Differentiation

6 methods · 11 questions

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A maximum-area or shortest-distance question is solved by turning the geometry into a single-variable function. Pick one free variable, use the geometric constraint to eliminate every other variable, write the target quantity as a function of that one variable, and set its derivative to zero.

Trigger: 'maximum possible volume/area' or 'minimum distance' with a curve or region.

Instances:
(i) use the given volume to substitute for one dimension, leaving the surface area as a function of one variable to differentiate;
(ii) express coordinates of a rectangle's vertices in terms of a single parameter $p$ to write the area as $A (p)$ ;
(iii) to find the maximum or minimum of a vertical distance, write the difference in $y$ -coordinates as a function of $x$ and set its derivative to zero.

Linked questions (3)

NSAA 2018 Section 1, Part E (Mathematics questions) · Q89

The dimensions of a solid cuboid, in cm, are

x

2 x

and

y

The volume of the cuboid is

576 cm^{3}

.

At this volume, the surface area of the cuboid has its maximum value.

What is the area, in

cm^{2}

, of the face that has the largest area?

A $2 (288)^{\frac{2}{3}}$
B $72$
C $96$
D $432$
E $4 (144)^{\frac{2}{3}}$

Worked solution

Answer:C

The volume fixes

y

in terms of

x

2 x^{2} y = 576 ⟹ y = \frac{288}{x ^{2}} .

The surface area is made of three pairs of faces, so

S = 2 (2 x^{2} + x y + 2 x y) = 4 x^{2} + 6 x y = 4 x^{2} + \frac{1728}{x},

a function of

x

alone.

The stationary value described in the question is where

\frac{d S}{d x} = 0

\frac{d S}{d x} = 8 x - \frac{1728}{x ^{2}} = 0 ⟹ 8 x^{3} = 1728 ⟹ x^{3} = 216 ⟹ x = 6.

The dimensions are then

x = 6

2 x = 12

and

y = \frac{288}{36} = 8

, so the faces have areas

6 \times 12 = 72, 6 \times 8 = 48, 12 \times 8 = 96.

The largest face has area

96

.

The answer is C.

ENGAA 2016 Section 1, Part B (Mathematics questions) · Q49

A cursor starts at the point

(0, 10)

and moves parallel to the

x

-axis in the negative direction.

What is the minimum distance parallel to the

y

-axis between the cursor and the graph of

y = 4 x^{3} - 12 x^{2} - 36 x - 15

A $0$
B $5$
C $25$
D $69$
E $133$

Worked solution

Answer:B

The cursor sweeps along the ray

y = 10

with

x \leq 0

, so the distance measured parallel to the

y

-axis at abscissa

x

∣10 - f (x) ∣

, where

f (x) = 4 x^{3} - 12 x^{2} - 36 x - 15.

Differentiate to find how large

f

can be on that ray:

f^{'} (x) = 12 x^{2} - 24 x - 36 = 12 (x - 3) (x + 1) .

The stationary points are

x = - 1

and

x = 3

, and only

x = - 1

satisfies

x \leq 0

. There

f^{'}

changes from positive to negative, so

x = - 1

is a local maximum, with

f (- 1) = - 4 - 12 + 36 - 15 = 5.

x \leq 0

the curve rises from

- \infty

up to that value and then falls back to

f (0) = - 15

, so the greatest value of

f

on the ray is

5

.

Since

f (x) \leq 5 < 10

throughout, the gap equals

10 - f (x)

, and it is smallest where

f

is largest:

10 - 5 = 5.

The answer is B.

ENGAA 2021 Section 1, Part B (Mathematics questions) · Q39

P QR S

is a rectangle.

P

and

Q

lie on the

x

-axis.

Q

and

R

lie on the line

x = 15

S

lies on the curve

y = x

What is the maximum possible area of the rectangle?

Cartesian axes, with the vertical axis labelled y and the horizontal axis labelled x, and 0 marked at the origin on both axes. A curve labelled y = the square root of x rises from the origin and flattens as it goes to the right. A vertical line labelled x = 15 crosses the curve; the point where this line meets the x-axis is labelled Q, and the point where it meets the horizontal dashed line is labelled R. A point P is marked on the x-axis to the left of Q. A dashed vertical segment rises from P to a point S on the curve, and a dashed horizontal segment runs from S to the right until it reaches R on the line x = 15. No lengths or angles are marked on the diagram.

A $55$
B $105$
C $50$
D $255$
E $100$
F $125$

Worked solution

Answer:B

Let

P = (p, 0)

with

0 \leq p \leq 15

. Then

Q = (15, 0)

, and

S

is the point of the curve above

P

, so

S = (p, p)

and

R = (15, p)

.

The rectangle has base

P Q = 15 - p

and height

p

, so its area is

A (p) = (15 - p) p = 15 p^{1/2} - p^{3/2} .

Differentiate:

A^{'} (p) = \frac{15}{2} p^{- 1/2} - \frac{3}{2} p^{1/2} = \frac{15 - 3 p}{2 p} .

For

0 < p < 15

the denominator is positive, so

A^{'} (p) = 0

gives

p = 5

. Moreover

A^{'} (p) > 0

for

p < 5

and

A^{'} (p) < 0

for

p > 5

, so

p = 5

is a maximum, and the endpoints

p = 0

and

p = 15

both give area

0

.

The maximum area is

A (5) = (15 - 5) 5 = 105 .

The answer is B.