Hard A-Level Optimisation Questions

Challenging, exam-style A-Level Optimisation questions with worked solutions. Stretch yourself on the hardest optimisation, differentiation, volume, area problems.

optimisationdifferentiationvolumeareacost
A-Level34 questionsStep-by-step solutions
Question 1
10 markschallenging
A cylinder is inscribed in a sphere of radius RR. Using calculus, which statement correctly gives the cylinder height hh that maximises the volume, with fully valid reasoning?
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Worked solution

  1. Set up the geometry

    sphere radius R, cylinder height h, radius ρ\text{sphere radius }R,\ \text{cylinder height }h,\ \text{radius }\rho

    The cylinder's two ends touch the sphere, and by symmetry the sphere's centre is level with the middle of the cylinder.

  2. Use symmetry

    each end is h2 from the centre\text{each end is }\tfrac{h}{2}\text{ from the centre}

    Because the cylinder is centred in the sphere, the top and bottom are each a distance h/2h/2 from the centre.

  3. Apply Pythagoras

    (h2)2+ρ2=R2\left(\tfrac{h}{2}\right)^2+\rho^2=R^2

    The radius, the half-height and the sphere radius form a right-angled triangle.

  4. Make ρ2\rho^2 the subject

    ρ2=R2−h24\rho^2=R^2-\tfrac{h^2}{4}

    This removes ρ\rho so the volume can be written using hh only.

  5. Write the volume

    V=πρ2hV=\pi\rho^2 h

    The cylinder volume is the base area times the height.

  6. Substitute

    V=π(R2h−h34)V=\pi\left(R^2 h-\tfrac{h^3}{4}\right)

    Replacing ρ2\rho^2 gives the volume as a cubic in hh.

  7. Differentiate

    dVdh=π(R2−3h24)\frac{dV}{dh}=\pi\left(R^2-\tfrac{3h^2}{4}\right)

    Differentiating term by term gives the gradient of the volume curve.

  8. Set the derivative to zero

    R2−3h24=0R^2-\tfrac{3h^2}{4}=0

    At the maximum the gradient is zero, giving an equation for hh.

  9. Rearrange

    h2=4R23h^2=\tfrac{4R^2}{3}

    Solving for h2h^2 isolates the square of the height.

  10. Take the positive root

    h=2R3h=\frac{2R}{\sqrt{3}}

    A height must be positive, so we discard the negative square root.

  11. Rationalise the surd

    h=233Rh=\frac{2\sqrt{3}}{3}R

    Multiplying top and bottom by 3\sqrt{3} writes the answer without a surd in the denominator.

  12. Differentiate again

    d2Vdh2=−3πh2\frac{d^2V}{dh^2}=-\frac{3\pi h}{2}

    The second derivative tells us whether this stationary point is a maximum or a minimum.

  13. Check the sign

    −3πh2<0 (h>0)-\frac{3\pi h}{2}<0\ (h>0)

    For a positive height the second derivative is negative, so the point is a maximum.

  14. Find the corresponding radius

    ρ2=R2−h24=2R23\rho^2=R^2-\frac{h^2}{4}=\frac{2R^2}{3}

    Substituting the optimal height back into the Pythagoras relation gives the matching radius, confirming the cylinder is valid.

  15. State the conclusion

    h=2R3 (maximum)h=\frac{2R}{\sqrt{3}}\ (\text{maximum})

    So the cylinder of greatest volume has height 2R3\dfrac{2R}{\sqrt3}, and this is genuinely a maximum.

Answer
h=2R3h=\dfrac{2R}{\sqrt{3}}, a maximum because d2Vdh2=−3πh2<0\dfrac{d^2V}{dh^2}=-\dfrac{3\pi h}{2}<0
Question 2
10 markschallenging
A closed cylindrical can must hold 256π cm3256 \pi\,\text{cm}^3. The metal for the top and bottom costs twice as much per unit area as the metal for the curved side. Find the minimum total cost (in cost units), giving your answer as a multiple of π\pi.
Show worked solution

Worked solution

  1. Plan the approach

    optimise C(r)\text{optimise }C(r)

    The strategy is to write the quantity we want as a function of a single variable using the given constraint, then use differentiation to find and classify the turning point.

  2. Set up the cost per area

    top+base: 2 per cm2, side: 1 per cm2\text{top+base: }2\,\text{per cm}^2,\ \text{side: }1\,\text{per cm}^2

    The metal for the two circular ends costs twice as much per square centimetre as the metal for the curved side.

  3. Write the fixed volume

    πr2h=256π\pi r^2 h=256 \pi

    The can must hold 256π cm3256 \pi\,\text{cm}^3, linking the radius rr and height hh.

  4. Make hh the subject

    h=256ππr2=256r2h=\frac{256 \pi}{\pi r^2}=\frac{256}{r^2}

    Dividing by πr2\pi r^2 writes the height in terms of the radius.

  5. Write the cost function

    C=2 (2πr2)+1 (2πrh)=4πr2+2πrhC=2\,(2\pi r^2)+1\,(2\pi r h)=4\pi r^2+2\pi r h

    The end pieces have total area 2πr22\pi r^2 costed at 22 each, and the curved side has area 2πrh2\pi r h costed at 11.

  6. Substitute for hh

    C=4πr2+512πrC=4\pi r^2+\frac{512 \pi}{r}

    Replacing hh makes the cost a function of rr only, ready to differentiate.

  7. Note the aim

    minimise C(r)\text{minimise } C(r)

    We differentiate to find the radius giving the cheapest can.

  8. Rewrite using index notation

    C=4π(r3+128)rC=\frac{4 \pi \left(r^{3} + 128\right)}{r}

    Before differentiating, write any fractions as negative powers so that the power rule can be applied to every term.

  9. Differentiate with respect to rr

    dCdr=8π(r3−64)r2\frac{dC}{dr}=\frac{8 \pi \left(r^{3} - 64\right)}{r^{2}}

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

  10. Set the derivative equal to zero

    8π(r3−64)r2=0\frac{8 \pi \left(r^{3} - 64\right)}{r^{2}}=0

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

  11. Clear fractions and rearrange

    8πr3−512π=08 \pi r^{3} - 512 \pi=0

    Multiplying through to remove the fraction leaves a simpler polynomial equation that is easier to solve.

  12. Solve for rr

    r=4r=4

    Rearranging gives the value of rr at the stationary point.

  13. Differentiate again

    d2Cdr2=8π+1024πr3\frac{d^2C}{dr^2}=8 \pi + \frac{1024 \pi}{r^{3}}

    The second derivative tells us whether the stationary point is a peak or a trough.

  14. Confirm it is a minimum

    d2Cdr2∣r=4=24π>0\left.\frac{d^2C}{dr^2}\right|_{r=4}=24 \pi>0

    The second derivative is positive here, so by the second-derivative test the stationary point is a minimum.

  15. Find the height

    h=16 cmh=16\,cm

    Substituting r=4r=4 into the constraint gives the height, which we need for the final quantity.

  16. Work out the minimum cost

    C=192πC=192 \pi

    Finally substitute r=4r=4 back in to obtain the minimum cost.

  17. State and interpret the result

    Copt=192πC_{\text{opt}}=192 \pi

    This is the required optimal value; the working above shows both that it is a genuine turning point and that it is the correct type (maximum or minimum).

Answer
C=192πC=192 \pi
Question 3
10 markschallenging
A closed cylindrical can must hold 108π cm3108 \pi\,\text{cm}^3. The metal for the top and bottom costs twice as much per unit area as the metal for the curved side. Find the minimum total cost (in cost units), giving your answer as a multiple of π\pi.
Show worked solution

Worked solution

  1. Plan the approach

    optimise C(r)\text{optimise }C(r)

    The strategy is to write the quantity we want as a function of a single variable using the given constraint, then use differentiation to find and classify the turning point.

  2. Set up the cost per area

    top+base: 2 per cm2, side: 1 per cm2\text{top+base: }2\,\text{per cm}^2,\ \text{side: }1\,\text{per cm}^2

    The metal for the two circular ends costs twice as much per square centimetre as the metal for the curved side.

  3. Write the fixed volume

    πr2h=108π\pi r^2 h=108 \pi

    The can must hold 108π cm3108 \pi\,\text{cm}^3, linking the radius rr and height hh.

  4. Make hh the subject

    h=108ππr2=108r2h=\frac{108 \pi}{\pi r^2}=\frac{108}{r^2}

    Dividing by πr2\pi r^2 writes the height in terms of the radius.

  5. Write the cost function

    C=2 (2πr2)+1 (2πrh)=4πr2+2πrhC=2\,(2\pi r^2)+1\,(2\pi r h)=4\pi r^2+2\pi r h

    The end pieces have total area 2πr22\pi r^2 costed at 22 each, and the curved side has area 2πrh2\pi r h costed at 11.

  6. Substitute for hh

    C=4πr2+216πrC=4\pi r^2+\frac{216 \pi}{r}

    Replacing hh makes the cost a function of rr only, ready to differentiate.

  7. Note the aim

    minimise C(r)\text{minimise } C(r)

    We differentiate to find the radius giving the cheapest can.

  8. Rewrite using index notation

    C=4π(r3+54)rC=\frac{4 \pi \left(r^{3} + 54\right)}{r}

    Before differentiating, write any fractions as negative powers so that the power rule can be applied to every term.

  9. Differentiate with respect to rr

    dCdr=8π(r3−27)r2\frac{dC}{dr}=\frac{8 \pi \left(r^{3} - 27\right)}{r^{2}}

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

  10. Set the derivative equal to zero

    8π(r3−27)r2=0\frac{8 \pi \left(r^{3} - 27\right)}{r^{2}}=0

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

  11. Clear fractions and rearrange

    8πr3−216π=08 \pi r^{3} - 216 \pi=0

    Multiplying through to remove the fraction leaves a simpler polynomial equation that is easier to solve.

  12. Solve for rr

    r=3r=3

    Rearranging gives the value of rr at the stationary point.

  13. Differentiate again

    d2Cdr2=8π+432πr3\frac{d^2C}{dr^2}=8 \pi + \frac{432 \pi}{r^{3}}

    The second derivative tells us whether the stationary point is a peak or a trough.

  14. Confirm it is a minimum

    d2Cdr2∣r=3=24π>0\left.\frac{d^2C}{dr^2}\right|_{r=3}=24 \pi>0

    The second derivative is positive here, so by the second-derivative test the stationary point is a minimum.

  15. Find the height

    h=12 cmh=12\,cm

    Substituting r=3r=3 into the constraint gives the height, which we need for the final quantity.

  16. Work out the minimum cost

    C=108πC=108 \pi

    Finally substitute r=3r=3 back in to obtain the minimum cost.

  17. State and interpret the result

    Copt=108πC_{\text{opt}}=108 \pi

    This is the required optimal value; the working above shows both that it is a genuine turning point and that it is the correct type (maximum or minimum).

Answer
C=108πC=108 \pi
Question 4
10 markschallenging
A closed cylindrical can must hold 32π cm332 \pi\,\text{cm}^3. The metal for the top and bottom costs twice as much per unit area as the metal for the curved side. Find the minimum total cost (in cost units), giving your answer as a multiple of π\pi.
Show worked solution

Worked solution

  1. Plan the approach

    optimise C(r)\text{optimise }C(r)

    The strategy is to write the quantity we want as a function of a single variable using the given constraint, then use differentiation to find and classify the turning point.

  2. Set up the cost per area

    top+base: 2 per cm2, side: 1 per cm2\text{top+base: }2\,\text{per cm}^2,\ \text{side: }1\,\text{per cm}^2

    The metal for the two circular ends costs twice as much per square centimetre as the metal for the curved side.

  3. Write the fixed volume

    πr2h=32π\pi r^2 h=32 \pi

    The can must hold 32π cm332 \pi\,\text{cm}^3, linking the radius rr and height hh.

  4. Make hh the subject

    h=32ππr2=32r2h=\frac{32 \pi}{\pi r^2}=\frac{32}{r^2}

    Dividing by πr2\pi r^2 writes the height in terms of the radius.

  5. Write the cost function

    C=2 (2πr2)+1 (2πrh)=4πr2+2πrhC=2\,(2\pi r^2)+1\,(2\pi r h)=4\pi r^2+2\pi r h

    The end pieces have total area 2πr22\pi r^2 costed at 22 each, and the curved side has area 2πrh2\pi r h costed at 11.

  6. Substitute for hh

    C=4πr2+64πrC=4\pi r^2+\frac{64 \pi}{r}

    Replacing hh makes the cost a function of rr only, ready to differentiate.

  7. Note the aim

    minimise C(r)\text{minimise } C(r)

    We differentiate to find the radius giving the cheapest can.

  8. Rewrite using index notation

    C=4π(r3+16)rC=\frac{4 \pi \left(r^{3} + 16\right)}{r}

    Before differentiating, write any fractions as negative powers so that the power rule can be applied to every term.

  9. Differentiate with respect to rr

    dCdr=8π(r3−8)r2\frac{dC}{dr}=\frac{8 \pi \left(r^{3} - 8\right)}{r^{2}}

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

  10. Set the derivative equal to zero

    8π(r3−8)r2=0\frac{8 \pi \left(r^{3} - 8\right)}{r^{2}}=0

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

  11. Clear fractions and rearrange

    8πr3−64π=08 \pi r^{3} - 64 \pi=0

    Multiplying through to remove the fraction leaves a simpler polynomial equation that is easier to solve.

  12. Solve for rr

    r=2r=2

    Rearranging gives the value of rr at the stationary point.

  13. Differentiate again

    d2Cdr2=8π+128πr3\frac{d^2C}{dr^2}=8 \pi + \frac{128 \pi}{r^{3}}

    The second derivative tells us whether the stationary point is a peak or a trough.

  14. Confirm it is a minimum

    d2Cdr2∣r=2=24π>0\left.\frac{d^2C}{dr^2}\right|_{r=2}=24 \pi>0

    The second derivative is positive here, so by the second-derivative test the stationary point is a minimum.

  15. Find the height

    h=8 cmh=8\,cm

    Substituting r=2r=2 into the constraint gives the height, which we need for the final quantity.

  16. Work out the minimum cost

    C=48πC=48 \pi

    Finally substitute r=2r=2 back in to obtain the minimum cost.

  17. State and interpret the result

    Copt=48πC_{\text{opt}}=48 \pi

    This is the required optimal value; the working above shows both that it is a genuine turning point and that it is the correct type (maximum or minimum).

Answer
C=48πC=48 \pi
Question 5
10 markschallenging
An open-topped box with a square base is to be made using 192 cm2192\,\text{cm}^2 of card. Find the maximum possible volume of the box.
Show worked solution

Worked solution

  1. Plan the approach

    optimise V(x)\text{optimise }V(x)

    The strategy is to write the quantity we want as a function of a single variable using the given constraint, then use differentiation to find and classify the turning point.

  2. Name the dimensions

    square base side x, height h, open top\text{square base side }x,\ \text{height }h,\ \text{open top}

    The box has a square base of side xx and height hh, with no lid. We must use exactly the given amount of material.

  3. Write the surface-area constraint

    x2+4xh=192x^2+4xh=192

    An open box uses one square base (x2x^2) and four rectangular sides (4xh4xh); together they must equal the 192 cm2192\,\text{cm}^2 of material.

  4. Make hh the subject

    h=192−x24xh=\frac{192-x^2}{4x}

    Rearranging the constraint gives the height in terms of xx, so the volume becomes a function of one variable.

  5. Write the volume

    V=x2hV=x^2 h

    The volume of the box is the base area x2x^2 times the height hh.

  6. Substitute for hh

    V=x2⋅192−x24x=−x34+48xV=x^2\cdot\frac{192-x^2}{4x}=- \frac{x^{3}}{4} + 48 x

    Substituting and cancelling one factor of xx leaves a neat cubic in xx.

  7. State the range for xx

    0<x<1920<x<\sqrt{192}

    The side must be positive and small enough that the height stays positive, giving this interval.

  8. Differentiate with respect to xx

    dVdx=48−3x24\frac{dV}{dx}=48 - \frac{3 x^{2}}{4}

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

  9. Set the derivative equal to zero

    48−3x24=048 - \frac{3 x^{2}}{4}=0

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

  10. Clear fractions and rearrange

    192−3x2=0192 - 3 x^{2}=0

    Multiplying through to remove the fraction leaves a simpler polynomial equation that is easier to solve.

  11. Solve for xx

    x=8x=8

    Rearranging gives the value of xx at the stationary point.

  12. Differentiate again

    d2Vdx2=−3x2\frac{d^2V}{dx^2}=- \frac{3 x}{2}

    The second derivative tells us whether the stationary point is a peak or a trough.

  13. Confirm it is a maximum

    d2Vdx2∣x=8=−12<0\left.\frac{d^2V}{dx^2}\right|_{x=8}=-12<0

    The second derivative is negative here, so by the second-derivative test the stationary point is a maximum.

  14. Find the height

    h=4 cmh=4\,cm

    Substituting x=8x=8 into the constraint gives the height, which we need for the final quantity.

  15. Work out the maximum volume

    V=256V=256

    Finally substitute x=8x=8 back in to obtain the maximum volume.

  16. State and interpret the result

    Vopt=256V_{\text{opt}}=256

    This is the required optimal value; the working above shows both that it is a genuine turning point and that it is the correct type (maximum or minimum).

Answer
V=256 cm3V=256\,\text{cm}^3

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