Lesson 4.8
4.8 Using calculus to find maxima and minima in simple problems Quiz: AQA Further Maths, Unit 4
20 questions
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Lesson 4.8, Using calculus to find maxima and minima in simple problems: 20 multiple choice questions for the AQA GCSE Further Maths (8365), Unit 4: Calculus, written with Revision Ninja.
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The 20 questions
-
A rectangle has perimeter 40 and width x. What is its area in terms of x?
- x(20 + x)
- x(20 - x)
- x(40 - x)
- 2x(20 - x)
-
What value of x maximises A = x(20 - x)?
- 5
- 15
- 20
- 10
-
What is the maximum area of the rectangle with perimeter 40?
- 50
- 400
- 200
- 100
-
A rectangle has area 16 and sides x and 16/x. Which x minimises its perimeter?
- 8
- 2
- 4
- 16
-
What is the minimum perimeter of a rectangle of area 16?
- 32
- 4
- 16
- 8
-
For C = x^2 + 16/x with x > 0, what value of x gives the minimum cost?
- x = 2
- x = 4
- x = 1
- x = 8
-
For C = x^2 + 16/x, what does d^2C/dx^2 > 0 at x = 2 show?
- A point of inflection
- A maximum
- Only a zero gradient
- A minimum
-
For x > 0, which value of x minimises x + 1/x?
- 0
- 1
- -1
- 2
-
What is the minimum value of x + 1/x for x > 0?
- 2
- 1
- 0
- -2
-
A ball has height h = 12t - 3t^2. At what time is the height greatest?
- t = 1
- t = 4
- t = 6
- t = 2
-
What is the greatest height of the ball with h = 12t - 3t^2?
- 36
- 24
- 6
- 12
-
A rectangle has its base on the x-axis and top corners on y = 9 - x^2, giving area A = 2x(9 - x^2). Which x maximises A?
- sqrt(3)
- sqrt(6)
- 3
- 1.5
-
What is the maximum area for the rectangle described with A = 2x(9 - x^2)?
- 12 sqrt(3)
- 6 sqrt(3)
- 36
- 18 sqrt(3)
-
What is the minimum value of x + 4/x for x > 0?
- 2
- 4
- 5
- 3
-
Two positive numbers have sum 20 and product as large as possible. What are they?
- 10 and 10
- 8 and 12
- 4 and 16
- 5 and 15
-
A cylinder has volume 16 pi, with surface area S = 2 pi r^2 + 32 pi / r. Which r minimises S?
- 2
- 1
- 4
- 8
-
For the cylinder with S = 2 pi r^2 + 32 pi / r, what is the minimum surface area?
- 48 pi
- 16 pi
- 24 pi
- 32 pi
-
A rectangle has perimeter 24 and width W. Its area is A = W(12 - W). What width gives the maximum area?
- 6
- 4
- 3
- 12
-
What is the largest value of y = -x^2 + 4x + 1?
- 1
- 9
- 5
- 4
-
Profit is P = 100x - 2x^2 for x units. Which x maximises profit?
- 100
- 50
- 12.5
- 25
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