The curve has equation y=x-2 x for 0 x 2π. Find the coordinates of its minimum point, giving your answer in exact form.

State what we are looking for: stationary points where (dy)/(dx)=0 Recall the derivatives we need: (d)/(dx)(x)=1, (d)/(dx)( x)= x Differentiate the function: (dy)/(dx)=1-2 x Set the gradient equal to zero: 1-2 x=0 Rearrange for x: x=12 Recall where cosine is positive: x>0 in the 1st and 4th quadrants Solve on 0 x 2π: x=π3 or x=5π3 Find the second derivative: d^2ydx^2=2 x Test x=π3: 2π3=√(3)>0 Test x=5π3: 25π3=-√(3)<0 Select the minimum: x=π3 Find the y-coordinate: y=π3-2π3 Evaluate exactly: y=π3-2·√(3)2=π3-√(3) State the coordinates: (π3, π3-√(3)) Sanity check the value: π3 1.05, √(3) 1.73 y -0.68 Answer: The minimum point is at ((π)/(3), (π)/(3)-√(3)).

A cuboid has a square base of side a and total surface area 150 cm^2. Find its maximum possible volume.

State what we maximise: V=a^2h Write the surface constraint: 2a^2+4ah=150 Make h the subject: h=150-2a^24a Substitute into the volume: V=a^2·150-2a^24a Simplify: V=150a-2a^34 State the domain: 0<a<√(75) Plan the method: find the stationary value of V: (dV)/(da)=0 Differentiate V with respect to a: (dV)/(da)=(75)/(2) - 3 a^22 Set the derivative equal to zero: 3 (25 - a^2)2=0 Solve for a: a=5 Find the second derivative: (d^2V)/(da^2)=- 3 a Confirm it is a maximum: .(d^2V)/(da^2)|_a=5=-15<0 Calculate the maximum value of V: V=125 Confirm the maximum: V=0 at a=0 and a=√(75) Interpret the result: a=5 V=125 Answer: The maximum volume is 125 cm^3, when a=5 cm.

An open box is made from a rectangular sheet measuring 8 cm by 5 cm by cutting a square of side x from each corner and folding up the sides. Find the maximum volume of the box.

Understand the situation: cut x from each corner of an 8×5 sheet Find the base dimensions: base=(8-2x)×(5-2x) Find the height: height=x Form the volume: V=x(8-2x)(5-2x) Expand: V=4x^3-26x^2+40x State the domain: 0<x<2.5 Plan the method: find the stationary value of V: (dV)/(dx)=0 Differentiate V with respect to x: (dV)/(dx)=12 x^2 - 52 x + 40 Set the derivative equal to zero: 4 (3 x^2 - 13 x + 10)=0 Solve for x: x=1, (10)/(3) Reject values outside the sensible range: x=1 Find the second derivative: (d^2V)/(dx^2)=24 x - 52 Confirm it is a maximum: .(d^2V)/(dx^2)|_x=1=-28 2.5 rejected; V=0 at the ends Interpret the result: x=1 V=18 Answer: The maximum volume is 18 cm^3, when x=1 cm.

Hard A-Level Increasing, decreasing, stationary points Questions

Challenging, exam-style A-Level Increasing, decreasing, stationary points questions with worked solutions. Stretch yourself on the hardest exponential, product rule, maximum, logarithm problems.

exponentialproduct rulemaximumlogarithmminimumquotient rule

A-Level34 questionsStep-by-step solutions

Question 1

9 markschallenging

The curve has equation

y=x-2\sin x

for

0\leq x\leq 2\pi

. Find the coordinates of its minimum point, giving your answer in exact form.

Show worked solution

Worked solution

State what we are looking for
$\text{stationary points where }\frac{dy}{dx}=0$
We want the minimum of the curve, which is one of its stationary points. So we begin by finding where the gradient is zero.
Recall the derivatives we need
$\frac{d}{dx}(x)=1,\quad \frac{d}{dx}(\sin x)=\cos x$
Before differentiating, remember the two standard results we need: the derivative of $x$ is $1$ , and the derivative of $\sin x$ is $\cos x$ . These come from the trigonometric differentiation topic.
Differentiate the function
$\frac{dy}{dx}=1-2\cos x$
Applying those rules term by term, the derivative of $x$ is $1$ and the derivative of $-2\sin x$ is $-2\cos x$ .
Set the gradient equal to zero
$1-2\cos x=0$
At a stationary point the gradient is zero, so we set the derivative to zero and solve for $x$ .
Rearrange for $\cos x$
$\cos x=\tfrac{1}{2}$
Isolating $\cos x$ gives a standard trig equation we can solve on the given interval.
Recall where cosine is positive
$\cos x>0\text{ in the 1st and 4th quadrants}$
Because $\tfrac{1}{2}$ is positive, the solutions lie in the quadrants where cosine is positive. This tells us to expect one solution near the start of the interval and one near the end.
Solve on $0\leq x\leq 2\pi$
$x=\tfrac{\pi}{3}\ \text{or}\ x=\tfrac{5\pi}{3}$
Cosine equals $\tfrac{1}{2}$ at $\tfrac{\pi}{3}$ in the first quadrant and at $\tfrac{5\pi}{3}$ in the fourth quadrant within one full turn.
Find the second derivative
$\frac{d^{2}y}{dx^{2}}=2\sin x$
To decide which solution is the minimum we use the second derivative test, so we differentiate the gradient function again.
Test $x=\tfrac{\pi}{3}$
$2\sin\tfrac{\pi}{3}=\sqrt{3}>0$
The second derivative is positive here, so the curve bends upwards — this point is a minimum.
Test $x=\tfrac{5\pi}{3}$
$2\sin\tfrac{5\pi}{3}=-\sqrt{3}<0$
The second derivative is negative here, so this point is a maximum, not the minimum we want.
Select the minimum
$x=\tfrac{\pi}{3}$
Only $x=\tfrac{\pi}{3}$ gives a minimum, so that is the $x$ -coordinate we use.
Find the $y$ -coordinate
$y=\tfrac{\pi}{3}-2\sin\tfrac{\pi}{3}$
Substitute $x=\tfrac{\pi}{3}$ into the original equation $y=x-2\sin x$ .
Evaluate exactly
$y=\tfrac{\pi}{3}-2\cdot\tfrac{\sqrt{3}}{2}=\tfrac{\pi}{3}-\sqrt{3}$
Using the exact value $\sin\tfrac{\pi}{3}=\tfrac{\sqrt{3}}{2}$ keeps the answer exact, as required.
State the coordinates
$\left(\tfrac{\pi}{3},\ \tfrac{\pi}{3}-\sqrt{3}\right)$
Combining the results, the minimum point is at $\left(\tfrac{\pi}{3},\ \tfrac{\pi}{3}-\sqrt{3}\right)$ .
Sanity check the value
$\tfrac{\pi}{3}\approx 1.05,\ \sqrt{3}\approx 1.73\Rightarrow y\approx -0.68$
A quick decimal check gives $y\approx -0.68$ , which is below the $x$ -axis and sensible for a minimum of this curve. This confirms the exact answer is reasonable.

Answer

The minimum point is at

\left(\dfrac{\pi}{3},\ \dfrac{\pi}{3}-\sqrt{3}\right)

Question 2

9 markschallenging

The curve

y=x^{3}+ax^{2}+bx+2

has stationary points at

x=1

and

x=3

. Find the value of

a

Show worked solution

Worked solution

Differentiate the curve
$\frac{dy}{dx}=3x^{2}+2ax+b$
Stationary points occur where the gradient is zero, so we differentiate first. The constants $a$ and $b$ just come along for the ride.
Use the first stationary point
$\left.\frac{dy}{dx}\right|_{x=1}=0$
We are told $x=1$ is a stationary point, so the gradient there must be zero.
Substitute $x=1$
$3+2a+b=0$
Putting $x=1$ into the gradient function and setting it to zero gives our first equation.
Use the second stationary point
$\left.\frac{dy}{dx}\right|_{x=3}=0$
Likewise the gradient is zero at the second stationary point $x=3$ .
Substitute $x=3$
$27+6a+b=0$
Putting $x=3$ into the gradient function and setting it to zero gives our second equation.
Subtract the equations
$(27+6a+b)-(3+2a+b)=0$
Subtracting eliminates $b$ so we can find $a$ first. This is the standard method for two simultaneous equations.
Simplify
$24+4a=0$
The $b$ terms cancel and the constants and $a$ -terms combine neatly.
Solve for $a$
$a=-6$
Rearranging gives the value of $a$ .
Substitute back to find $b$
$3+2(-6)+b=0$
Putting $a=-6$ into the first equation lets us find $b$ as well.
Solve for $b$
$b=9$
Rearranging gives $b=9$ .
Write the gradient function
$\frac{dy}{dx}=3x^{2}-12x+9$
With $a$ and $b$ known, the gradient function is fully determined.
Factorise to check
$3x^{2}-12x+9=3(x-1)(x-3)$
Factorising confirms the gradient is zero exactly at $x=1$ and $x=3$ , matching the question.
Find the second derivative
$\frac{d^{2}y}{dx^{2}}=6x-12$
To classify the points (a useful check) we differentiate once more.
Test $x=1$
$6(1)-12=-6<0\Rightarrow\text{maximum}$
A negative second derivative at $x=1$ means the curve bends downwards there, so it is a maximum.
Test $x=3$ and conclude
$6(3)-12=6>0\Rightarrow\text{minimum};\ a=-6$
At $x=3$ the second derivative is positive, so that point is a minimum. The required value is $a=-6$ (with $b=9$ ).

Answer

a=-6

(and

b=9

Question 3

9 markschallenging

A cuboid has a square base of side

a

and total surface area

150\,\text{cm}^{2}

. Find its maximum possible volume.

Show worked solution

Worked solution

State what we maximise
$V=a^{2}h$
The cuboid has a square base of side $a$ and height $h$ , so its volume is $V=a^{2}h$ .
Write the surface constraint
$2a^{2}+4ah=150$
Two square faces and four rectangular sides give the fixed surface area $150$ .
Make $h$ the subject
$h=\dfrac{150-2a^{2}}{4a}$
Rearranging the surface-area equation gives the height in terms of $a$ .
Substitute into the volume
$V=a^{2}\cdot\dfrac{150-2a^{2}}{4a}$
Replacing $h$ leaves the volume as a function of $a$ only.
Simplify
$V=\dfrac{150a-2a^{3}}{4}$
Tidying gives a clean cubic in $a$ to differentiate.
State the domain
$0<a<\sqrt{75}$
The height must be positive, so $2a^{2}<150$ , i.e. $a<\sqrt{75}$ .
Plan the method
$\text{find the stationary value of }V:\ \frac{dV}{da}=0$
Now that V is written with the single variable $a$ , the plan is to differentiate, set the derivative to zero to find the optimum, and use the second derivative to confirm it is the maximum.
Differentiate V with respect to $a$
$\frac{dV}{da}=\frac{75}{2} - \frac{3 a^{2}}{2}$
To find the largest value we look for a stationary point of V, so we differentiate and will set the result to zero.
Set the derivative equal to zero
$\frac{3 \left(25 - a^{2}\right)}{2}=0$
At a maximum or minimum the rate of change is zero. Setting $\frac{dV}{da}=0$ gives the equation to solve.
Solve for $a$
$a=5$
Solving the equation gives the candidate value(s) of $a$ for the optimum.
Find the second derivative
$\frac{d^2V}{da^2}=- 3 a$
The second derivative test confirms whether we have a maximum or a minimum, which we must check rather than assume.
Confirm it is a maximum
$\left.\frac{d^2V}{da^2}\right|_{a=5}=-15<0$
Because the second derivative is negative here, the curve bends downwards, so this is a maximum.
Calculate the maximum value of V
$V=125$
Substitute the optimal $a$ back into V to get the actual largest value.
Confirm the maximum
$V=0\text{ at }a=0\text{ and }a=\sqrt{75}$
The volume is zero at both ends of the range, so the turning point gives the maximum.
Interpret the result
$a=5\Rightarrow V=125$
A base of side $5\,\text{cm}$ (height $5\,\text{cm}$ , a cube) gives the greatest volume, $125\,\text{cm}^{3}$ .

Answer

The maximum volume is

125\,\text{cm}^{3}

, when

a=5\,\text{cm}

Question 4

9 markschallenging

An open box is made from a rectangular sheet measuring

8\,\text{cm}

5\,\text{cm}

by cutting a square of side

x

from each corner and folding up the sides. Find the maximum volume of the box.

Show worked solution

Worked solution

Understand the situation
$\text{cut }x\text{ from each corner of an }8\times5\text{ sheet}$
We cut a square of side $x$ from each corner of the sheet and fold up the flaps to make an open box.
Find the base dimensions
$\text{base}=(8-2x)\times(5-2x)$
Removing $x$ from both ends of each side gives base dimensions $8-2x$ and $5-2x$ .
Find the height
$\text{height}=x$
The folded flaps have height equal to the size of the cut, $x$ .
Form the volume
$V=x(8-2x)(5-2x)$
Volume is length times width times height, giving this product.
Expand
$V=4x^{3}-26x^{2}+40x$
Multiplying out gives a cubic in $x$ ready to differentiate.
State the domain
$0<x<2.5$
The shorter side sets the limit: $5-2x>0$ means $x<2.5$ .
Plan the method
$\text{find the stationary value of }V:\ \frac{dV}{dx}=0$
Now that V is written with the single variable $x$ , the plan is to differentiate, set the derivative to zero to find the optimum, and use the second derivative to confirm it is the maximum.
Differentiate V with respect to $x$
$\frac{dV}{dx}=12 x^{2} - 52 x + 40$
To find the largest value we look for a stationary point of V, so we differentiate and will set the result to zero.
Set the derivative equal to zero
$4 \left(3 x^{2} - 13 x + 10\right)=0$
At a maximum or minimum the rate of change is zero. Setting $\frac{dV}{dx}=0$ gives the equation to solve.
Solve for $x$
$x=1,\ \frac{10}{3}$
Solving the equation gives the candidate value(s) of $x$ for the optimum.
Reject values outside the sensible range
$x=1$
A length or physical quantity must be positive and lie inside the allowed range, so we keep only the value that makes sense in this context.
Find the second derivative
$\frac{d^2V}{dx^2}=24 x - 52$
The second derivative test confirms whether we have a maximum or a minimum, which we must check rather than assume.
Confirm it is a maximum
$\left.\frac{d^2V}{dx^2}\right|_{x=1}=-28<0$
Because the second derivative is negative here, the curve bends downwards, so this is a maximum.
Calculate the maximum value of V
$V=18$
Substitute the optimal $x$ back into V to get the actual largest value.
Reject the value outside the range and check ends
$x=\tfrac{10}{3}>2.5\ \text{rejected};\ V=0\text{ at the ends}$
One solution lies outside the allowed range, and the volume is zero at both ends, so the remaining turning point gives the maximum.
Interpret the result
$x=1\Rightarrow V=18$
Cutting squares of side $1\,\text{cm}$ gives the largest box, of volume $18\,\text{cm}^{3}$ .

Answer

The maximum volume is

18\,\text{cm}^{3}

, when

x=1\,\text{cm}

Question 5

9 markschallenging

A piece of wire of length

20\,\text{cm}

is cut into two pieces. One piece is bent into a circle and the other into a square. Find the length

x

of the piece used for the circle that minimises the total area enclosed.

Show worked solution

Worked solution

Set up the two pieces
$x+\text{(square piece)}=20$
Let $x$ be the length of wire used for the circle; the remaining $20-x$ is used for the square.
Area of the circle
$\text{radius}=\dfrac{x}{2\pi},\ \text{area}=\dfrac{x^{2}}{4\pi}$
The circle's circumference is $x$ , so its radius is $\tfrac{x}{2\pi}$ and its area is $\pi r^{2}=\tfrac{x^{2}}{4\pi}$ .
Area of the square
$\text{side}=\dfrac{20-x}{4},\ \text{area}=\dfrac{(20-x)^{2}}{16}$
The square's perimeter is $20-x$ , so each side is $\tfrac{20-x}{4}$ and its area is the side squared.
Form the total area
$A=\dfrac{x^{2}}{4\pi}+\dfrac{(20-x)^{2}}{16}$
Adding the two areas gives the total as a function of $x$ .
State the domain
$0<x<20$
Each piece must have positive length, so $x$ is between $0$ and $20$ .
Note the aim
$\text{minimise }A$
We look for the stationary point of $A$ that gives the smallest total area.
Plan the method
$\text{find the stationary value of }A:\ \frac{dA}{dx}=0$
Now that A is written with the single variable $x$ , the plan is to differentiate, set the derivative to zero to find the optimum, and use the second derivative to confirm it is the minimum.
Differentiate A with respect to $x$
$\frac{dA}{dx}=\frac{4 x + \pi \left(x - 20\right)}{8 \pi}$
To find the smallest value we look for a stationary point of A, so we differentiate and will set the result to zero.
Set the derivative equal to zero
$\frac{4 x + \pi \left(x - 20\right)}{8 \pi}=0$
At a maximum or minimum the rate of change is zero. Setting $\frac{dA}{dx}=0$ gives the equation to solve.
Solve for $x$
$x=\frac{20 \pi}{\pi + 4}$
Solving the equation gives the candidate value(s) of $x$ for the optimum.
Find the second derivative
$\frac{d^2A}{dx^2}=\frac{\pi + 4}{8 \pi}$
The second derivative test confirms whether we have a maximum or a minimum, which we must check rather than assume.
Confirm it is a minimum
$\left.\frac{d^2A}{dx^2}\right|_{x=\frac{20 \pi}{\pi + 4}}=\frac{\pi + 4}{8 \pi}>0$
Because the second derivative is positive here, the curve bends upwards, so this is a minimum.
Calculate the minimum value of A
$A=\frac{100}{\pi + 4}$
Substitute the optimal $x$ back into A to get the actual smallest value.
Confirm the minimum
$\frac{d^{2}A}{dx^{2}}=\frac{1}{2\pi}+\frac{1}{8}>0$
The second derivative is positive for all $x$ , so the turning point is a minimum.
Interpret the result
$x=\dfrac{20\pi}{\pi+4}\,\text{cm}$
This length of wire for the circle gives the smallest possible total area.

Answer

The total area is least when

x=\dfrac{20\pi}{\pi+4}\,\text{cm}

is used for the circle.

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