Answer
The linear function whose contour map shown in Figure 24 (contour interval $m=6$) is:
$f\left( {x,y} \right) = 2x + 6y + 6$
If the contour interval is $m=3$, the linear function becomes:
$f\left( {x,y} \right) = x + 3y + 3$
Work Step by Step
A linear function has the form: $f\left( {x,y} \right) = Ax + By + C$
Consider the line for $c=6$. We notice that the origin $\left( {0,0} \right)$ is located in the line. So,
$6 = A\left( 0 \right) + B\left( 0 \right) + C$
$C = 6$
Consider the line for $c=0$. We have the points $\left( { - 3,0} \right)$ and $\left( {0, - 1} \right)$ located in the line. So,
$\begin{array}{l}
0 = A\left( { - 3} \right) + B\left( 0 \right) + 6\\
0 = - 3A + 6\\
A = 2
\end{array}$
$\begin{array}{l}
0 = A\left( 0 \right) + B\left( { - 1} \right) + 6\\
0 = - B + 6\\
B = 6
\end{array}$
Thus, the linear function is $f\left( {x,y} \right) = 2x + 6y + 6$
If the curve labeled $c=6$ is relabeled $c=3$, then we still have the origin $\left( {0,0} \right)$ located in the line. So,
$3 = A\left( 0 \right) + B\left( 0 \right) + C$
$C = 3$
Consider the line for $c=0$. We have the points $\left( { - 3,0} \right)$ and $\left( {0, - 1} \right)$ located in the line. So,
$\begin{array}{l}
0 = A\left( { - 3} \right) + B\left( 0 \right) + 3\\
0 = - 3A + 3\\
A = 1
\end{array}$
$\begin{array}{l}
0 = A\left( 0 \right) + B\left( { - 1} \right) + 3\\
0 = - B + 3\\
B = 3
\end{array}$
Thus, the new linear function is $f\left( {x,y} \right) = x + 3y + 3$
