Optimisation problems · Grade 10 · 4 units
In Optimisation problems (Grade 10 · 4 units) you practise 5 skills: Optimising with numbers · Optimising shapes · Optimising on graphs · Optimising profit and cost · Optimising on a closed interval. Solve an optimisation problem in five steps: choose a variable, write the quantity you want to make largest or smallest as a function of it, state the interval, differentiate and set the derivative to zero, and check the ends of the interval too. The answer is the quantity the question asks about, which is not always the variable.
For example: a rectangle with a perimeter of 20. If one side is x, the other side is ten minus x, and the area is x times ten minus x. The derivative is ten minus twice x, which is zero at x equals five. To the left of five the derivative is positive and to the right negative, so it is a maximum: the largest area is 25.
Tip: Write in words what you want to make largest or smallest before you write a function.
What you learn
Optimising with numbers
For example: Two positive numbers add up to 14 . What is the largest possible product of the two?
Answer and solution
The answer: 49
One number is x , and the other is 14 − x . The product: f(x) = x(14 − x) = −x² + 14x The domain: 0 < x < 14 f′(x) = −2x + 14 = 0 x = 7 For x < 7 the derivative is positive, and for x > 7 it is negative. This is a maximum point. The largest product is f(7) = 7 × 7 = 49
Optimising shapes
For example: The perimeter of a rectangle is 16 . What is the largest possible area of the rectangle?
Answer and solution
The answer: 16
One side is x , and the other side is 8 − x . The area: S(x) = x(8 − x) = −x² + 8x The domain: 0 < x < 8 S′(x) = −2x + 8 = 0 x = 4 For x < 4 the derivative is positive, and for x > 4 it is negative. This is a maximum point. The rectangle is a square, and the largest area is S(4) = 4 × 4 = 16
Optimising on graphs
For example: Which point on the line y = x + 6 is closest to the origin?
Answer and solution
The answer: (−3, 3)
The distance is smallest where its square is smallest, so look for the minimum of the distance squared: f(x) = x² + (x + 6)² = 2x² + 12x + 36 The domain: any number. f′(x) = 4x + 12 = 0 x = −3 For x < −3 the derivative is negative, and for x > −3 it is positive. This is a minimum point. y = 3 . The point is (−3, 3)
Optimising profit and cost
For example: The profit of a factory from making and selling x items is P(x) = −x² + 40x − 30 . How many items give the largest profit?
Answer and solution
The answer: 20
The domain: x > 0 P′(x) = −2x + 40 = 0 x = 20 For x < 20 the derivative is positive, and for x > 20 it is negative. This is a maximum point. The largest profit comes from making 20 items.
Optimising on a closed interval
For example: Find the largest value of the function f(x) = −x² + 2x + 4 on the interval −5 ≤ x ≤ −3 .
Answer and solution
The answer: −11
f′(x) = −2x + 2 = 0 x = 1 . This point is outside the interval, so the function only rises or only falls on it, and the largest value is at an end: f(−5) = −31 f(−3) = −11 . The largest value is −11 .
Not on the site yet
Parts of this topic that are not built yet: Writing the function from an open story · Optimisation with triangles in a circle · Optimisation with trigonometry.
The exercises are made by rules and the answers are worked out in code, so there are new exercises every time. The topic follows the Israeli Ministry of Education’s high school maths curriculum at 4 units (grades 10 to 12).