Year 11 Methods (Unit 1 & 2) Applications Of Differentiation Of Polynomials

Applications To Maximum And Minimum Problems

Q: How do I set up an optimisation problem?

Write the quantity to optimise as the objective, use the constraint to express it in one variable, state a sensible domain, then differentiate.

Q: How do I know if I have a maximum or a minimum?

Solve f prime of x equals zero, then apply the first derivative sign test. Positive then negative is a maximum, and negative then positive is a minimum.

Q: Why do I need a domain?

The variable is a real length or quantity, so it has physical limits. Solutions of f prime of x equals zero outside that range are rejected.

Q: Should I give the dimension or the maximum value?

Whatever the question asks. Some want the optimal x, and others want the maximum or minimum value f of x. Often you state both.

Q: Can I use the second derivative to check?

Not in Year 11 Methods. Use the first derivative sign test. The second derivative test belongs to Year 12.

Q: How do I eliminate the second variable?

Rearrange the constraint, such as a fixed perimeter, area or volume, to make one variable the subject, then substitute it into the objective.

20 practice questions 2 video lessons Theory + worked examples

Master maximum and minimum problems for Queensland Year 11 Mathematical Methods (QCAA). Optimisation uses calculus to find the greatest or least value a quantity can take — its local or global maximum or minimum.

You will learn to build a function from the problem, differentiate and solve where the gradient is zero, check the nature of that result with a sign test, and answer questions about maximum area, volume or profit.

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Practice questions

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 2), an optimisation problem asks for the largest or smallest value of a quantity in a context — a maximum area, volume or profit, or a minimum length or cost. You build a polynomial objective from the information, differentiate, solve $f'(x)=0$, confirm max or min with a first-derivative sign test, and answer what was asked.

The quantity to be made as large or as small as possible is the objective. It often starts with two variables, so a constraint (a fixed perimeter, area, or volume) is used to eliminate one, leaving the objective as a function of a single variable.

The optimum occurs at a stationary point of the objective, where $f'(x)=0$. Whether it is a maximum or a minimum is confirmed with the first-derivative sign test. Because the variable is a real length or quantity, it also has a sensible domain (for example \(0

Optimise in one variable. Use the constraint to write the objective with a single variable, then differentiate, solve \(f'(x)=0\), and confirm the type by a sign test.

The maximum area is the peak of the objective curve, where $f'(x)=0$.

For an open box, volume peaks at an interior stationary point within \(0

Write the objective $f(x)$ in one variable, then find the stationary value:

\[f'(x)=0\]

f^{'} (x) = 0

Confirm the nature with the first-derivative sign test:

\[f'(x^-)>0,\ f'(x^+)<0\ \Rightarrow\ ext{maximum}\]

f^{'} (x^{-}) > 0, f^{'} (x^{+}) < 0 \Rightarrow max

Answer the question asked. Some problems want the optimal dimension \(x\); others want the maximum or minimum value \(f(x)\). Read carefully and give that quantity, within the valid domain.

How to solve an optimisation problem

Build the objective: write the quantity to optimise, and use the constraint to express it with one variable; state a sensible domain.
Differentiate and solve: find $f'(x)$, set $f'(x)=0$, and solve for the variable, keeping only values in the domain.
Confirm and answer: use a first-derivative sign test to check maximum or minimum, then state the quantity asked (the dimension or the optimal value).

Example 1 — Maximum area, fixed perimeter

A rectangle has perimeter $40$ m. Find the width that maximises its area, and the maximum area.

Solution

Objective — width $w$, so length $=20-w$; area:

$A$	$=$	$w(20-w)$
	$=$	$20w-w^2$

Domain: \(0

Differentiate and solve $A'(w)=0$:

$A'(w)$	$=$	$20-2w$
$20-2w$	$=$	$0$
$w$	$=$	$10$

Sign test around $w=10$:

$A'(9)$	$=$	$20-18=2>0$
$A'(11)$	$=$	$20-22=-2<0$

$+\rightarrow-$, so $w=10$ gives a maximum.

Maximum area:

$A(10)$	$=$	$10(20-10)$
	$=$	$100$

Width $10$ m gives the maximum area $100$ m$^2$ (a square).

A = 100

Example 2 — Maximum volume of an open box

An open box is made from a $12$ cm square of card by cutting a square of side $x$ from each corner and folding up. Find $x$ for maximum volume.

Solution

Objective — base $(12-2x)$ each way, height $x$:

$V$	$=$	$x(12-2x)^2$
	$=$	$4x^3-48x^2+144x$

Domain: \(0

Differentiate and solve $V'(x)=0$:

$V'(x)$	$=$	$12x^2-96x+144$
$12(x^2-8x+12)$	$=$	$0$
$12(x-2)(x-6)$	$=$	$0$
$x$	$=$	$2 \ \ (x=6 \text{ rejected})$

Sign test around $x=2$:

$V'(1)$	$=$	$12(1-8+12)=60>0$
$V'(3)$	$=$	$12(9-24+12)=-36<0$

$+\rightarrow-$, so $x=2$ gives a maximum.

Maximum volume:

$V(2)$	$=$	$2(12-4)^2=2(64)$
	$=$	$128$

Cut $x=2$ cm; the maximum volume is $128$ cm$^3$.

V = 128

Example 3 — Maximum profit

A stall sells $x$ items at price $(60-x)$ dollars each, with costs $100+20x$ dollars. Find the number of items that maximises profit and the maximum profit.

Solution

Objective — profit $=$ revenue $-$ cost:

$P$	$=$	$x(60-x)-(100+20x)$
	$=$	$60x-x^2-100-20x$
	$=$	$-x^2+40x-100$

Domain: \(0

Differentiate and solve $P'(x)=0$:

$P'(x)$	$=$	$-2x+40$
$-2x+40$	$=$	$0$
$x$	$=$	$20$

Sign test around $x=20$:

$P'(19)$	$=$	$-38+40=2>0$
$P'(21)$	$=$	$-42+40=-2<0$

$+\rightarrow-$, so $x=20$ gives a maximum.

Maximum profit:

$P(20)$	$=$	$-(20)^2+40(20)-100$
	$=$	$-400+800-100$
	$=$	$300$

Selling $20$ items gives the maximum profit $\$300$.

P = 300

Example 4 — Constructing the objective from a constraint

An open-top box has a square base of side $x$ cm and is made from $300$ cm$^2$ of material. Find $x$ for maximum volume, and that volume.

Solution

Constraint — base plus four sides use $300$ cm$^2$:

$x^2+4xh$	$=$	$300$
$h$	$=$	$\dfrac{300-x^2}{4x}$

Objective — volume in one variable:

$V$	$=$	$x^2h=x^2\cdot\dfrac{300-x^2}{4x}$
	$=$	$\dfrac{300x-x^3}{4}$
	$=$	$75x-\dfrac{x^3}{4}$

Domain: \(0

Differentiate and solve $V'(x)=0$:

$V'(x)$	$=$	$75-\dfrac{3x^2}{4}$
$75-\dfrac{3x^2}{4}$	$=$	$0$
$x^2$	$=$	$100$
$x$	$=$	$10$

Sign test around $x=10$:

$V'(9)$	$=$	$75-\tfrac{3(81)}{4}=14.25>0$
$V'(11)$	$=$	$75-\tfrac{3(121)}{4}=-15.75<0$

$+\rightarrow-$, so $x=10$ gives a maximum.

Maximum volume:

$V(10)$	$=$	$75(10)-\dfrac{(10)^3}{4}$
	$=$	$750-250$
	$=$	$500$

A base of $x=10$ cm gives the maximum volume $500$ cm$^3$.

V = 500

Common pitfalls

Optimising with two variables. Use the constraint to eliminate one variable first, so the objective is a function of a single variable before you differentiate.

Reaching for the second derivative. Confirm maximum or minimum with the first-derivative sign test — test the sign of $f'$ each side of the stationary value.

Ignoring the domain. Reject solutions of $f'(x)=0$ that fall outside the sensible range (a length cannot be negative, and here $x=6$ collapses the box).

Answering the wrong quantity. Check whether the question wants the optimal dimension $x$ or the maximum/minimum value $f(x)$, and give exactly that.

Frequently asked questions

How do I set up an optimisation problem?

Write the quantity to optimise as the objective, use the constraint to express it in one variable, state a sensible domain, then differentiate.

How do I know if I have a maximum or a minimum?

Solve $f'(x)=0$, then apply the first-derivative sign test: $+\rightarrow-$ is a maximum and $-\rightarrow+$ is a minimum.

Why do I need a domain?

The variable is a real length or quantity, so it has physical limits; solutions of $f'(x)=0$ outside that range are rejected.

Should I give the dimension or the maximum value?

Whatever the question asks. Some want the optimal $x$; others want the maximum or minimum value $f(x)$. Often you state both.

Can I use the second derivative to check?

Not in Year 11 Methods — use the first-derivative sign test. The second-derivative test belongs to Year 12.

How do I eliminate the second variable?

Rearrange the constraint (fixed perimeter, area or volume) to make one variable the subject, then substitute it into the objective.

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Applications Of Differentiation To Kinematics

More in Applications Of Differentiation Of Polynomials

\(A\)	\(=\)	\(w(20-w)\)
	\(=\)	\(20w-w^2\)

\(A'(w)\)	\(=\)	\(20-2w\)
\(20-2w\)	\(=\)	\(0\)
\(w\)	\(=\)	\(10\)

\(A'(9)\)	\(=\)	\(20-18=2>0\)
\(A'(11)\)	\(=\)	\(20-22=-2<0\)

\(A(10)\)	\(=\)	\(10(20-10)\)
	\(=\)	\(100\)

\(V\)	\(=\)	\(x(12-2x)^2\)
	\(=\)	\(4x^3-48x^2+144x\)

\(V'(x)\)	\(=\)	\(12x^2-96x+144\)
\(12(x^2-8x+12)\)	\(=\)	\(0\)
\(12(x-2)(x-6)\)	\(=\)	\(0\)
\(x\)	\(=\)	\(2 \ \ (x=6 \text{ rejected})\)

\(V'(1)\)	\(=\)	\(12(1-8+12)=60>0\)
\(V'(3)\)	\(=\)	\(12(9-24+12)=-36<0\)

\(P\)	\(=\)	\(x(60-x)-(100+20x)\)
	\(=\)	\(60x-x^2-100-20x\)
	\(=\)	\(-x^2+40x-100\)

\(P'(x)\)	\(=\)	\(-2x+40\)
\(-2x+40\)	\(=\)	\(0\)
\(x\)	\(=\)	\(20\)

\(P(20)\)	\(=\)	\(-(20)^2+40(20)-100\)
	\(=\)	\(-400+800-100\)
	\(=\)	\(300\)

\(V\)	\(=\)	\(x^2h=x^2\cdot\dfrac{300-x^2}{4x}\)
	\(=\)	\(\dfrac{300x-x^3}{4}\)
	\(=\)	\(75x-\dfrac{x^3}{4}\)

\(V'(x)\)	\(=\)	\(75-\dfrac{3x^2}{4}\)
\(75-\dfrac{3x^2}{4}\)	\(=\)	\(0\)
\(x^2\)	\(=\)	\(100\)
\(x\)	\(=\)	\(10\)

\(V'(9)\)	\(=\)	\(75-\tfrac{3(81)}{4}=14.25>0\)
\(V'(11)\)	\(=\)	\(75-\tfrac{3(121)}{4}=-15.75<0\)

\(V(10)\)	\(=\)	\(75(10)-\dfrac{(10)^3}{4}\)
	\(=\)	\(750-250\)
	\(=\)	\(500\)

\(V(2)\)	\(=\)	\(2(12-4)^2=2(64)\)
	\(=\)	\(128\)

\(P'(19)\)	\(=\)	\(-38+40=2>0\)
\(P'(21)\)	\(=\)	\(-42+40=-2<0\)

\(x^2+4xh\)	\(=\)	\(300\)
\(h\)	\(=\)	\(\dfrac{300-x^2}{4x}\)

Applications To Maximum And Minimum Problems

📖 Theory

How to solve an optimisation problem

Common pitfalls

Frequently asked questions

Theory