IB Mathematics Analysis and Approaches Topic AHL…

Implicit Differentiation and Rates: When You Cannot Isolate y

Most real curves and constraints tie $x$ and $y$ together so tightly that you can never write $y = f(x)$, yet they still have gradients you need.
When several quantities change together in time, one chain-rule move turns a static equation into a live statement about how fast each is moving.
Optimisation asks for the best possible value, and the best answer is not always where the gradient is zero: sometimes it sits at the very edge of what is allowed.
All three skills run on one engine you already own: the chain rule.

Differentiating When y Is Tangled Up With x

In SL calculus every function came pre-solved as $y = f(x)$, so you differentiated the right-hand side and were done.
But a relation like $x^2 + y^2 = 25$ cannot be written as a single function of $x$, since each $x$ has two $y$ values.
1. Isolating gives $y = \pm\sqrt{25 - x^2}$, which is two functions glued together, not one.
Implicit differentiation lets us find $\frac{dy}{dx}$ directly from the untangled equation, without ever isolating $y$.
The trick is to treat $y$ as an unknown function of $x$ the whole time we differentiate.

Hint

Before differentiating, silently read every $y$ as a function of $x$ whose formula you do not know.
That mindset is what forces the extra $\frac{dy}{dx}$ to appear on each $y$-term.

The One Idea: Every y Carries a dy/dx

Recall the chain rule: to differentiate a function of $y$, differentiate with respect to $y$ then multiply by $\frac{dy}{dx}$.
Take the term $y^2$ and differentiate it with respect to $x$, remembering $y$ itself depends on $x$. $$\frac{d}{dx}\left(y^2\right) = \frac{d}{dy}\left(y^2\right)\cdot\frac{dy}{dx} = 2y\,\frac{dy}{dx}$$
The $2y$ is the ordinary derivative, and the $\frac{dy}{dx}$ is the chain-rule factor that appears because $y$ is not the variable we are differentiating by.
In general $\frac{d}{dx}\big(g(y)\big) = g'(y)\,\frac{dy}{dx}$, so a $\frac{dy}{dx}$ is tacked onto every term built from $y$.

Note

The HL formula booklet (Topic 5) gives the chain rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$ and the product and quotient rules.
Implicit differentiation is a technique built from those rules, not a separate booklet formula, so you must recognise when to apply it.

Worked Derivation: The Circle

Start with the relation $x^2 + y^2 = r^2$, where $r$ is a fixed constant.
Differentiate every term with respect to $x$. $$\frac{d}{dx}\left(x^2\right) + \frac{d}{dx}\left(y^2\right) = \frac{d}{dx}\left(r^2\right)$$
1. The $x^2$ term gives $2x$, and the constant $r^2$ gives $0$.
The $y^2$ term gives $2y\,\frac{dy}{dx}$ from the chain rule above. $$2x + 2y\,\frac{dy}{dx} = 0$$
Now isolate $\frac{dy}{dx}$ with ordinary algebra. $$\frac{dy}{dx} = -\frac{x}{y}$$
Read this geometrically: at the point $(x,y)$ the radius from the origin has gradient $\frac{y}{x}$.
1. The tangent gradient $-\frac{x}{y}$ is the negative reciprocal of $\frac{y}{x}$.
2. That is precisely the condition for the tangent to be perpendicular to the radius, which every circle obeys.

Common Mistake

The single most common error is dropping the $\frac{dy}{dx}$ when differentiating a $y$-term.
Writing $\frac{d}{dx}(y^2)=2y$ is wrong: the correct result is $2y\,\frac{dy}{dx}$.
A safe check: after differentiating, every term that came from $y$ should carry a factor of $\frac{dy}{dx}$.

A Second Example With a Product Term

Differentiate the relation $x^2 + xy + y^2 = 7$ with respect to $x$.
The middle term $xy$ is a product, so it needs the product rule: $\frac{d}{dx}(xy) = y + x\,\frac{dy}{dx}$. $$2x + \left(y + x\,\frac{dy}{dx}\right) + 2y\,\frac{dy}{dx} = 0$$
Gather every term containing $\frac{dy}{dx}$ on one side. $$\left(x + 2y\right)\frac{dy}{dx} = -\left(2x + y\right)$$
Divide to finish. $$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$$
Notice the answer usually contains both $x$ and $y$, which is normal for an implicit gradient.

Example question

The curve $x^3 + y^3 = 9xy$ passes through the point $(2,4)$. Find the gradient of the tangent to the curve at $(2,4)$, giving an exact value.

Solution

Differentiate implicitly: $$3x^2 + 3y^2\frac{dy}{dx} = 9\left(y + x\frac{dy}{dx}\right)$$
Collect terms: $\left(3y^2 - 9x\right)\frac{dy}{dx} = 9y - 3x^2$, so $\frac{dy}{dx} = \dfrac{3y - x^2}{y^2 - 3x}$.
Substitute $(2,4)$: $$\frac{dy}{dx} = \dfrac{3(4) - 2^2}{4^2 - 3(2)} = \dfrac{12 - 4}{16 - 6} = \dfrac{8}{10}$$
The gradient is $\dfrac{4}{5}$.

Tip

After differentiating, gather every $\frac{dy}{dx}$ term on one side and factorise before dividing.
This keeps the algebra clean when the gradient depends on both $x$ and $y$.

Related Rates: Linking How Fast Things Change

When two or more quantities are linked by an equation and all change with time, their rates are linked too.
The move is to differentiate the linking equation with respect to time $t$, treating each variable as a function of $t$.

Note

The linking equation itself contains no time, but differentiating it with respect to $t$ turns each variable into a moving quantity.
Every changing quantity then picks up its own rate factor, such as $\frac{dr}{dt}$ or $\frac{dA}{dt}$.

The General Method

Write the equation that links the quantities.
Identify which quantities vary with time and which are constants.
Differentiate both sides with respect to $t$, putting a $\frac{d(\ )}{dt}$ factor on each variable via the chain rule.
Substitute the known numerical values LAST, only after differentiating.
Solve for the unknown rate and state units.

Common Mistake

Never substitute the given values before differentiating.
- If you set $r = 5$ first, the equation becomes $A = 25\pi$, a constant, and differentiating gives $\frac{dA}{dt} = 0$.
- Freezing the variable destroys the very change you are trying to measure, so always differentiate first, substitute last.

Example

A circular ripple has area $A = \pi r^2$, and its radius grows with time.
Differentiate both sides with respect to $t$, treating $r$ as a function of $t$.

$$\frac{dA}{dt} = \pi\cdot 2r\cdot\frac{dr}{dt} = 2\pi r\,\frac{dr}{dt}$$

Suppose the radius grows at a steady $\frac{dr}{dt} = 2\ \mathrm{cm\,s^{-1}}$, and we want $\frac{dA}{dt}$ when $r = 5\ \mathrm{cm}$.

$$\frac{dA}{dt} = 2\pi (5)(2) = 20\pi \approx 62.8\ \mathrm{cm^2\,s^{-1}}$$

A steady $\frac{dr}{dt}$ does NOT give a steady $\frac{dA}{dt}$.
Because $\frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}$ contains $r$, the area grows faster and faster as the circle gets bigger.

Example

A $5\ \mathrm{m}$ ladder leans against a vertical wall, with its foot $x$ from the wall and its top $y$ up the wall.
By Pythagoras the link is $x^2 + y^2 = 25$.
The foot is pulled away from the wall at $\frac{dx}{dt} = 0.5\ \mathrm{m\,s^{-1}}$; find how fast the top slides down when $x = 3\ \mathrm{m}$.
Differentiate the link with respect to $t$.

$$2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0 \quad\Longrightarrow\quad x\,\frac{dx}{dt} + y\,\frac{dy}{dt} = 0$$

When $x = 3$, the constraint gives $y = \sqrt{25 - 9} = 4$.
Substitute the numbers now.

$$3(0.5) + 4\,\frac{dy}{dt} = 0 \quad\Longrightarrow\quad \frac{dy}{dt} = -\frac{1.5}{4} = -0.375\ \mathrm{m\,s^{-1}}$$

The top slides down at $0.375\ \mathrm{m\,s^{-1}}$, and the minus sign confirms $y$ is decreasing.

Example question

Air is pumped into a spherical balloon at a rate of $100\ \mathrm{cm^3\,s^{-1}}$. Find the rate at which the radius is increasing at the instant when $r = 5\ \mathrm{cm}$. The volume of a sphere is $V = \frac{4}{3}\pi r^3$.

Solution

Differentiate with respect to $t$: $$\frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt}$$
Substitute $\frac{dV}{dt} = 100$ and $r = 5$: $100 = 4\pi (5)^2\,\frac{dr}{dt} = 100\pi\,\frac{dr}{dt}$.
Solve: $$\frac{dr}{dt} = \dfrac{1}{\pi} \approx 0.318\ \mathrm{cm\,s^{-1}}$$

Exam technique

Implicit differentiation, the related-rates method and the optimisation procedure are techniques, not formulas, so none of them appear in the formula booklet.
What the booklet does give you is the chain rule and the product and quotient rules in Topic 5, which is the machinery every one of these techniques runs on.
These questions appear often on Paper 2 and Paper 3, where a GDC may be used to solve the resulting equations or to classify a turning point.
On Paper 1 you must show the differentiation and algebra by hand for exact answers.

Optimisation: Finding the Best Value, Edges Included

Optimisation asks for a maximum or minimum of some quantity subject to a constraint.

Note

A zero gradient only locates candidate points, it does not prove which one is the true maximum or minimum.
An optimisation answer is trustworthy only once you have also checked the endpoints of the allowed domain.

The General Method

Define a variable and write the quantity to optimise (the objective).
Use the constraint to write the objective in terms of ONE variable.
Differentiate and set the derivative equal to zero to find stationary points.
Classify each stationary point using the second derivative or a sign test.
Check the ENDPOINTS of the allowed domain, because the true optimum may sit there instead.
1. On a closed interval, compare the objective at every stationary point and at both endpoints, then pick the winner.

Where the optimum lives	How you find it	When it happens
Interior of domain	Solve derivative = 0, classify	Stationary point lies inside the allowed interval
Endpoint of domain	Evaluate objective at the boundary	No valid stationary point, or an endpoint value beats it

Tip

Classify a stationary point with the second derivative: negative gives a maximum and positive gives a minimum.
If the second derivative is awkward, a sign check of the first derivative on each side works just as well.

Example

A rectangle has its base on the $x$-axis and its upper corners on the line $y = 12 - 3x$, in the first quadrant.
Its width is $x$ and its height is $y = 12 - 3x$, so the area is $A = x(12 - 3x) = 12x - 3x^2$.
The height stays positive only for $0 \le x \le 4$, which sets the domain.
Differentiate and set to zero: $\frac{dA}{dx} = 12 - 6x = 0$, giving $x = 2$.
Classify: $\frac{d^2A}{dx^2} = -6 < 0$, so $x = 2$ is a maximum.
Check endpoints: $A(0) = 0$ and $A(4) = 0$, both smaller than $A(2) = 2(6) = 12$.
So the maximum area is $12$ square units at the interior point $x = 2$.

Example question

An open-top box has a square base of side $x\ \mathrm{cm}$ and volume $1000\ \mathrm{cm^3}$. Find the value of $x$ that minimises the surface area, and confirm it is a minimum. Give answers to 3 s.f.

Solution

Height from volume: $$h = \dfrac{1000}{x^2}$$
Surface area (base plus four sides): $$S = x^2 + 4xh = x^2 + \dfrac{4000}{x}$$
Differentiate and set to zero: $$\frac{dS}{dx} = 2x - \dfrac{4000}{x^2} = 0$, so $x^3 = 2000$$
Solve: $$x = \sqrt[3]{2000} \approx 12.6\ \mathrm{cm}$$
Confirm minimum: $\frac{d^2S}{dx^2} = 2 + \dfrac{8000}{x^3} > 0$, so it is a minimum.

Here the box optimum sits inside its domain, so the zero-gradient point is the genuine answer.
Keep this case in mind, because the next example shows when that comfortable situation breaks down.

Example

A farmer builds a rectangular pen against a straight wall using $20\ \mathrm{m}$ of fencing for the three non-wall sides.
The wall is only $8\ \mathrm{m}$ long, so the side along the wall satisfies $0 \le x \le 8$.
Let $x$ be the side along the wall; the two widths use the rest, so each is $\frac{20 - x}{2}$.
The area is $A = x\cdot\frac{20 - x}{2} = 10x - \frac{x^2}{2}$.
Set the derivative to zero: $\frac{dA}{dx} = 10 - x = 0$, giving $x = 10$.
But $x = 10$ lies OUTSIDE the domain $[0, 8]$, so it is not allowed.
On $[0,8]$ the derivative $10 - x$ is positive throughout, so $A$ is increasing across the whole domain.
The maximum therefore occurs at the right endpoint $x = 8$, giving $A = 8\cdot\frac{12}{2} = 48\ \mathrm{m^2}$.
Had you trusted $\frac{dA}{dx} = 0$ alone, you would have missed the real answer entirely.

Example question

The profit, in thousands of dollars, from producing $x$ units is $P(x) = 24 - \left(x + \dfrac{16}{x}\right)$ for $1 \le x \le 3$. Find the value of $x$ in this domain that gives the maximum profit.

Solution

Differentiate: $$P'(x) = -1 + \dfrac{16}{x^2}$$
Set $P'(x) = 0$: $x^2 = 16$, so $x = 4$, which lies OUTSIDE $[1,3]$ and is rejected.
On $[1,3]$, $x < 4$ gives $\dfrac{16}{x^2} > 1$, so $P'(x) > 0$ and $P$ is increasing throughout.
So the maximum is at the endpoint $x = 3$, where $P(3) = 24 - \left(3 + \dfrac{16}{3}\right) = \dfrac{47}{3} \approx 15.7$ thousand dollars.

Active recall

Why does differentiating $y^3$ with respect to $x$ produce a factor of $\frac{dy}{dx}$?
For $x^2 + y^2 = r^2$, how does $\frac{dy}{dx} = -\frac{x}{y}$ show the tangent is perpendicular to the radius?
In a related-rates problem, why must you substitute the given values only after differentiating?
If the radius of a circle grows at a constant rate, why does its area grow at an increasing rate?
When can the maximum of a quantity occur at an endpoint rather than where the derivative is zero?

Implicit Differentiation and Rates: When You Cannot Isolate y

Most real curves and constraints tie $x$ and $y$ together so tightly that you can never write $y = f(x)$, yet they still have gradients you need.
When several quantities change together in time, one chain-rule move turns a static equation into a live statement about how fast each is moving.
Optimisation asks for the best possible value, and the best answer is not always where the gradient is zero: sometimes it sits at the very edge of what is allowed.
All three skills run on one engine you already own: the chain rule.

Differentiating When y Is Tangled Up With x

In SL calculus every function came pre-solved as $y = f(x)$, so you differentiated the right-hand side and were done.
But a relation like $x^2 + y^2 = 25$ cannot be written as a single function of $x$, since each $x$ has two $y$ values.
1. Isolating gives $y = \pm\sqrt{25 - x^2}$, which is two functions glued together, not one.
Implicit differentiation lets us find $\frac{dy}{dx}$ directly from the untangled equation, without ever isolating $y$.
The trick is to treat $y$ as an unknown function of $x$ the whole time we differentiate.

Hint

Before differentiating, silently read every $y$ as a function of $x$ whose formula you do not know.
That mindset is what forces the extra $\frac{dy}{dx}$ to appear on each $y$-term.

The One Idea: Every y Carries a dy/dx

Recall the chain rule: to differentiate a function of $y$, differentiate with respect to $y$ then multiply by $\frac{dy}{dx}$.
Take the term $y^2$ and differentiate it with respect to $x$, remembering $y$ itself depends on $x$. $$\frac{d}{dx}\left(y^2\right) = \frac{d}{dy}\left(y^2\right)\cdot\frac{dy}{dx} = 2y\,\frac{dy}{dx}$$
The $2y$ is the ordinary derivative, and the $\frac{dy}{dx}$ is the chain-rule factor that appears because $y$ is not the variable we are differentiating by.
In general $\frac{d}{dx}\big(g(y)\big) = g'(y)\,\frac{dy}{dx}$, so a $\frac{dy}{dx}$ is tacked onto every term built from $y$.

Note

The HL formula booklet (Topic 5) gives the chain rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$ and the product and quotient rules.
Implicit differentiation is a technique built from those rules, not a separate booklet formula, so you must recognise when to apply it.

Worked Derivation: The Circle

Start with the relation $x^2 + y^2 = r^2$, where $r$ is a fixed constant.
Differentiate every term with respect to $x$. $$\frac{d}{dx}\left(x^2\right) + \frac{d}{dx}\left(y^2\right) = \frac{d}{dx}\left(r^2\right)$$
1. The $x^2$ term gives $2x$, and the constant $r^2$ gives $0$.
The $y^2$ term gives $2y\,\frac{dy}{dx}$ from the chain rule above. $$2x + 2y\,\frac{dy}{dx} = 0$$
Now isolate $\frac{dy}{dx}$ with ordinary algebra. $$\frac{dy}{dx} = -\frac{x}{y}$$
Read this geometrically: at the point $(x,y)$ the radius from the origin has gradient $\frac{y}{x}$.
1. The tangent gradient $-\frac{x}{y}$ is the negative reciprocal of $\frac{y}{x}$.
2. That is precisely the condition for the tangent to be perpendicular to the radius, which every circle obeys.

Common Mistake

The single most common error is dropping the $\frac{dy}{dx}$ when differentiating a $y$-term.
Writing $\frac{d}{dx}(y^2)=2y$ is wrong: the correct result is $2y\,\frac{dy}{dx}$.
A safe check: after differentiating, every term that came from $y$ should carry a factor of $\frac{dy}{dx}$.

A Second Example With a Product Term

Differentiate the relation $x^2 + xy + y^2 = 7$ with respect to $x$.
The middle term $xy$ is a product, so it needs the product rule: $\frac{d}{dx}(xy) = y + x\,\frac{dy}{dx}$. $$2x + \left(y + x\,\frac{dy}{dx}\right) + 2y\,\frac{dy}{dx} = 0$$
Gather every term containing $\frac{dy}{dx}$ on one side. $$\left(x + 2y\right)\frac{dy}{dx} = -\left(2x + y\right)$$
Divide to finish. $$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$$
Notice the answer usually contains both $x$ and $y$, which is normal for an implicit gradient.

Example question

The curve $x^3 + y^3 = 9xy$ passes through the point $(2,4)$. Find the gradient of the tangent to the curve at $(2,4)$, giving an exact value.

Solution

Differentiate implicitly: $$3x^2 + 3y^2\frac{dy}{dx} = 9\left(y + x\frac{dy}{dx}\right)$$
Collect terms: $\left(3y^2 - 9x\right)\frac{dy}{dx} = 9y - 3x^2$, so $\frac{dy}{dx} = \dfrac{3y - x^2}{y^2 - 3x}$.
Substitute $(2,4)$: $$\frac{dy}{dx} = \dfrac{3(4) - 2^2}{4^2 - 3(2)} = \dfrac{12 - 4}{16 - 6} = \dfrac{8}{10}$$
The gradient is $\dfrac{4}{5}$.

Tip

After differentiating, gather every $\frac{dy}{dx}$ term on one side and factorise before dividing.
This keeps the algebra clean when the gradient depends on both $x$ and $y$.

Related Rates: Linking How Fast Things Change

When two or more quantities are linked by an equation and all change with time, their rates are linked too.
The move is to differentiate the linking equation with respect to time $t$, treating each variable as a function of $t$.

Note

The linking equation itself contains no time, but differentiating it with respect to $t$ turns each variable into a moving quantity.
Every changing quantity then picks up its own rate factor, such as $\frac{dr}{dt}$ or $\frac{dA}{dt}$.

The General Method

Write the equation that links the quantities.
Identify which quantities vary with time and which are constants.
Differentiate both sides with respect to $t$, putting a $\frac{d(\ )}{dt}$ factor on each variable via the chain rule.
Substitute the known numerical values LAST, only after differentiating.
Solve for the unknown rate and state units.

Common Mistake

Never substitute the given values before differentiating.
- If you set $r = 5$ first, the equation becomes $A = 25\pi$, a constant, and differentiating gives $\frac{dA}{dt} = 0$.
- Freezing the variable destroys the very change you are trying to measure, so always differentiate first, substitute last.

Example

A circular ripple has area $A = \pi r^2$, and its radius grows with time.
Differentiate both sides with respect to $t$, treating $r$ as a function of $t$.

$$\frac{dA}{dt} = \pi\cdot 2r\cdot\frac{dr}{dt} = 2\pi r\,\frac{dr}{dt}$$

Suppose the radius grows at a steady $\frac{dr}{dt} = 2\ \mathrm{cm\,s^{-1}}$, and we want $\frac{dA}{dt}$ when $r = 5\ \mathrm{cm}$.

$$\frac{dA}{dt} = 2\pi (5)(2) = 20\pi \approx 62.8\ \mathrm{cm^2\,s^{-1}}$$

A steady $\frac{dr}{dt}$ does NOT give a steady $\frac{dA}{dt}$.
Because $\frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}$ contains $r$, the area grows faster and faster as the circle gets bigger.

Example

A $5\ \mathrm{m}$ ladder leans against a vertical wall, with its foot $x$ from the wall and its top $y$ up the wall.
By Pythagoras the link is $x^2 + y^2 = 25$.
The foot is pulled away from the wall at $\frac{dx}{dt} = 0.5\ \mathrm{m\,s^{-1}}$; find how fast the top slides down when $x = 3\ \mathrm{m}$.
Differentiate the link with respect to $t$.

$$2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0 \quad\Longrightarrow\quad x\,\frac{dx}{dt} + y\,\frac{dy}{dt} = 0$$

When $x = 3$, the constraint gives $y = \sqrt{25 - 9} = 4$.
Substitute the numbers now.

$$3(0.5) + 4\,\frac{dy}{dt} = 0 \quad\Longrightarrow\quad \frac{dy}{dt} = -\frac{1.5}{4} = -0.375\ \mathrm{m\,s^{-1}}$$

The top slides down at $0.375\ \mathrm{m\,s^{-1}}$, and the minus sign confirms $y$ is decreasing.

Example question

Solution

Differentiate with respect to $t$: $$\frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt}$$
Substitute $\frac{dV}{dt} = 100$ and $r = 5$: $100 = 4\pi (5)^2\,\frac{dr}{dt} = 100\pi\,\frac{dr}{dt}$.
Solve: $$\frac{dr}{dt} = \dfrac{1}{\pi} \approx 0.318\ \mathrm{cm\,s^{-1}}$$

Exam technique

Implicit differentiation, the related-rates method and the optimisation procedure are techniques, not formulas, so none of them appear in the formula booklet.
What the booklet does give you is the chain rule and the product and quotient rules in Topic 5, which is the machinery every one of these techniques runs on.
These questions appear often on Paper 2 and Paper 3, where a GDC may be used to solve the resulting equations or to classify a turning point.
On Paper 1 you must show the differentiation and algebra by hand for exact answers.

Optimisation: Finding the Best Value, Edges Included

Optimisation asks for a maximum or minimum of some quantity subject to a constraint.

Note

A zero gradient only locates candidate points, it does not prove which one is the true maximum or minimum.
An optimisation answer is trustworthy only once you have also checked the endpoints of the allowed domain.

The General Method

Define a variable and write the quantity to optimise (the objective).
Use the constraint to write the objective in terms of ONE variable.
Differentiate and set the derivative equal to zero to find stationary points.
Classify each stationary point using the second derivative or a sign test.
Check the ENDPOINTS of the allowed domain, because the true optimum may sit there instead.
1. On a closed interval, compare the objective at every stationary point and at both endpoints, then pick the winner.

Where the optimum lives	How you find it	When it happens
Interior of domain	Solve derivative = 0, classify	Stationary point lies inside the allowed interval
Endpoint of domain	Evaluate objective at the boundary	No valid stationary point, or an endpoint value beats it

Tip

Classify a stationary point with the second derivative: negative gives a maximum and positive gives a minimum.
If the second derivative is awkward, a sign check of the first derivative on each side works just as well.

Example

A rectangle has its base on the $x$-axis and its upper corners on the line $y = 12 - 3x$, in the first quadrant.
Its width is $x$ and its height is $y = 12 - 3x$, so the area is $A = x(12 - 3x) = 12x - 3x^2$.
The height stays positive only for $0 \le x \le 4$, which sets the domain.
Differentiate and set to zero: $\frac{dA}{dx} = 12 - 6x = 0$, giving $x = 2$.
Classify: $\frac{d^2A}{dx^2} = -6 < 0$, so $x = 2$ is a maximum.
Check endpoints: $A(0) = 0$ and $A(4) = 0$, both smaller than $A(2) = 2(6) = 12$.
So the maximum area is $12$ square units at the interior point $x = 2$.

Example question

An open-top box has a square base of side $x\ \mathrm{cm}$ and volume $1000\ \mathrm{cm^3}$. Find the value of $x$ that minimises the surface area, and confirm it is a minimum. Give answers to 3 s.f.

Solution

Height from volume: $$h = \dfrac{1000}{x^2}$$
Surface area (base plus four sides): $$S = x^2 + 4xh = x^2 + \dfrac{4000}{x}$$
Differentiate and set to zero: $$\frac{dS}{dx} = 2x - \dfrac{4000}{x^2} = 0$, so $x^3 = 2000$$
Solve: $$x = \sqrt[3]{2000} \approx 12.6\ \mathrm{cm}$$
Confirm minimum: $\frac{d^2S}{dx^2} = 2 + \dfrac{8000}{x^3} > 0$, so it is a minimum.

Here the box optimum sits inside its domain, so the zero-gradient point is the genuine answer.
Keep this case in mind, because the next example shows when that comfortable situation breaks down.

Example

A farmer builds a rectangular pen against a straight wall using $20\ \mathrm{m}$ of fencing for the three non-wall sides.
The wall is only $8\ \mathrm{m}$ long, so the side along the wall satisfies $0 \le x \le 8$.
Let $x$ be the side along the wall; the two widths use the rest, so each is $\frac{20 - x}{2}$.
The area is $A = x\cdot\frac{20 - x}{2} = 10x - \frac{x^2}{2}$.
Set the derivative to zero: $\frac{dA}{dx} = 10 - x = 0$, giving $x = 10$.
But $x = 10$ lies OUTSIDE the domain $[0, 8]$, so it is not allowed.
On $[0,8]$ the derivative $10 - x$ is positive throughout, so $A$ is increasing across the whole domain.
The maximum therefore occurs at the right endpoint $x = 8$, giving $A = 8\cdot\frac{12}{2} = 48\ \mathrm{m^2}$.
Had you trusted $\frac{dA}{dx} = 0$ alone, you would have missed the real answer entirely.

Example question

The profit, in thousands of dollars, from producing $x$ units is $P(x) = 24 - \left(x + \dfrac{16}{x}\right)$ for $1 \le x \le 3$. Find the value of $x$ in this domain that gives the maximum profit.

Solution

Differentiate: $$P'(x) = -1 + \dfrac{16}{x^2}$$
Set $P'(x) = 0$: $x^2 = 16$, so $x = 4$, which lies OUTSIDE $[1,3]$ and is rejected.
On $[1,3]$, $x < 4$ gives $\dfrac{16}{x^2} > 1$, so $P'(x) > 0$ and $P$ is increasing throughout.
So the maximum is at the endpoint $x = 3$, where $P(3) = 24 - \left(3 + \dfrac{16}{3}\right) = \dfrac{47}{3} \approx 15.7$ thousand dollars.

Active recall

Why does differentiating $y^3$ with respect to $x$ produce a factor of $\frac{dy}{dx}$?
For $x^2 + y^2 = r^2$, how does $\frac{dy}{dx} = -\frac{x}{y}$ show the tangent is perpendicular to the radius?
In a related-rates problem, why must you substitute the given values only after differentiating?
If the radius of a circle grows at a constant rate, why does its area grow at an increasing rate?
When can the maximum of a quantity occur at an endpoint rather than where the derivative is zero?

IB Mathematics Analysis and Approaches AHL 5.14—implicit Functions, Related Rates, Optimisation Notes

IB Mathematics Analysis and Approaches AHL 5.14—implicit Functions, Related Rates, Optimisation Notes

Implicit Differentiation and Rates: When You Cannot Isolate y

Differentiating When y Is Tangled Up With x

The One Idea: Every y Carries a dy/dx

Worked Derivation: The Circle

A Second Example With a Product Term

Related Rates: Linking How Fast Things Change

The General Method

Optimisation: Finding the Best Value, Edges Included

The General Method

Implicit Differentiation and Rates: When You Cannot Isolate y

Differentiating When y Is Tangled Up With x

The One Idea: Every y Carries a dy/dx

Worked Derivation: The Circle

A Second Example With a Product Term

Related Rates: Linking How Fast Things Change

The General Method

Optimisation: Finding the Best Value, Edges Included

The General Method

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

IB Mathematics Analysis and Approaches AHL 5.14—implicit Functions, Related Rates, Optimisation Notes

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

IB Mathematics Analysis and Approaches AHL 5.14—implicit Functions, Related Rates, Optimisation Notes

Implicit Differentiation and Rates: When You Cannot Isolate y

Differentiating When y Is Tangled Up With x

The One Idea: Every y Carries a dy/dx

Worked Derivation: The Circle

A Second Example With a Product Term

Related Rates: Linking How Fast Things Change

The General Method

Optimisation: Finding the Best Value, Edges Included

The General Method

Implicit Differentiation and Rates: When You Cannot Isolate y

Differentiating When y Is Tangled Up With x

The One Idea: Every y Carries a dy/dx

Worked Derivation: The Circle

A Second Example With a Product Term

Related Rates: Linking How Fast Things Change

The General Method

Optimisation: Finding the Best Value, Edges Included

The General Method