Solving Optimization Problems
From setup to the winning number
- You've reduced the problem to one variable. Now solve the optimization.
- The plan: differentiate the objective, find critical points, and confirm which is the max or min.
- Then answer the actual question — often the dimensions or the optimal value, with units.
- It's the extremum machinery of this unit, applied to a story.
The solving steps
- 1. Differentiate the single-variable objective and set it to zero to find critical points.
- 2. Confirm max vs. min with the first or second derivative test.
- 3. Check the endpoints of the domain too, if the domain is closed or bounded.
- 4. Answer the question asked — plug the optimal variable back in.
The area has a single peak
y = ax² + bx
The area parabola opens downward with a single peak — the derivative is zero at the vertex, a maximum.
For $A(x)=40x-2x^2$, solve $A'(x)=40-4x=0$ for the critical $x$.
$40-4x=0\Rightarrow x=10$.
Order the optimization solving steps.
Differentiate, find, confirm, answer.
Confirm, don't assume
- A critical point is only a candidate — you must show it's the max (or min) you want.
- Second Derivative Test: if $A''(x)<0$ at the critical point, it's a maximum.
- Or check the sign of $A'$ around it, or compare against domain endpoints.
- Skipping the confirmation is a common lost-mark; the AP expects a justification.
Since $A''(x)=-4<0$, the critical point is a...
Concave down ⇒ maximum.
You should justify that a critical point is truly the maximum, not just assume it.
A derivative test or endpoint check is expected.
Answer the actual question
- If asked for the maximum area, report the area value; if asked for dimensions, report the lengths.
- Include units and check the answer is physically reasonable.
- Reread the prompt — "what dimensions?" and "what's the largest area?" want different final numbers.
- A number without the right label loses the point.
With $x=10$ and $A(x)=40x-2x^2$, find the maximum area.
$A(10)=400-200=200$.
If the question asks for the dimensions, you should report...
Report what is asked — dimensions, not the area, here.
Two common slips: (1) reporting the critical $x$-value when the question wants the optimized quantity (or vice versa) — reread what's asked. (2) Skipping the justification that your critical point is truly the max/min. Always confirm with a derivative test or endpoint check.
Maximize area $A(x)=40x-2x^2$ (from the fence setup), $0
- $A'(x)=40-4x=0\Rightarrow x=10$.
- $A''(x)=-4<0$ → concave down → a maximum ✓.
- Then $y=40-2(10)=20$, and $A=10\cdot20=200\ \text{m}^2$. Dimensions $10\times20$, max area $200\ \text{m}^2$.
To solve an optimization: differentiate the one-variable objective, set it to zero for critical points, confirm the max/min with a derivative test (and check domain endpoints), then answer the exact question with units. Don't report the wrong quantity, and always justify.