Optimisation Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Optimisation questions. See exactly how to solve problems on optimisation, differentiation, area, cost.

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A-Level70 questionsStep-by-step solutions
Question 1
3 markseasy
The area A m2A\,\text{m}^2 of a rectangular region is modelled by A=40x−x2A = 40x - x^2, where xx metres is one of the side lengths. Find the maximum possible area.

Worked solution

  1. Write down the quantity to be maximised

    A=40x−x2A = 40x - x^2

    The area is already written as a function of the single variable xx, so we can differentiate it straight away.

  2. Differentiate with respect to xx

    dAdx=40−2x\frac{dA}{dx}=40 - 2 x

    Differentiating gives the gradient function. At a turning point the gradient is zero, so this is the expression we set to zero.

  3. Set the derivative equal to zero

    40−2x=040 - 2 x=0

    At a maximum or minimum the curve is momentarily flat, so its gradient is zero. This equation locates the stationary point(s).

  4. Solve for xx

    x=20x=20

    Rearranging gives the value of xx at the stationary point.

  5. Confirm it is a maximum with the second derivative

    d2Adx2=−2<0\frac{d^2A}{dx^2}=-2<0

    Differentiating a second time gives a negative value, so by the second-derivative test the stationary point is a maximum (a peak).

  6. Work out the maximum area

    A=400A=400

    Finally substitute x=20x=20 back in to obtain the maximum area.

Answer
A=400 m2A=400\,\text{m}^2
Question 2
3 markseasy
The area A m2A\,\text{m}^2 of a rectangular region is modelled by A=60x−x2A = 60x - x^2, where xx metres is one of the side lengths. Find the maximum possible area.

Worked solution

  1. Write down the quantity to be maximised

    A=60x−x2A = 60x - x^2

    The area is already written as a function of the single variable xx, so we can differentiate it straight away.

  2. Differentiate with respect to xx

    dAdx=60−2x\frac{dA}{dx}=60 - 2 x

    Differentiating gives the gradient function. At a turning point the gradient is zero, so this is the expression we set to zero.

  3. Set the derivative equal to zero

    60−2x=060 - 2 x=0

    At a maximum or minimum the curve is momentarily flat, so its gradient is zero. This equation locates the stationary point(s).

  4. Solve for xx

    x=30x=30

    Rearranging gives the value of xx at the stationary point.

  5. Confirm it is a maximum with the second derivative

    d2Adx2=−2<0\frac{d^2A}{dx^2}=-2<0

    Differentiating a second time gives a negative value, so by the second-derivative test the stationary point is a maximum (a peak).

  6. Work out the maximum area

    A=900A=900

    Finally substitute x=30x=30 back in to obtain the maximum area.

Answer
A=900 m2A=900\,\text{m}^2
Question 3
3 markseasy
The area A m2A\,\text{m}^2 of a rectangular region is modelled by A=24x−x2A = 24x - x^2, where xx metres is one of the side lengths. Find the maximum possible area.

Worked solution

  1. Write down the quantity to be maximised

    A=24x−x2A = 24x - x^2

    The area is already written as a function of the single variable xx, so we can differentiate it straight away.

  2. Differentiate with respect to xx

    dAdx=24−2x\frac{dA}{dx}=24 - 2 x

    Differentiating gives the gradient function. At a turning point the gradient is zero, so this is the expression we set to zero.

  3. Set the derivative equal to zero

    24−2x=024 - 2 x=0

    At a maximum or minimum the curve is momentarily flat, so its gradient is zero. This equation locates the stationary point(s).

  4. Solve for xx

    x=12x=12

    Rearranging gives the value of xx at the stationary point.

  5. Confirm it is a maximum with the second derivative

    d2Adx2=−2<0\frac{d^2A}{dx^2}=-2<0

    Differentiating a second time gives a negative value, so by the second-derivative test the stationary point is a maximum (a peak).

  6. Work out the maximum area

    A=144A=144

    Finally substitute x=12x=12 back in to obtain the maximum area.

Answer
A=144 m2A=144\,\text{m}^2
Question 4
3 markseasy
The area A m2A\,\text{m}^2 of a rectangular region is modelled by A=30x−x2A = 30x - x^2, where xx metres is one of the side lengths. Find the maximum possible area.

Worked solution

  1. Write down the quantity to be maximised

    A=30x−x2A = 30x - x^2

    The area is already written as a function of the single variable xx, so we can differentiate it straight away.

  2. Differentiate with respect to xx

    dAdx=30−2x\frac{dA}{dx}=30 - 2 x

    Differentiating gives the gradient function. At a turning point the gradient is zero, so this is the expression we set to zero.

  3. Set the derivative equal to zero

    30−2x=030 - 2 x=0

    At a maximum or minimum the curve is momentarily flat, so its gradient is zero. This equation locates the stationary point(s).

  4. Solve for xx

    x=15x=15

    Rearranging gives the value of xx at the stationary point.

  5. Confirm it is a maximum with the second derivative

    d2Adx2=−2<0\frac{d^2A}{dx^2}=-2<0

    Differentiating a second time gives a negative value, so by the second-derivative test the stationary point is a maximum (a peak).

  6. Work out the maximum area

    A=225A=225

    Finally substitute x=15x=15 back in to obtain the maximum area.

Answer
A=225 m2A=225\,\text{m}^2
Question 5
3 markseasy
The running cost CC (in pounds) of a machine is modelled by C=x2−20x+150C = x^2 - 20x + 150, where xx is the speed setting. Find the minimum cost.

Worked solution

  1. Write down the quantity to be minimised

    C=x2−20x+150C = x^2 - 20x + 150

    The cost is given directly as a function of xx, so we differentiate to find the lowest point of the curve.

  2. Differentiate with respect to xx

    dCdx=2x−20\frac{dC}{dx}=2 x - 20

    Differentiating gives the gradient function. At a turning point the gradient is zero, so this is the expression we set to zero.

  3. Set the derivative equal to zero

    2x−20=02 x - 20=0

    At a maximum or minimum the curve is momentarily flat, so its gradient is zero. This equation locates the stationary point(s).

  4. Solve for xx

    x=10x=10

    Rearranging gives the value of xx at the stationary point.

  5. Confirm it is a minimum with the second derivative

    d2Cdx2=2>0\frac{d^2C}{dx^2}=2>0

    Differentiating a second time gives a positive value, so by the second-derivative test the stationary point is a minimum (a trough).

  6. Work out the minimum cost

    C=50C=50

    Finally substitute x=10x=10 back in to obtain the minimum cost.

Answer
C=50 poundsC=50\,\text{pounds}

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