NDA Maths · Application of Derivatives

Optimisation — Word Problems & AM-GM

Maximise or minimise a real quantity: model it, reduce to one variable using the constraint, then either set the derivative to zero or — often faster — apply AM-GM.

All Application of Derivatives notes NDA Maths strategy

Why this matters

Optimisation word problems look intimidating but follow one recipe, and a large share collapse to a single AM-GM step that beats calculus. Geometric set-ups (max area/volume) recur every year.

Concept 1 of 3

The optimisation recipe

Intuition

Every optimisation problem is the same four steps: write the quantity to optimise, use the constraint to get it in one variable, set the derivative to zero, and confirm it's a max or min. The hard part is the modelling, not the calculus.

Definition

1. Express the target $Q$ and the constraint. 2. Eliminate a variable so $Q=Q(t)$ in one variable. 3. Solve $Q'(t)=0$ for critical $t$ . 4. Verify with $Q''$ or the first-derivative test (and check the domain's endpoints). Classic set-ups: a cylinder/box of fixed volume with least surface, the shortest distance from a point to a curve, $\sum (x-a_j)^2$ minimised at the mean.

Worked example

A closed cylinder of volume

V

has least surface area when its height equals what?

$S=2\pi r^2+2\pi rh$ , with $V=\pi r^2h\Rightarrow h=\dfrac{V}{\pi r^2}$ .
$S(r)=2\pi r^2+\dfrac{2V}{r}$ ; $S'(r)=4\pi r-\dfrac{2V}{r^2}=0\Rightarrow V=2\pi r^3$ , giving $h=2r$ .

Answer:Height

=2r

(equals the diameter).

Practice this conceptself-check · 4 quick reps

Try it yourself

Where is

f(x)=\sum_{j=1}^{n}(x-a_j)^2

minimised?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Step 2 of the recipe?
2.
How to confirm a min?
3.
$\sum(x-a_j)^2$ is least at?
4.
Closed cylinder least surface: $h=$ ?

From the bank · past-year question

Example 1Application of DerivativesMODERATE

A cylindrical jar without a lid has to be constructed using a given surface area of a metal sheet. If the capacity of the jar is to be maximum, then the diameter of the jar must be

k

times the height of the jar. The value of

k

[Q89 · Sep · 2017]

Drill 12 more on the optimisation recipe

Concept 2 of 3

Geometric maxima (area, volume, inscribed figures)

Intuition

Geometry problems carry standard results worth knowing outright: the largest-area triangle in a circle is equilateral, a fixed-perimeter rectangle is largest as a square, and many max-area set-ups land on a 60° or equilateral configuration. Recognise the figure, then run the recipe or quote the result.

Definition

Model the area/volume, use the geometric constraint (perimeter, radius, given side), reduce to one variable (often an angle $\theta$ ), and optimise. Useful facts: max-area triangle inscribed in a circle of radius $R$ is equilateral (area $\tfrac{3\sqrt3}{4}R^2$ ); fixed-perimeter rectangle ⇒ square; a sector of fixed perimeter has max area at a specific radius/angle.

Worked example

A wire of length

20

cm is bent into a rectangle of maximum area. Find the area.

Perimeter $2(l+b)=20\Rightarrow l+b=10$ ; area $A=l(10-l)$ .
$A'(l)=10-2l=0\Rightarrow l=5$ (a square); $A=25$ .

Answer:

25

cm² (a

5\times5

square).

Practice this conceptself-check · 4 quick reps

Try it yourself

What is the maximum area of a triangle inscribed in a circle of radius

R

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Max-area triangle in a circle is?
2.
Fixed-perimeter max-area rectangle is?
3.
Wire $20$ cm, max rectangle area?
4.
Variable to reduce to in many geometry problems?

From the bank · past-year question

Example 2Application of DerivativesMODERATE

A flower-bed in the form of a sector has been fenced by a wire of 40 m length. If the flower-bed has the greatest possible area, then what is the radius of the sector?

[Q93 · Sep · 2018]

Drill 10 more on geometric maxima (area, volume, inscribed figures)

Concept 3 of 3

AM-GM: the calculus-free shortcut

Intuition

When you want the minimum of a sum or the maximum of a product under a fixed-sum or fixed-product constraint, AM-GM gives the answer in one line — no derivatives. Equality (the optimum) occurs when the terms are equal.

Definition

For positive terms, AM $\ge$ GM: $\dfrac{u+v}{2}\ge\sqrt{uv}$ , equality at $u=v$ . So a sum with fixed product is minimised, and a product with fixed sum is maximised, when the terms are equal. Examples: $a^2x+b^2y$ with $xy=c^2$ has min $2abc$ ; $x+y=k\Rightarrow xy$ is max at $x=y=k/2$ .

AM-GM

\frac{u+v}{2}\ge\sqrt{uv}\quad(u,v>0),\ \text{equality at } u=v

Worked example

x+y=20

, what is the maximum of

P=xy

Product with fixed sum is greatest when $x=y$ .
$x=y=10\Rightarrow P=100$ .

Answer:

100

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the minimum of

a^2x+b^2y

subject to

xy=c^2

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
AM-GM equality holds when?
2.
$x+y=k$ : $xy$ is max at?
3.
Min of $a^2x+b^2y$ with $xy=c^2$ ?
4.
AM-GM beats calculus for which problems?

From the bank · past-year question

Example 3Application of DerivativesMODERATE

What is the minimum value of

a^2x+b^2y

where

xy=c^2

[Q93 · Apr · 2019]

Drill 5 more on am-gm: the calculus-free shortcut

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

AM-GM: the calculus-free shortcut
AM-GM
$\frac{u+v}{2}\ge\sqrt{uv}\quad(u,v>0),\ \text{equality at } u=v$

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Application of DerivativesEASY

What is the maximum value of the function

f(x)=\frac{1}{\tan x+\cot x}

, where

0<x<\frac{\pi}{2}

[Q66 · Apr · 2022]

Example 2Application of DerivativesEASY

Let

l

be the length and

b

be the breadth of a rectangle such that

l+b=k

. What is the maximum area of the rectangle?

[Q100 · Apr · 2020]

Example 3Application of DerivativesMODERATE

Given that

4x^2+y^2=9

What is the maximum value of

xy

[Q48 · Sep · 2023]

Example 4Application of DerivativesEASY

The maximum value of

\dfrac{\ln x}{x}

[Q71 · Sep · 2017]

Example 5Application of DerivativesMODERATE

Consider the following information for the items that follow: Three sides of a trapezium are each equal to 6 cm. Let alpha in (0, pi/2) be the angle between a pair of adjacent sides.

What is the maximum area of the trapezium?

[Q78 · Apr · 2018]

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