereas $P(0,8)=32$, so the maximum is $38$.

If $(0,8)$ were optimal, $8k=38$ would give $k=\frac{19}{4}$. At the other candidate vertex $\big(\frac{47}{8},\frac{17}{8}\big)$,

Therefore this vertex gives a greater value than $(0,8)$, so $k=\frac{19}{4}$ is rejected. The only possible value is