题目

Problem

The region $R_1$ , shown shaded in Figure 2, is defined by the inequalities

\begin{align*} 0\le y\le2,\qquad y\le10-2x,\qquad y\le kx, \end{align*}

Figure 2

where $k$ is a constant.

The line $x=a$ , where $a$ is a constant, passes through the intersection of the lines $y=2$ and $y=kx$ .

Given that the area of $R_1$ is $\dfrac{27}{4}$ square units,

(a) find

(i) the value of $a$

(ii) the value of $k$

(4)

(b) Define the region $R_2$ , also shown shaded in Figure 2, using inequalities.

(2)

解答

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先找几个关键横坐标： $y=10-2x$ 与 $x$ 轴相交于 $x=5$ ，与 $y=2$ 相交于 $x=4$ 。区域 $R_1$ 可以看成底边在 $x$ 轴上的一个多边形，面积等于一个大梯形减去左上角小三角形，也可直接算成 $9-a$ 。

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For $y=10-2x$ , the $x$ -intercept is found from

\begin{align*} 0=&\,10-2x\\ x=&\,5. \end{align*}

The line $y=10-2x$ meets $y=2$ when

\begin{align*} 2=&\,10-2x\\ x=&\,4. \end{align*}

Since $x=a$ passes through the intersection of $y=2$ and $y=kx$ ,

\begin{align*} ka=2. \end{align*}

The area of $R_1$ is made of:

\begin{align*} \text{left triangle} =&\, \frac12(a)(2)=a,\\ \text{rectangle} =&\,2(4-a),\\ \text{right triangle} =&\,\frac12(1)(2)=1. \end{align*}

\begin{align*} a+2(4-a)+1=&\,\frac{27}{4}\\ 9-a=&\,\frac{27}{4}\\ a=&\,\frac94. \end{align*}

Now use $ka=2$ :

\begin{align*} k\left(\frac94\right)=&\,2\\ k=&\,\frac89. \end{align*}

Therefore

\begin{align*} a=\frac94,\qquad k=\frac89. \end{align*}

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$R_2$ 在直线 $y=\frac89x$ 的上方、直线 $y=10-2x$ 的下方，并且在 $x=\frac94$ 的右侧。

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The region $R_2$ is defined by

\begin{align*} y&\ge \frac89x,\\ y&\le 10-2x,\\ x&\ge \frac94. \end{align*}