What dimensions use the least material? What price brings the most revenue? These questions ask for the largest or smallest value of a quantity, which is exactly what the tools from Curve Sketching and Critical Points and Absolute Extrema find. The new skill is building the function from words.
Key idea
Optimization method. (1) Draw a picture and name the variables. (2) Write the quantity to optimize. (3) Use the constraint to write it as a function of one variable. (4) Find the realistic domain. (5) Find the critical numbers and confirm the max or min (closed interval method, or the first or second derivative test). (6) Answer the question that was asked, with units.
Why it works
A real problem has many variables tied together by a constraint (a fixed amount of fence, a fixed volume). Using the constraint to eliminate all but one variable turns the problem into finding the max or min of a function of one variable, which we know how to do. The domain matters because only physically possible values count: a length cannot be negative, and a cut cannot be wider than half the sheet.
Worked example: The best box
Squares of side $x$ are cut from the corners of a 12-in by 12-in sheet, and the sides are folded up. What $x$ gives the largest volume?
- The base is $(12 - 2x)$ by $(12 - 2x)$ and the height is $x$: $V(x) = x(12 - 2x)^2$, with $0 \le x \le 6$.
- $$V'(x) = (12 - 2x)^2 - 4x(12 - 2x) = (12 - 2x)(12 - 6x)$$.
- Critical numbers: $x = 6$ (an endpoint) and $x = 2$.
- Compare: $V(0) = 0$, $V(2) = 2 \cdot 64 = 128$, $V(6) = 0$.
Answer: Cut 2-inch squares; the volume is 128 in³.
Worked example: Fence along a river
A farmer has 600 m of fence for a rectangular field beside a river; no fence is needed along the river. What is the largest possible area?
- Let $x$ be each side perpendicular to the river and $y$ the side parallel to it: $2x + y = 600$, so $y = 600 - 2x$.
- $A(x) = x(600 - 2x) = 600x - 2x^2$ for $0 < x < 300$.
- $A'(x) = 600 - 4x = 0$ at $x = 150$; $A'' = -4 < 0$, a max.
- $y = 300$ and $A = 150 \cdot 300 = 45{,}000$.
Answer: $45{,}000$ m² (a 150 m by 300 m field)
Worked example: The thriftiest can
A closed cylindrical can must hold $16\pi$ cm³. What radius uses the least metal?
- Surface area $S = 2\pi r^2 + 2\pi rh$; volume $\pi r^2h = 16\pi$, so $h = \frac{16}{r^2}$.
- $S(r) = 2\pi r^2 + \frac{32\pi}{r}$ for $r > 0$.
- $S'(r) = 4\pi r - \frac{32\pi}{r^2} = 0$ gives $r^3 = 8$, $r = 2$.
- $S'' = 4\pi + \frac{64\pi}{r^3} > 0$: a min. Then $h = 4$.
Answer: $r = 2$ cm and $h = 4$ cm: the height equals the diameter.
Common mistake
Optimizing the wrong quantity (for example the perimeter when the area was asked), or ignoring the domain and reporting a critical number that is impossible, like a negative length. Check: reread the question at the end, make sure the answer is in the domain, and confirm it is a max or a min.