M171 Calculus I · Unit 4: Optimization & the integral · on Exam 4
§3.4 Applied optimization
Study notes for Applied optimization in M171 Calculus I at Montana State University: the big ideas, key formulas, a worked example, common mistakes and an exam tip. The same topic has endless practice questions and flashcards in Mathub.
Big ideas
- Procedure: draw a picture and name variables; write the quantity to optimize (objective); write the constraint; use the constraint to make the objective a function of one variable; state the domain; find critical numbers; check endpoints or use a derivative test; answer the actual question with units.
- Classic set-ups: fencing a rectangle (perimeter constraint, area objective), open box from a sheet (cut corners), minimizing material for a can, shortest distance from a point to a curve.
Key formulas
Rectangle, fixed perimeter P
$$A(x) = x\left(\tfrac{P}{2}-x\right),\quad \text{max at } x = \tfrac P4$$
Three-sided fence, length F
$$A(x) = x(F-2x),\quad \text{max at } x = \tfrac F4,\ A = \tfrac{F^2}{8}$$
Worked example
A farmer has 200 ft of fence for a rectangular pen against a straight river (no fence on the river side). Maximize the area.
Show the solution
Let $x$ be the side perpendicular to the river: $2x + y = 200$, so $y = 200-2x$ and $A = x(200-2x) = 200x - 2x^2$ on $0\le x\le100$. $A' = 200 - 4x = 0 \Rightarrow x = 50$, $y = 100$. $A'' = -4 < 0$ so it is a max: $A = 5000$ ft$^2$.
Common mistakes
- Optimizing the constraint instead of the objective.
- Not stating the domain, so an endpoint answer gets missed.
Exam tipThe last line of your answer should restate what was asked, with units: "The maximum area is 5000 ft² when the pen is 50 ft by 100 ft."
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