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Maximum and Minimum Problems

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AQA Level 2 Further Maths practice questions, each with a hint and a full worked solution.

Have a pen and paper handy, to work out your answers.

Question 13 marks

A company makes x thousand games consoles each month.

The monthly profit, \pounds P thousand, is given by

P=96x-2x^3

Use calculus to work out the maximum value of P.

Hint

Differentiate, set \dfrac{dP}{dx}=0 to find x, then substitute that x back into P.

Worked solution
  1. \dfrac{dP}{dx}=96-6x^2
  2. At a maximum, 96-6x^2=0
  3. x^2=16, so x=4 (the number of consoles cannot be negative)
  4. P=96\times4-2\times4^3=384-128=256
  5. The question asks for the maximum value of P, not x
  6. Answer: 256 (a profit of £256 thousand)

Question 22 marks

A farmer uses 120 m of fencing to make a rectangular pen.

The pen is divided into two parts by another fence of length x m, parallel to the two sides of length x m, as shown in the diagram.

All of the 120 m of fencing is used, including the dividing fence.

The pen is x m wide and y m long.

The total area of the pen is A m^2.

Work out an expression for A in terms of x only.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Count how many lengths of x and y the fencing makes, use this to write y in terms of x, then substitute into A=xy.

Worked solution
  1. The fencing is three widths and two lengths: 3x+2y=120
  2. y=\dfrac{120-3x}{2}=60-1.5x
  3. A=xy=x(60-1.5x)
  4. Answer: A=60x-\frac32x^2

Question 3Challenge6 marks

A closed box is a cuboid.

The base of the box is 2x cm long and x cm wide.

The height of the box is h cm.

The total surface area of the box is 432 cm^2.

The volume of the box is V cm^3.

(a)

Work out an expression for V in terms of x only.

Simplify your answer.

3 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

(b)

Use calculus to work out the maximum volume of the box.

3 marks

Hint

Write the surface area as six faces to get h in terms of x, substitute into V=2x\times x\times h, then differentiate and set \dfrac{dV}{dx}=0.

Worked solution

Part (a)

  1. Surface area: two faces 2x\times x, two faces 2x\times h, two faces x\times h
  2. 2(2x^2)+2(2xh)+2(xh)=4x^2+6xh
  3. 4x^2+6xh=432, so h=\dfrac{432-4x^2}{6x}
  4. V=2x\times x\times h=2x^2\times\dfrac{432-4x^2}{6x}
  5. V=\dfrac{x(432-4x^2)}{3}
  6. Answer: V=144x-\frac43x^3

Part (b)

  1. \dfrac{dV}{dx}=144-4x^2
  2. Set 144-4x^2=0, so x^2=36 and x=6 (a length is positive)
  3. \dfrac{d^2V}{dx^2}=-8x=-48<0, so this is a maximum
  4. V=144\times6-\frac43\times6^3=864-288=576
  5. Answer: 576 cm^3

More on this topic: Maximum and Minimum Problems worksheet with full solutions

All AQA Level 2 Further Maths practice questions

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