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 single \(\max(S-K,0)\) hinge produces one kink; two kinks in the graph mean you need two hinge terms whose slopes cancel appropriately after \(S=20\).
Step 1: Read the shape of \(N(S)\) off the graph. The plotted curve is flat at 0 for \(S \leq 10\), rises in a straight line of slope 1 from \(N=0\) at \(S=10\) up to \(N=10\) at \(S=20\), and then stays flat at 10 for all \(S \geq 20\). This is a classic piecewise linear ramp with a kink (change of slope) at \(S=10\) and another kink at \(S=20\).
Step 2: Recall what a single hinge function \(H_K(S)=\max(S-K,0)\) looks like. \(H_K(S)\) is 0 for \(S \leq K\) and rises with slope 1 for \(S > K\). It has exactly one kink, at \(S=K\). Since \(N(S)\) has two kinks, it must be built from two hinge functions added or subtracted so that their slopes combine correctly in each region.
Step 3: Test option (A), \(H_{10}(S)-H_{20}(S)\), region by region. For \(S \leq 10\): \(H_{10}(S)=0\) and \(H_{20}(S)=0\), so the expression gives 0, matching the flat part of the graph. For \(10 < S \leq 20\): \(H_{10}(S)=S-10\) and \(H_{20}(S)=0\) (since \(S \leq 20\)), so the expression gives \(S-10\), which rises from 0 at \(S=10\) to 10 at \(S=20\), matching the ramp exactly. For \(S > 20\): \(H_{10}(S)=S-10\) and \(H_{20}(S)=S-20\), so the expression gives \((S-10)-(S-20)=10\), a constant, matching the flat top of the graph. Every region matches, so \(N(S)=H_{10}(S)-H_{20}(S)\).
Step 4: Rule out the other options by finding a region where each one fails. Option (B), \(H_{10}(S)-2H_{20}(S)\): for \(S > 20\) this gives \((S-10)-2(S-20)=30-S\), which keeps decreasing as \(S\) grows, contradicting the flat plateau at 10 seen for \(S \geq 20\). Option (C), \(-H_{10}(S)+H_{20}(S)\): for \(10 < S \leq 20\) this gives \(-(S-10)+0=10-S\), which decreases from 0 to -10 as \(S\) goes from 10 to 20, but the graph rises (positive slope) in that interval, so this is wrong. Option (D), \(H_{15}(S)-H_{20}(S)\): for \(10 < S \leq 15\) both hinge terms are 0, giving \(N(S)=0\) on that whole stretch, but the actual graph already starts rising immediately after \(S=10\), so this option fails to match the graph between \(S=10\) and \(S=15\).
Step 5: Conclude. Only option (A) reproduces the graph in every region. \[ \boxed{N(S) = H_{10}(S) - H_{20}(S)} \]
