For positive real numbers \(S\) and \(K\), the function \(H_K(S)\) is defined as: \[ H_K(S) = \max(S-K, 0). \] The max function is defined as: \[ \max(a,b) = \begin{cases} a, & \text{when } a > b \\ b, & \text{when } a \leq b \end{cases} \] The graph below shows the plot of a function \(N(S)\) versus \(S\). \(N(S)\) can be expressed as ______.
Show Hint
A ramp that starts rising at one point and flattens out at another can always be built as the difference of two shifted \(\max(S-K,0)\) functions.
Step 1: Read the key points off the graph.
From the plot, \(N(S)=0\) for \(S\) from \(0\) to \(10\), then \(N(S)\) rises in a straight line from \(0\) at \(S=10\) to \(10\) at \(S=20\), and after that \(N(S)\) stays flat at \(10\) all the way to \(S=30\). This is the shape of a capped ramp that starts rising at \(S=10\) and flattens out at \(S=20\).
Step 2: Recall what \(H_K(S)\) looks like on its own.
\[ H_K(S)=\max(S-K,0) \]
This is \(0\) for \(S\leq K\), and then rises with slope \(1\) for \(S>K\), with no upper limit. So \(H_{10}(S)\) is \(0\) up to \(S=10\) and then climbs forever with slope \(1\), and \(H_{20}(S)\) is \(0\) up to \(S=20\) and then also climbs forever with slope \(1\).
Step 3: Combine \(H_{10}(S)\) and \(H_{20}(S)\) to build a capped ramp.
Consider \(H_{10}(S)-H_{20}(S)\). For \(S\leq10\): both terms are \(0\), so the difference is \(0\). For \(10<S\leq20\): \(H_{10}(S)=S-10\) and \(H_{20}(S)=0\), so the difference is \(S-10\), a line of slope \(1\) rising from \(0\) at \(S=10\) to \(10\) at \(S=20\). For \(S>20\): \(H_{10}(S)=S-10\) and \(H_{20}(S)=S-20\), so the difference is \[ (S-10)-(S-20)=10 \] a flat value of \(10\) for all larger \(S\).
Step 4: Compare with the graph.
This is exactly the shape read off the graph: flat at \(0\) up to \(S=10\), a ramp of slope \(1\) up to \(S=20\), then flat at \(10\). So \(N(S)=H_{10}(S)-H_{20}(S)\).
Step 5: Rule out the other options.
\(H_{10}(S)-2H_{20}(S)\) would fall back below \(10\) for large \(S\) instead of staying flat, since the second term grows twice as fast once \(S>20\). \(-H_{10}(S)+H_{20}(S)\) is negative for \(10<S<20\) and never rises to \(10\), so it does not match a graph that goes up. \(H_{15}(S)-H_{20}(S)\) starts rising only at \(S=15\), not at \(S=10\), so it does not match the graph's starting point.
Final Answer:
The correct expression is \(H_{10}(S)-H_{20}(S)\).
\[ \boxed{H_{10}(S)-H_{20}(S)} \]
