A-Level Optimisation Practice Questions

Free A-Level Optimisation practice questions with full step-by-step worked solutions. Covers optimisation, differentiation, area, cost. Practise exam-style problems and check your method.

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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.
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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
Which graph shows C=x+9xC = x + \dfrac{9}{x} for x>0x>0, including its minimum point?
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Worked solution

  1. Consider the behaviour

    C=x+9xC = x + \frac{9}{x}

    For very small xx the term 9/x9/x is huge, and for large xx the term xx dominates, so the curve dips in the middle.

  2. Locate the minimum

    dCdx=1−9x2=0⇒x=3,  C=6\frac{dC}{dx}=1-\frac{9}{x^2}=0\Rightarrow x=3,\;C=6

    Setting the derivative to zero gives x=3x=3, and substituting back gives C=6C=6.

  3. Match the curve

    minimum at (3,6)\text{minimum at }(3,6)

    The correct graph is a positive curve with a single lowest point at (3,6)(3,6).

Answer
Curve with a minimum at (3,6)(3,6)
Question 3
5 marksintermediate
The surface area of a box is S=2x2+108xS=2x^2+\dfrac{108}{x} for x>0x>0. Which graph correctly shows SS against xx, including its minimum point?
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Worked solution

  1. Consider the two terms

    S=2x2+108xS=2x^2+\frac{108}{x}

    For small xx the term 108x\dfrac{108}{x} is very large, and for large xx the term 2x22x^2 dominates, so the curve has a dip.

  2. Differentiate

    dSdx=4x−108x2\frac{dS}{dx}=4x-\frac{108}{x^2}

    Using the power rule on each term gives the gradient function.

  3. Find the stationary point

    4x=108x2⇒x3=27⇒x=34x=\frac{108}{x^2}\Rightarrow x^3=27\Rightarrow x=3

    Setting the derivative to zero and rearranging gives x3=27x^3=27, so x=3x=3.

  4. Find the minimum value

    S=2(9)+1083=54S=2(9)+\frac{108}{3}=54

    Substituting x=3x=3 gives the least surface area of 5454.

  5. Confirm it is a minimum

    d2Sdx2=4+216x3>0\frac{d^2S}{dx^2}=4+\frac{216}{x^3}>0

    The second derivative is positive for x>0x>0, confirming a minimum (a valley).

  6. Match the graph

    minimum at (3,54)\text{minimum at }(3,54)

    The correct graph is a positive curve that dips to a single lowest point at (3,54)(3,54).

Answer
Curve with a minimum at (3,54)(3,54)
Question 4
7 markshard
A closed cylinder has a fixed volume VV. Using calculus, which statement correctly gives the ratio of height to radius that minimises the surface area, together with a valid justification?
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Worked solution

  1. Write the volume constraint

    πr2h=V\pi r^2 h=V

    The volume is fixed, so the radius rr and height hh are linked by this equation.

  2. Make hh the subject

    h=Vπr2h=\frac{V}{\pi r^2}

    Dividing by πr2\pi r^2 lets us later replace hh so the surface area uses one variable.

  3. Write the surface area

    S=2πr2+2πrhS=2\pi r^2+2\pi r h

    Two circular ends plus the curved side give the total surface area of a closed cylinder.

  4. Substitute for hh

    S=2πr2+2VrS=2\pi r^2+\frac{2V}{r}

    Replacing hh turns SS into a function of rr only.

  5. Rewrite in index form

    S=2πr2+2Vr−1S=2\pi r^2+2V r^{-1}

    Writing 2Vr\frac{2V}{r} as 2Vr−12Vr^{-1} makes the power rule easy to apply.

  6. Differentiate

    dSdr=4πr−2Vr2\frac{dS}{dr}=4\pi r-\frac{2V}{r^2}

    This is the gradient of the surface-area curve; it is zero at the turning point.

  7. Set the derivative to zero

    4πr=2Vr24\pi r=\frac{2V}{r^2}

    At the minimum the gradient is zero, giving this equation to solve.

  8. Solve for rr

    r3=V2π ⇒ V=2πr3r^3=\frac{V}{2\pi}\ \Rightarrow\ V=2\pi r^3

    Rearranging gives the special radius; it is convenient to write VV in terms of rr.

  9. Find hh in terms of rr

    h=Vπr2=2πr3πr2=2rh=\frac{V}{\pi r^2}=\frac{2\pi r^3}{\pi r^2}=2r

    Substituting V=2πr3V=2\pi r^3 into the height formula shows the height is exactly twice the radius.

  10. Check with the second derivative

    d2Sdr2=4π+4Vr3>0\frac{d^2S}{dr^2}=4\pi+\frac{4V}{r^3}>0

    For a positive radius this is positive, so the second-derivative test confirms a minimum.

  11. State the conclusion

    h=2r (minimum)h=2r\ (\text{minimum})

    So the surface area is least when the height equals twice the radius, and it is genuinely a minimum.

Answer
h=2rh=2r, and this is a minimum because d2Sdr2=4π+4Vr3>0\dfrac{d^2S}{dr^2}=4\pi+\dfrac{4V}{r^3}>0
Question 5
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

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