Lessons 9 · 1 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 11, Parts 1 and 2, 1st edition. ZAMIN NASHR, Tashkent, 2018
9
Differential calculus in extremal (optimisation) problems
Textbook, Part 1: pp. 50–55
GoalSolve geometric, physical and economic extremal problems by forming a function and using the derivative.
New words
extremal (optimisation) problem · ekstremal masalaacceleration · tezlanishrevenue / profit · daromaddomain · aniqlanish sohasi
Explanation
An extremal problem asks for the greatest or least value of some quantity. Procedure: 1) choose a variable (say x) and write its admissible values (the domain); 2) express the quantity to optimise as a function of that variable, eliminating other variables using the conditions; 3) find the derivative and the stationary points; 4) decide the greatest or least value by the sign of the derivative or by the values at the endpoints; 5) state the answer in the language of the problem. In physics velocity is v = s′ and acceleration is a = v′ = s″ (m/s²), so for “greatest speed” you study v and for “greatest acceleration” you study a. In economics the revenue or profit is a function P(x) of the number x of items and its maximum comes from P′(x) = 0.
Worked examples
A rectangular garden next to a wall is to be fenced on three sides with 80 m of netting (the wall side is not fenced). Let each side perpendicular to the wall be x m; the side parallel to the wall is 80 − 2x m, and the area is S(x) = x(80 − 2x), 0 < x < 40. S′ = 80 − 4x = 0 ⇒ x = 20; S′ changes from + to −, so this is a maximum: S(20) = 20 · 40 = 800 m².
The law of motion is s(t) = −t³ + 12t² (m). v(t) = −3t² + 24t, a(t) = −6t + 24. The greatest speed occurs when v′ = a = 0: t = 4 s (a changes from + to −). v(4) = −48 + 96 = 48 m/s. The distance covered by then is s(4) = −64 + 192 = 128 m.
Class activity
“Design the box”: pairs make open boxes from A4 paper with different corner sizes x (cutting is done by or with an adult; be careful with scissors), compute the volumes into a table and guess which x gives the greatest volume. Then compare the result found with the derivative to the table.
Practice
1
Two positive numbers have sum 20. What is the greatest possible value of their product?
100
2
A point moves by s(t) = t³ − 9t² + 24t (m). At what time (s) is the acceleration zero?
3
3
If a café sets the price of a sandwich at x thousand so‘m, it sells 120 − 10x sandwiches a day. At the price giving the greatest daily revenue, what is the revenue in thousand so‘m?
360
4
Why does the stationary point of the area function S(x) = x(80 − 2x) give a greatest value and not a least value?
S′ = 80 − 4x is positive for x < 20 (S increases) and negative for x > 20 (S decreases). At the ends S(0) = S(40) = 0, so the interior point is a maximum.