Solving Profit-Target Inequalities

Why Profit-Target Inequalities Need Care

An option profit question may ask for more than a break-even price. It may ask when a position earns at least a specified amount, loses more than a specified amount, or exactly reaches a target. These questions require solving an inequality, not just finding one price. The answer is a range of possible terminal underlying prices, and whether an endpoint belongs in that range depends on whether the question says “greater than” or “greater than or equal to.”

The key step is to split the calculation at the strike. On one side, the option expires without exercise and profit is constant; on the other, the profit changes linearly with the terminal price. Work each piece separately, then combine the results. This approach applies to a purchased call, written call, purchased put, and written put.

Model and definitions: Let $S_T$ be the underlying asset price at expiration, with $S_T\geq 0$, and let $K$ be the strike. Let $c$ and $p$ be the call and put premiums per share paid at initiation. Profit is measured per share at expiration as payoff minus premium for a purchased option, or premium minus payoff for a written option. This convention does not accumulate the initial premium with interest and excludes fees and taxes. A target $g$ is also measured per share. If a problem specifies a different profit convention, such as carrying the premium forward with interest, these formulas must be adjusted.

Profit Functions and the Strike Split

The payoff is determined by whether exercising the option at expiration produces a positive payment. A call payoff is $\max(S_T-K,0)$, and a put payoff is $\max(K-S_T,0)$. Subtracting the premium for a purchased position, or subtracting the payoff from the premium for a written position, gives the following profit functions.

\begin{aligned} \text{Long call:}\quad \Pi_{LC}(S_T)&= \begin{cases} -c, & 0\leq S_T\leq K,\\ S_T-K-c, & S_T>K; \end{cases}\\[4pt] \text{Written call:}\quad \Pi_{SC}(S_T)&= \begin{cases} c, & 0\leq S_T\leq K,\\ c-(S_T-K)=K+c-S_T, & S_T>K; \end{cases}\\[4pt] \text{Long put:}\quad \Pi_{LP}(S_T)&= \begin{cases} K-S_T-p, & 0\leq S_T<K,\\ -p, & S_T\geq K; \end{cases}\\[4pt] \text{Written put:}\quad \Pi_{SP}(S_T)&= \begin{cases} p-(K-S_T)=S_T+p-K, & 0\leq S_T<K,\\ p, & S_T\geq K. \end{cases} \end{aligned}

At $S_T=K$, the option payoff is zero. The formulas on either side meet at the same profit: $-c$ for a long call, $c$ for a written call, $-p$ for a long put, and $p$ for a written put. The strike changes the slope of the profit graph, but does not create a jump in profit.

Method: To solve $\Pi(S_T)>g$, $\Pi(S_T)=g$, or $\Pi(S_T)<g$, first write the profit formula on each side of $K$. Solve the requested comparison on each piece, retain only prices that belong to that piece and satisfy $S_T\geq0$, and combine the resulting ranges. Check the strike and every boundary in the final answer.

Why the Piecewise Method Works

The payoff formulas use a maximum, so each profit expression is affine on either side of the strike. For example, for a long call, if $S_T\leq K$, then $S_T-K\leq0$, so $\max(S_T-K,0)=0$. Profit is therefore $-c$. If $S_T>K$, then $S_T-K>0$, so the payoff is $S_T-K$, and profit is $S_T-K-c$. This proves the two branches directly from the payoff definition.

On each branch, solving a profit comparison is ordinary algebra. The restriction to that branch still matters: a solution obtained from the $S_T>K$ formula cannot include prices below $K$. Combining the valid branch results gives the complete range. This also explains why a position can have a flat region: in that region, profit is constant, so it is either entirely above the target, entirely below it, or equal to it throughout.

Identify the target and units.
Confirm whether the target is per share or for a whole position. Convert it to per-share profit before solving if a contract multiplier is given.

Partition at the strike.
Use the no-exercise and exercise branches, including the strike in a branch where the payoff is zero.

Solve on each branch.
Use the requested strict or non-strict comparison and intersect each algebraic result with its branch.

Combine and verify.
Report the full price range, respect $S_T\geq0$, and test the strike and any equality boundary in the original profit formula.

Worked Applications

Worked Example: A Long Call Above a Positive Profit Target

An investor purchases one call for a premium of $4.50 per share with strike $K=\$52$. At expiration, for which terminal prices is the profit greater than $2.25 per share, equal to $2.25, or less than $2.25?

Method. Split at $S_T=52$ and solve all three comparisons on each branch. This avoids treating the break-even price as if it were automatically the boundary for every profit target.

For $0\leq S_T\leq52$, the call expires without exercise, so $\Pi_{LC}=-4.50$. Since $-4.50<2.25$, every price in this branch gives profit below the target; none gives profit equal to or greater than it.

For $S_T>52$, profit is $S_T-52-4.50=S_T-56.50$. Solving the strict comparison gives $S_T-56.50>2.25$, hence $S_T>58.75$. Equality gives $S_T-56.50=2.25$, hence $S_T=58.75$. Below the target on this branch, $S_T-56.50<2.25$, so $S_T<58.75$, subject to $S_T>52$.

Combining the branches, profit is greater than $2.25 for $S_T>58.75$, equal to $2.25 at $S_T=58.75$, and below $2.25 for $0\leq S_T<58.75$. At the boundary, the profit is $58.75-52-4.50=\$2.25$. At the strike, it is $-\$4.50$, confirming the branch result.

Worked Example: A Written Put With a Negative Target

A trader writes a put with strike $K=\$40$ and receives a premium of $3 per share. Find all terminal prices for which the profit is greater than $-\$5.50$, equal to $-\$5.50$, or less than $-\$5.50$.

Method. Solve on the exercise and no-exercise branches, then combine them. A negative target is still a profit comparison: it specifies a loss threshold, not a change in the profit formula.

For $0\leq S_T<40$, the put payoff is $40-S_T$, so the writer’s profit is $3-(40-S_T)=S_T-37$. Thus $S_T-37>-5.50$ means $S_T>31.50$; equality means $S_T=31.50$; and below the target means $S_T<31.50$. Intersect these results with $0\leq S_T<40$.

For $S_T\geq40$, the put expires without exercise and profit is $3. Since $3>-5.50$, all prices in this branch give profit greater than the target. None give equality or profit below it.

Therefore, profit is greater than $-\$5.50$ for $S_T>31.50$, equal to $-\$5.50$ at $S_T=31.50$, and below $-\$5.50$ for $0\leq S_T<31.50$. At $31.50$, the put payoff is $40-31.50=\$8.50$, and the writer’s profit is $3-8.50=-\$5.50$. At the strike, profit is $3, consistent with the combined range.

Worked Example: A Long Put and a Target Gain

A purchased put has strike $K=\$70$ and premium $p=\$6$ per share. Determine the terminal prices at which profit exceeds $8 per share, equals $8, or is less than $8.

Method. First solve where the put is exercised, then compare the constant profit on the other branch. This prevents extending the decreasing profit formula beyond the strike.

For $0\leq S_T<70$, profit is $70-S_T-6=64-S_T$. The condition $64-S_T>8$ gives $S_T<56$. Equality $64-S_T=8$ gives $S_T=56$, and profit below the target requires $S_T>56$, within this branch.

For $S_T\geq70$, the put expires without exercise and profit is $-\$6$, which is below $8. Combining the branches, profit exceeds $8 for $0\leq S_T<56$, equals $8 at $S_T=56$, and is below $8 for $S_T>56$.

Check the boundary: at $S_T=56$, the payoff is $70-56=\$14$, so profit is $14-6=\$8$. At $S_T=70$, profit is $-\$6$, as the stated below-target range requires.

Worked Example: A Written Call Near Its Target Boundary

A call writer receives a premium of $5 per share for a call with strike $K=\$90$. Find the terminal-price ranges for profit greater than $1, equal to $1, and less than $1.

Method. Use the constant-profit branch below the strike and the decreasing-profit branch above it. The common mistake here is to reverse the inequality when rearranging a decreasing profit expression without changing its direction.

For $0\leq S_T\leq90$, the call expires without exercise and profit is $5, which is greater than $1. For $S_T>90$, profit is $5-(S_T-90)=95-S_T$. Then $95-S_T>1$ implies $-S_T>-94$, so $S_T<94$ after reversing the inequality when multiplying by $-1$. Equality gives $S_T=94$, and profit is below $1 when $S_T>94$.

Combining the branches, profit is greater than $1 for $0\leq S_T<94$, equal to $1 at $S_T=94$, and below $1 for $S_T>94$. At $S_T=94$, the writer’s loss on exercise is $94-90=\$4$, offset by the $5 premium, leaving $1 profit. At the strike, profit is $5.

Endpoint Checks and Common Traps

A useful final check is to substitute the strike and the target boundary into the original profit function. At the strike, the payoff is zero. At an equality boundary, profit must equal the specified target exactly. These checks often reveal a misplaced endpoint or a sign error.

Confusing target profit with break-even. Break-even is the special target $g=0$. For another target, solve $\Pi(S_T)=g$; do not reuse the zero-profit price without adjustment.
Forgetting the flat branch. A purchased call below its strike always has profit $-c$, while a written put above its strike always has profit $p$. Compare that constant with the target before combining intervals.
Including the equality boundary in a strict range. If profit equals the target at $S_T=b$, then the range for profit strictly greater than the target excludes $b$.
Failing to reverse an inequality. For a written call above its strike, profit decreases as $S_T$ increases. Dividing or multiplying an inequality by a negative number reverses its direction.
Ignoring price restrictions. A terminal asset price cannot be negative in this model. Intersect any algebraic solution with $S_T\geq0$ and with the branch on which its formula was derived.

When the target equals the profit on a flat branch, equality holds over an interval rather than at a single price. For example, if a long call’s target is $-c$, then every $S_T$ from zero through $K$ gives exactly that profit; above $K$, profit is greater. This is why the complete piecewise comparison matters.

Key takeaway: A profit-target question is solved by comparing the target with each piece of the profit function, not by reading one threshold in isolation. Keep the branch restrictions, strictness of the inequality, and zero lower bound on $S_T$ in the final answer.

Check Your Understanding

For each position, state the full terminal-price ranges for profit greater than, equal to, and less than the specified target. Unless noted otherwise, amounts are per share and use the expiration-profit convention above.

A long call has strike $K=\$44$, premium $c=\$2.50$, and target profit $-\$2.50$. (Answer: Greater for $S_T>44$; equal for $0\leq S_T\leq44$; less for no terminal prices.)
A long put has strike $K=\$35$, premium $p=\$4$, and target profit $-\$1$. (Answer: Greater for $0\leq S_T<32$; equal at $S_T=32$; less for $S_T>32$.)
A written call has strike $K=\$120$, premium $c=\$7$, and target profit $-\$4$. (Answer: Greater for $0\leq S_T<131$; equal at $S_T=131$; less for $S_T>131$.)
A written put has strike $K=\$25$, premium $p=\$2$, and target profit $-\$10$. (Answer: Greater for $S_T>13$; equal at $S_T=13$; less for $0\leq S_T<13$.)
An investor buys one put contract with multiplier 100, strike $K=\$68$, and premium $p=\$3.20$ per share. Find the terminal-price ranges where the contract profit is greater than, equal to, or less than $1,080. (Answer: Per-share target is $1080/100=\$10.80$; greater for $0\leq S_T<54$; equal at $S_T=54$; less for $S_T>54$.)

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