What calculus is examined in VCE Math Methods?

Differentiation from first principles, the four standard differentiation rules (sum, product, quotient, chain), differentiation of polynomial, exponential, logarithmic and trig functions, antidifferentiation (the reverse of differentiation), definite integrals and area under curves, applications including kinematics (displacement, velocity, acceleration), optimisation (finding maxima and minima), and average rate of change versus instantaneous rate of change.

What are the differentiation rules I need to memorise?

Four rules. (1) Sum rule, $\frac{d}{dx}[f + g] = f' + g'$. (2) Product rule, $\frac{d}{dx}[fg] = f'g + fg'$. (3) Quotient rule, $\frac{d}{dx}[f/g] = (f'g - fg') / g^2$. (4) Chain rule, $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$. Plus the standard derivatives, $\frac{d}{dx}(x^n) = nx^{n-1}$, $\frac{d}{dx}(e^x) = e^x$, $\frac{d}{dx}(\ln x) = 1/x$, $\frac{d}{dx}(\sin x) = \cos x$, $\frac{d}{dx}(\cos x) = -\sin x$, $\frac{d}{dx}(\tan x) = \sec^2 x$.

What is the difference between antidifferentiation and integration?

Antidifferentiation is the inverse of differentiation, producing a family of functions $F(x) + c$ where $c$ is the constant of integration. Definite integration uses antidifferentiation to compute a number, $\int_a^b f(x)\,dx = F(b) - F(a)$, by the Fundamental Theorem of Calculus. The definite integral is the signed area between the curve and the x-axis on $[a, b]$. The constant of integration disappears in definite integration but is essential in indefinite integration.

How does optimisation work in VCE Math Methods?

Optimisation problems ask you to maximise or minimise a quantity subject to a constraint. The standard recipe. (1) Write the quantity to be optimised in terms of one variable using the constraint. (2) Differentiate. (3) Set the derivative to zero to find stationary points. (4) Test the sign of the second derivative or the sign of the first derivative either side to classify as max or min. (5) Check endpoints if the domain is restricted. (6) State the answer with units. Almost every Section B of Paper 2 has at least one optimisation question.

What kinematics topics are examined?

Position (also called displacement, denoted $x(t)$), velocity $v(t) = x'(t)$, acceleration $a(t) = v'(t) = x''(t)$. The reverse process, from acceleration back to velocity, then to position, by antidifferentiation with given initial conditions to find the constants. Average speed equals total distance divided by total time. Average velocity equals displacement divided by time. Distance versus displacement matters when the object changes direction.

What CAS calculator skills do I need for calculus on Paper 2?

Five core operations. (1) Symbolic differentiate, often written $d/dx$ or $diff()$. (2) Symbolic integrate, $\int$ or $integral()$. (3) Solve, especially $solve(f'(x) = 0, x)$ to find stationary points. (4) Numerical value of a definite integral. (5) Find x for given y, $solve(f(x) = 5, x)$. Practice these until they are reflex. A CAS-fluent student saves 15-20 minutes on Paper 2.

VICMath Methods

VCE Math Methods calculus (differentiation and integration): the 2026 guide

A complete guide to VCE Math Methods Area of Study 3 (Calculus) for Units 3 and 4. Differentiation rules and applications, antidifferentiation and definite integrals, kinematics and optimisation, plus the Paper 1 by-hand technique that wins marks.

Generated by Claude Opus 4.822 min readVCAA-MM-AOS-3Updated 2026-06-21

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

What calculus is really asking
Differentiation
Antidifferentiation and integration
Applications
Common Paper 1 calculus traps
How calculus is examined
Practice strategy
Check your knowledge

What calculus is really asking

VCE Math Methods Area of Study 3 (Calculus) is the most heavily examined area of the course, accounting for roughly 35 percent of exam marks across the two papers. Calculus is also the most cumulative area, building on functions (AoS 1) and feeding into applications throughout Paper 2 Section B.

Calculus is examined in two flavours. Paper 1 (Tech-Free) tests by-hand differentiation and integration with exact-value trig and standard derivative rules. Paper 2 (Tech-Active, CAS) tests applications, modelling, optimisation, and kinematics where the CAS handles the algebra and you handle the interpretation.

Strong calculus students cap their study scores in the mid-40s. Weak calculus students cap in the mid-30s.

Differentiation

From first principles

The definition of the derivative:

f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

VCAA examines first-principles differentiation at least once per cycle, usually on Paper 1, for a simple polynomial.

Differentiate

f(x) = x^2 + 3x

from first principles

Step 1. Write the definition. Apply the limit definition so every method mark is visible.

f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

Step 2. Expand $f(x + h)$ . Substitute $(x+h)$ wherever $x$ appears in the rule.

f(x + h) = (x + h)^2 + 3(x + h) = x^2 + 2xh + h^2 + 3x + 3h

Step 3. Form the numerator $f(x + h) - f(x)$ . Cancel the $x^2$ and $3x$ terms so the numerator divides cleanly by $h$ .

f(x + h) - f(x) = 2xh + h^2 + 3h

Step 4. Divide by $h$ . This step is the whole point of first principles.

\frac{f(x + h) - f(x)}{h} = 2x + h + 3

Step 5. Take the limit as $h \to 0$ . Substitute $h = 0$ since the expression is now continuous in $h$ .

f'(x) = \lim_{h \to 0} (2x + h + 3) = 2x + 3

Answer. $f'(x) = 2x + 3$ .

The four rules

Sum rule: $\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$
Product rule: $\frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x)$
Quotient rule: $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$
Chain rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$

Standard derivatives (memorise)

$\frac{d}{dx}(x^n) = nx^{n-1}$
$\frac{d}{dx}(e^x) = e^x$
$\frac{d}{dx}(e^{kx}) = ke^{kx}$
$\frac{d}{dx}(\ln x) = \frac{1}{x}$
$\frac{d}{dx}(\sin x) = \cos x$
$\frac{d}{dx}(\cos x) = -\sin x$
$\frac{d}{dx}(\tan x) = \sec^2 x = \frac{1}{\cos^2 x}$

Differentiate

f(x) = x^2 e^{3x}

(Tech-Free, product rule)

Step 1. Identify the two factors. The product rule requires $u$ and $v$ with their derivatives.

Let $u = x^2$ and $v = e^{3x}$ .

Step 2. Differentiate each factor. Use the power rule for $u$ and the chain rule for $v$ .

u' = 2x, \qquad v' = 3e^{3x}

Step 3. Apply the product rule $f'(x) = u'v + uv'$ . Substitute the four pieces directly.

f'(x) = 2x \cdot e^{3x} + x^2 \cdot 3e^{3x}

Step 4. Factorise. VCAA marking guides reward a factored final form because it makes stationary points easier to read.

f'(x) = e^{3x}(2x + 3x^2) = x e^{3x}(2 + 3x)

Answer. $f'(x) = x e^{3x}(2 + 3x)$ .

Differentiate

h(x) = \sin(\cos(2x))

(chain rule on a composite of three)

Step 1. Name the three layers from outside in. Identifying layers stops sign errors creeping in.

The outer function is $\sin(\cdot)$ , the middle function is $\cos(\cdot)$ , and the inner function is $2x$ .

Step 2. Differentiate the outer layer, keeping its argument intact. The outer derivative is evaluated at the middle expression.

\frac{d}{dx}\sin(\cos(2x)) = \cos(\cos(2x)) \cdot \frac{d}{dx}[\cos(2x)]

Step 3. Differentiate the middle layer, keeping the inner argument intact.

\frac{d}{dx}\cos(2x) = -\sin(2x) \cdot \frac{d}{dx}(2x) = -2\sin(2x)

Step 4. Combine. Multiply the two derivatives, watching the negative sign from the cosine derivative.

h'(x) = \cos(\cos(2x)) \cdot (-2\sin(2x)) = -2\sin(2x)\cos(\cos(2x))

Answer. $h'(x) = -2\sin(2x)\cos(\cos(2x))$ .

Stationary points and the second derivative

A stationary point is where $f'(x) = 0$ . Classification.

Local minimum. $f''(x) > 0$ at the stationary point (concave up).
Local maximum. $f''(x) < 0$ (concave down).
Stationary point of inflection. $f''(x) = 0$ AND the sign of $f'$ does not change either side.

A point of inflection (not necessarily stationary) is where $f''(x) = 0$ AND $f''$ changes sign.

Classify the stationary points of

f(x) = x^3 - 3x^2 + 4

(Tech-Free)

Step 1. Differentiate. Use the power rule term by term.

f'(x) = 3x^2 - 6x = 3x(x - 2)

Step 2. Solve $f'(x) = 0$ . Factor and set each factor to zero so both stationary x-values are visible.

3x(x - 2) = 0 \implies x = 0 \text{ or } x = 2

Step 3. Compute the second derivative. A sign on $f''$ at each stationary point classifies it without a sign-line.

f''(x) = 6x - 6

Step 4. Test each stationary point. Substitute each x-value into $f''$ .

$f''(0) = -6 < 0$ , so $x = 0$ is a local maximum. $f''(2) = 6 > 0$ , so $x = 2$ is a local minimum.

Step 5. Find the y-coordinates. Substitute back into $f$ to give the full coordinates.

$f(0) = 4$ and $f(2) = 8 - 12 + 4 = 0$ .

Answer. Local max at $(0, 4)$ , local min at $(2, 0)$ .

Antidifferentiation and integration

Standard antiderivatives (memorise)

$\int x^n\,dx = \frac{x^{n+1}}{n+1} + c$ for $n \neq -1$
$\int \frac{1}{x}\,dx = \ln|x| + c$
$\int e^x\,dx = e^x + c$
$\int e^{kx}\,dx = \frac{1}{k}e^{kx} + c$
$\int \sin x\,dx = -\cos x + c$
$\int \cos x\,dx = \sin x + c$
$\int \sec^2 x\,dx = \tan x + c$

Definite integrals

By the Fundamental Theorem of Calculus:

\int_a^b f(x)\,dx = F(b) - F(a)

where $F'(x) = f(x)$ .

The definite integral represents the signed area between the curve and the x-axis on $[a, b]$ . Area below the x-axis contributes negatively.

For the total area between the curve and the x-axis (a common Paper 2 question):

A = \int_a^c |f(x)|\,dx

In practice, find the x-intercepts, then integrate each section separately taking absolute values.

Evaluate

\int_0^{\pi/2} \cos(2x)\,dx

(Tech-Free)

Step 1. Find the antiderivative. Use the reverse-chain factor $1/k$ for $\cos(kx)$ .

\int \cos(2x)\,dx = \frac{1}{2}\sin(2x) + c

Step 2. Apply the Fundamental Theorem of Calculus. Substitute the upper limit, then the lower, and subtract.

\int_0^{\pi/2} \cos(2x)\,dx = \left[\frac{1}{2}\sin(2x)\right]_0^{\pi/2}

Step 3. Evaluate. Use exact values $\sin(\pi) = 0$ and $\sin(0) = 0$ .

\frac{1}{2}\sin(\pi) - \frac{1}{2}\sin(0) = 0 - 0 = 0

Answer. $\int_0^{\pi/2} \cos(2x)\,dx = 0$ . The integrand is positive on $[0, \pi/4]$ and equally negative on $[\pi/4, \pi/2]$ , so the signed area cancels.

Find

\int \frac{2x}{x^2 + 5}\,dx

by recognition (Tech-Free)

Step 1. Recognise the structure. The numerator is the derivative of the denominator. That signals an $\ln$ antiderivative because $\frac{d}{dx}\ln|g(x)| = \frac{g'(x)}{g(x)}$ .

Here $g(x) = x^2 + 5$ and $g'(x) = 2x$ .

Step 2. State the standard form. Pattern match against $\int \frac{g'(x)}{g(x)}\,dx = \ln|g(x)| + c$ .

\int \frac{2x}{x^2 + 5}\,dx = \ln|x^2 + 5| + c

Step 3. Drop the absolute value if possible. Since $x^2 + 5 > 0$ for all real $x$ , the absolute value is unnecessary.

\int \frac{2x}{x^2 + 5}\,dx = \ln(x^2 + 5) + c

Answer. $\ln(x^2 + 5) + c$ .

Find the area enclosed by

y = x^2

and

y = 2x

(Tech-Free, area between curves)

Step 1. Find the intersection points. Set the two curves equal so the region's left and right limits are exact.

x^2 = 2x \implies x^2 - 2x = 0 \implies x(x - 2) = 0 \implies x = 0 \text{ or } x = 2

Step 2. Decide which curve is on top. Test a midpoint such as $x = 1$ .

At $x = 1$ : $y = x^2$ gives $1$ , while $y = 2x$ gives $2$ . So $y = 2x$ is on top on $[0, 2]$ .

Step 3. Set up the area integral. Subtract the lower curve from the upper across the interval of intersection.

A = \int_0^2 (2x - x^2)\,dx

Step 4. Integrate term by term.

\int (2x - x^2)\,dx = x^2 - \frac{x^3}{3} + c

Step 5. Evaluate the definite integral. Substitute the limits.

A = \left[x^2 - \frac{x^3}{3}\right]_0^2 = \left(4 - \frac{8}{3}\right) - 0 = \frac{12 - 8}{3} = \frac{4}{3}

Answer. The enclosed area is $\frac{4}{3}$ square units.

Applications

Kinematics

The standard variables for a particle moving in a straight line.

Position (also called displacement from origin), $x(t)$ .
Velocity, $v(t) = x'(t) = \frac{dx}{dt}$ .
Acceleration, $a(t) = v'(t) = x''(t)$ .

Going forward (position to velocity to acceleration) uses differentiation. Going backward (acceleration to velocity to position) uses antidifferentiation with initial conditions to find the constants.

Distance versus displacement. Displacement is the signed change in position. Distance is the total path length, accounting for direction changes. If a particle moves 5m right then 3m left, displacement is $+2$ m and distance is $8$ m.

Particle with

a(t) = 6t - 4

v(0) = -2

x(0) = 1

. Find

x(t)

(Tech-Free, kinematics)

Step 1. Integrate acceleration to get velocity. Antidifferentiation with a constant of integration $c_1$ .

v(t) = \int (6t - 4)\,dt = 3t^2 - 4t + c_1

Step 2. Use the initial velocity to find $c_1$ . Substitute $t = 0$ and $v(0) = -2$ .

v(0) = 0 - 0 + c_1 = -2 \implies c_1 = -2

So $v(t) = 3t^2 - 4t - 2$ .

Step 3. Integrate velocity to get position. A second constant $c_2$ appears.

x(t) = \int (3t^2 - 4t - 2)\,dt = t^3 - 2t^2 - 2t + c_2

Step 4. Use the initial position to find $c_2$ . Substitute $t = 0$ and $x(0) = 1$ .

x(0) = 0 - 0 - 0 + c_2 = 1 \implies c_2 = 1

Answer. $x(t) = t^3 - 2t^2 - 2t + 1$ .

Optimisation

The standard six-step recipe.

Read the problem and identify the quantity to be optimised (volume, area, cost, profit).
Express the quantity in terms of one variable using any constraint(s).
Differentiate.
Set the derivative to zero and solve for the stationary point(s).
Classify the stationary point as max or min using the second derivative or sign analysis.
Check the endpoints of the domain if it is restricted. The global max or min may be at an endpoint, not at a stationary point.

Minimise the surface area of a 1-litre cylindrical can (Tech-Active, optimisation)

Step 1. Write the quantity to optimise and the constraint. Surface area $A = 2\pi r^2 + 2\pi r h$ and the volume constraint $\pi r^2 h = 1000$ cm³ (since 1 litre is 1000 cm³).

Step 2. Reduce to one variable. Solve the constraint for $h$ and substitute.

h = \frac{1000}{\pi r^2}

A(r) = 2\pi r^2 + 2\pi r \cdot \frac{1000}{\pi r^2} = 2\pi r^2 + \frac{2000}{r}

Step 3. Differentiate. Rewrite the second term as $2000 r^{-1}$ for the power rule.

A'(r) = 4\pi r - \frac{2000}{r^2}

Step 4. Solve $A'(r) = 0$ . Set the derivative to zero and multiply through by $r^2$ .

4\pi r^3 = 2000 \implies r^3 = \frac{500}{\pi} \implies r = \left(\frac{500}{\pi}\right)^{1/3} \approx 5.42 \text{ cm}

Step 5. Classify with the second derivative.

A''(r) = 4\pi + \frac{4000}{r^3} > 0 \text{ for all } r > 0

So this is a minimum.

Step 6. State the matching $h$ with units. Substitute the optimal $r$ back into the constraint.

h = \frac{1000}{\pi (5.42)^2} \approx 10.84 \text{ cm}

Notice $h = 2r$ at the minimum, a standard result for the optimised cylinder.

Answer. The can is minimised when $r \approx 5.42$ cm and $h \approx 10.84$ cm.

Common Paper 1 calculus traps

Chain rule errors on $e^{f(x)}$: $\frac{d}{dx}(e^{3x^2}) = 6x \cdot e^{3x^2}$ , not just $e^{3x^2}$ or $6x e^{6x}$ . Always identify the inner function and apply the chain rule.
Missing the constant of integration: Indefinite integrals require $+ c$ . Definite integrals do not. Losing $c$ in an indefinite integral is a routine 1-mark penalty.
Sign error in $\int \sin$ versus $\int \cos$: $\int \sin x\,dx = -\cos x + c$ . Memorise this with the negative sign.
Forgetting to use absolute area for total distance or area: If asked for total distance (or total area enclosed), set up the integrand as $|v(t)|$ or $|f(x)|$ and integrate piecewise.
Endpoint neglect in optimisation: Always check the endpoints of the domain. The minimum cost or maximum volume might occur at the boundary, not at a stationary point.

How calculus is examined

In the VCE Math Methods exams:

Paper 1 (Tech-Free). Roughly 15-18 marks. Standard patterns include first-principles differentiation, apply product/quotient/chain rule to a complex function, evaluate a definite integral exactly, find a tangent line, set up an antiderivative with initial conditions.
Paper 2 Section A (Tech-Active, multiple choice). 4-6 multiple-choice questions including stationary point classification, signed area, average rate of change.
Paper 2 Section B (Tech-Active, extended response). Major modelling question almost always includes calculus. Common patterns: a function describes a curve (a hill, a river, a population), find the maximum or minimum, find the area between curves, find the average value over an interval.

Practice strategy

For VCE Math Methods calculus:

Term 1 and 2 of Year 12. Memorise the rules and standard derivatives and antiderivatives. Drill 20-30 routine differentiation and integration questions per week.
Term 3. Tackle optimisation and kinematics. These are SAC content and Paper 2 Section B content.
Term 4. Past VCAA Paper 1s under timed conditions. Score yourself separately on the calculus questions to identify gaps. The patterns repeat year to year.

See our VCE Math Methods practice questions for prompts modelled on VCAA past papers.

Check your knowledge

Eight questions, increasing in difficulty. Mark each as Tech-Free (Paper 1 style, by hand) or Tech-Active (Paper 2 style, CAS-permitted) and attempt them in those conditions.

Tech-Free (Paper 1 style)

Differentiate $f(x) = 5x^4 - 2x + \sqrt{x}$ .
Differentiate $g(x) = (3x + 1)e^{-2x}$ using the product rule.
Evaluate $\int_1^e \frac{1}{x}\,dx$ exactly.
Find the equation of the tangent to $y = x^3 - 4x$ at $x = 1$ .
Find $\int (2\sin(3x) + e^{-x})\,dx$ .

Tech-Active (Paper 2 style)

The position of a particle is $x(t) = t^3 - 6t^2 + 9t$ for $t \geq 0$ , in metres. Find the total distance travelled in the first 4 seconds.
Find the area enclosed between $y = 4 - x^2$ and $y = x + 2$ .
A rectangle has its base on the x-axis and its upper two vertices on the curve $y = 8 - x^2$ . Find the dimensions that maximise the rectangle's area.

Solutions to check-your-knowledge questions

These eight questions cover power-rule differentiation, the product rule, definite integrals of $1/x$ , tangent lines, antidifferentiation of trig and exponential terms, total distance from velocity, area between curves, and optimisation of a rectangle inscribed under a parabola.

Step 1: Q1. Differentiate $f(x) = 5x^4 - 2x + \sqrt{x}$

Rewrite $\sqrt{x} = x^{1/2}$ so every term is a power of $x$ , then differentiate term by term using the power rule.

f'(x) = 20x^3 - 2 + \frac{1}{2}x^{-1/2} = 20x^3 - 2 + \frac{1}{2\sqrt{x}}

Step 2: Q2. Differentiate $g(x) = (3x + 1)e^{-2x}$ using the product rule

Let $u = 3x + 1$ and $v = e^{-2x}$ , so $u' = 3$ and $v' = -2e^{-2x}$ . Apply $g'(x) = u'v + uv'$ then factorise.

g'(x) = 3e^{-2x} + (3x + 1)(-2e^{-2x}) = e^{-2x}(3 - 6x - 2) = e^{-2x}(1 - 6x)

Step 3: Q3. Evaluate $\displaystyle\int_1^e \frac{1}{x}\,dx$

The antiderivative of $\frac{1}{x}$ is $\ln|x|$ . Apply the fundamental theorem, noting that $\ln e = 1$ and $\ln 1 = 0$ .

\int_1^e \frac{1}{x}\,dx = [\ln|x|]_1^e = \ln e - \ln 1 = 1 - 0 = 1

Step 4: Q4. Find the equation of the tangent to $y = x^3 - 4x$ at $x = 1$

The gradient of the tangent equals $y'$ evaluated at the point. Find the point on the curve first, then use point-gradient form.

y'(x) = 3x^2 - 4, \quad y'(1) = -1 \quad \text{(gradient)}

y(1) = 1 - 4 = -3, \quad \text{so the point is } (1, -3)

y + 3 = -1(x - 1) \implies y = -x - 2

Step 5: Q5. Antidifferentiate $2\sin(3x) + e^{-x}$

Apply the reverse-chain factor to each term. For $\sin(3x)$ the inner derivative is $3$ , so divide by $3$ ; for $e^{-x}$ the inner derivative is $-1$ , so divide by $-1$ .

\int 2\sin(3x)\,dx = -\frac{2}{3}\cos(3x) + c, \qquad \int e^{-x}\,dx = -e^{-x} + c

Combined antiderivative: $-\dfrac{2}{3}\cos(3x) - e^{-x} + c$ .

Step 6: Q6. Total distance travelled by a particle with $x(t) = t^3 - 6t^2 + 9t$

The particle reverses direction where $v(t) = 0$ . Find those times, evaluate position at each turning point and the endpoints, then add the absolute distances between consecutive positions.

v(t) = 3t^2 - 12t + 9 = 3(t - 1)(t - 3) = 0 \implies t = 1,\ t = 3

Positions: $x(0) = 0$ , $x(1) = 4$ , $x(3) = 0$ , $x(4) = 4$ .

\text{Total distance} = |4 - 0| + |0 - 4| + |4 - 0| = 4 + 4 + 4 = 12 \text{ metres}

Step 7: Q7. Area between $y = 4 - x^2$ and $y = x + 2$

Find the intersection points by equating the two expressions, then integrate (top minus bottom) over the interval where the parabola lies above the line.

4 - x^2 = x + 2 \implies x^2 + x - 2 = 0 \implies (x + 2)(x - 1) = 0 \implies x = -2,\ 1

A = \int_{-2}^{1} (2 - x - x^2)\,dx = \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-2}^{1} = \frac{7}{6} + \frac{20}{6} = \frac{27}{6} = \frac{9}{2} \text{ square units}

Step 8: Q8. Maximum area of a rectangle inscribed under $y = 8 - x^2$

Place the upper-right vertex at $(x, 8 - x^2)$ with $0 < x < 2\sqrt{2}$ . The rectangle has width $2x$ and height $8 - x^2$ . Differentiate the area function and set it to zero.

A(x) = 2x(8 - x^2) = 16x - 2x^3 \implies A'(x) = 16 - 6x^2 = 0 \implies x^2 = \frac{8}{3},\ x = 2\sqrt{\frac{2}{3}}

Height: $8 - \frac{8}{3} = \frac{16}{3}$ . Width: $4\sqrt{\frac{2}{3}}$ .

Final answer: maximum area $= \dfrac{64\sqrt{6}}{9}$ square units (approximately $17.42$ square units).