For each value of $x, f(x)$ is defined to be the minimum value of the three numbers $2 x+2, \frac{1}{2} x+1$ and $-\frac{3}{4} x+7$. What is the maximum value of $f(x)$ ?
(A) $\frac{2}{3}$
(B) 2
(C) $\frac{17}{5}$
(D) $\frac{62}{11}$
(E) 7
For each value of $x, f(x)$ is defined to be the minimum value of the three numbers $2 x+2, \frac{1}{2} x+1$ and $-\frac{3}{4} x+7$. What is the maximum value of $f(x)$ ?
(A) $\frac{2}{3}$
(B) 2
(C) $\frac{17}{5}$
(D) $\frac{62}{11}$
(E) 7

## Solution

The three numbers $2 x+2, \frac{1}{2} x+1$ and $\frac{-3}{4} x+7$ can be viewed as the $y$-coordinates of points lying on the lines $y=2 x+2$ (1), $y=\frac{1}{2} x+1$ (2), and $y=\frac{-3}{4} x+7$ (3), respectively.

Draw all three lines on the same set of axes and find the points of intersection.

Subtract equation (2) from equation (1) to get

$$
\begin{aligned}
0 & =\frac{3}{2} x+1 \\
\frac{-3}{2} x & =1 \\
x & =\frac{-2}{3}
\end{aligned}
$$

Substitute $x=\frac{-2}{3}$ into equation (1) and get

$$
\begin{aligned}
& y=2\left(\frac{-2}{3}\right)+2 \\
& y=\frac{2}{3}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-310.jpg?height=651&width=748&top_left_y=214&top_left_x=1117)

Thus, the point of intersection of the lines $y=2 x+2$ and $y=\frac{1}{2} x+1$ is $\left(\frac{-2}{3}, \frac{2}{3}\right)$.

Similarly, we find $\left(\frac{20}{11}, \frac{62}{11}\right)$ as the point of intersection of the lines $y=2 x+2$ and $y=\frac{-3}{4} x+7$, and $\left(\frac{24}{5}, \frac{17}{5}\right)$ as the point of intersection of the lines $y=\frac{1}{2} x+1$ and $y=\frac{-3}{4} x+7$.

The minimum value of the three numbers $2 x+2, \frac{1}{2} x+1$, and $\frac{-3}{4} x+7$ is shown in the diagram as the smallest of the $y$-coordinates of points on the three lines for a given value of $x$. The maximum of these $y$-coordinates is $\frac{17}{5}$.