Setting up optimization problems
Word problems become calculus problems through one discipline: one objective function, one constraint, one derivative set to zero.
Four-step setup
First name the variables and draw a picture. Second, write the quantity to maximize or minimize as a function of one variable, using the constraint to eliminate the extra variable. Third, differentiate and set the derivative to zero to find critical numbers. Fourth, verify the answer with a test or the problem's context, then state it in words.
A = x y, \quad \text{constraint: } 2x + y = 120Worked example
A farmer has 120 meters of fence for a rectangular pen along a river, needing no fence on the river side. With x the two widths and y the length, the constraint is 2x + y = 120, so y = 120 - 2x. The area A(x) = x(120 - 2x) = 120x - 2x^2 has derivative 120 - 4x, zero at x = 30. Then y = 60 and the largest area is 1800 square meters.
A(x) = 120x - 2x^2, \quad A'(x) = 120 - 4x = 0Confirming the optimum
A critical number is only a candidate, so confirm it is the extremum you want. The second derivative test works here: A'' = -4 is negative, so x = 30 is a maximum. Context also helps, since area is zero at the fence running out, so the single interior critical point must be the peak.
A''(x) = -4 < 0Flashcard
Critical condition for the objective function
Check your understanding
Two positive numbers have product 36. What is their smallest possible sum, and when does it occur?