IB Mathematics Applications & Interpretation Topic SL…

Optimisation Turns A Real Goal Into A Stationary Point

Businesses want maximum profit, engineers want minimum material, farmers want maximum area for their fencing.
Each of these best values sits at a stationary point of a well-chosen function.
So optimisation is the payoff of the whole calculus topic: set up the right function, then find where its gradient is zero.

The Optimisation Routine

Write the quantity you want to optimise as a function of a single variable.
Use any constraint in the problem to remove the second variable.
Differentiate and solve $f'(x) = 0$ to find the candidate.
Check it is the maximum or minimum you want, using the sign test or your GDC.
Answer the actual question in context, with units.

Example

A farmer has $40\text{ m}$ of fencing for a rectangular pen. To maximise the area:

If the width is $x$, then the length is $20 - x$ because $2x + 2(20 - x) = 40$.
Area $A = x(20 - x) = 20x - x^2$.
Differentiate: $A'(x) = 20 - 2x$; solve $A'(x) = 0$ to get $x = 10$.
The largest area is $A = 10 \times 10 = 100\text{ m}^2$, a square.

Example question

Two positive numbers add to $20$. Find the largest possible value of their product.

Solution

Let the numbers be $x$ and $20 - x$.
Product $P = x(20 - x) = 20x - x^2$.
Differentiate and solve: $P'(x) = 20 - 2x = 0$, so $x = 10$.
Largest product $= 10 \times 10 = 100$.

Example question

A rectangular pen must enclose $25\text{ m}^2$. Find the dimensions that use the least fencing.

Solution

Width $x$, so length $\dfrac{25}{x}$ to keep the area $25$.
Perimeter $P = 2x + \dfrac{50}{x}$.
Differentiate: $P'(x) = 2 - \dfrac{50}{x^2}$; solve $P'(x) = 0$ to get $x^2 = 25$, $x = 5$.
A $5 \times 5$ square uses the least fencing, $P = 20\text{ m}$.

Example question

A square of side $x$ is cut from each corner of a $12\text{ cm}$ square card, and the sides are folded up into an open box. Find $x$ for the maximum volume.

Solution

Volume $V = x(12 - 2x)^2$ for $0 < x < 6$.
Graph $V$ on the GDC and find the maximum, or solve $V'(x) = 0$.
The maximum is at $x = 2$, giving $V = 2 \times 8^2 = 128\text{ cm}^3$.

Common Mistake

Check that a sensible domain applies, such as $x > 0$ for a length.
Confirm the stationary point is the type you want, a maximum or a minimum.
Answer the question that was asked, whether that is a dimension or the optimal value.

Exam technique

There is no optimisation formula; the method marks are for the one-variable function and solving $f'(x) = 0$.
State units and interpret the result in the context of the problem.
The GDC can find the optimum directly, but show the function you built and the derivative you set to zero.
Verify the optimum is a maximum or minimum rather than assuming it.

Active recall

What is the first step in any optimisation problem?
How do you reduce a two-variable problem to one variable?
After solving $f'(x) = 0$, what must you still check?
Two numbers sum to $20$; what makes their product largest?

Optimisation Turns A Real Goal Into A Stationary Point

Businesses want maximum profit, engineers want minimum material, farmers want maximum area for their fencing.
Each of these best values sits at a stationary point of a well-chosen function.
So optimisation is the payoff of the whole calculus topic: set up the right function, then find where its gradient is zero.

The Optimisation Routine

Write the quantity you want to optimise as a function of a single variable.
Use any constraint in the problem to remove the second variable.
Differentiate and solve $f'(x) = 0$ to find the candidate.
Check it is the maximum or minimum you want, using the sign test or your GDC.
Answer the actual question in context, with units.

Example

A farmer has $40\text{ m}$ of fencing for a rectangular pen. To maximise the area:

If the width is $x$, then the length is $20 - x$ because $2x + 2(20 - x) = 40$.
Area $A = x(20 - x) = 20x - x^2$.
Differentiate: $A'(x) = 20 - 2x$; solve $A'(x) = 0$ to get $x = 10$.
The largest area is $A = 10 \times 10 = 100\text{ m}^2$, a square.

Example question

Two positive numbers add to $20$. Find the largest possible value of their product.

Solution

Let the numbers be $x$ and $20 - x$.
Product $P = x(20 - x) = 20x - x^2$.
Differentiate and solve: $P'(x) = 20 - 2x = 0$, so $x = 10$.
Largest product $= 10 \times 10 = 100$.

Example question

A rectangular pen must enclose $25\text{ m}^2$. Find the dimensions that use the least fencing.

Solution

Width $x$, so length $\dfrac{25}{x}$ to keep the area $25$.
Perimeter $P = 2x + \dfrac{50}{x}$.
Differentiate: $P'(x) = 2 - \dfrac{50}{x^2}$; solve $P'(x) = 0$ to get $x^2 = 25$, $x = 5$.
A $5 \times 5$ square uses the least fencing, $P = 20\text{ m}$.

Example question

A square of side $x$ is cut from each corner of a $12\text{ cm}$ square card, and the sides are folded up into an open box. Find $x$ for the maximum volume.

Solution

Volume $V = x(12 - 2x)^2$ for $0 < x < 6$.
Graph $V$ on the GDC and find the maximum, or solve $V'(x) = 0$.
The maximum is at $x = 2$, giving $V = 2 \times 8^2 = 128\text{ cm}^3$.

Common Mistake

Check that a sensible domain applies, such as $x > 0$ for a length.
Confirm the stationary point is the type you want, a maximum or a minimum.
Answer the question that was asked, whether that is a dimension or the optimal value.

Exam technique

There is no optimisation formula; the method marks are for the one-variable function and solving $f'(x) = 0$.
State units and interpret the result in the context of the problem.
The GDC can find the optimum directly, but show the function you built and the derivative you set to zero.
Verify the optimum is a maximum or minimum rather than assuming it.

Active recall

What is the first step in any optimisation problem?
How do you reduce a two-variable problem to one variable?
After solving $f'(x) = 0$, what must you still check?
Two numbers sum to $20$; what makes their product largest?

IB Mathematics Applications & Interpretation SL 5.7—optimisation Notes

IB Mathematics Applications & Interpretation SL 5.7—optimisation Notes

Optimisation Turns A Real Goal Into A Stationary Point

The Optimisation Routine

Optimisation Turns A Real Goal Into A Stationary Point

The Optimisation Routine

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

IB Mathematics Applications & Interpretation SL 5.7—optimisation Notes

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

IB Mathematics Applications & Interpretation SL 5.7—optimisation Notes

Optimisation Turns A Real Goal Into A Stationary Point

The Optimisation Routine

Optimisation Turns A Real Goal Into A Stationary Point

The Optimisation Routine