Stationary Points, Maxima, and Minima Practice Questions — JAMB UTME

A curve is given by the equation $y = \frac{x}{x^2 + 4}$ . What is the maximum value of $y$ on this curve?

\frac{1}{4}

-\frac{1}{4}

2

0

Show answer & explanation

Answer: $\frac{1}{4}$

Answer

The maximum value of

y

on the curve is

\frac{1}{4}

.

To find the maximum value of

y = \frac{x}{x^2 + 4}

, differentiate

y

using the quotient rule to obtain

\frac{dy}{dx} = \frac{4 - x^2}{(x^2 + 4)^2}

. Setting the numerator to zero gives

x = \pm 2

. Since

\frac{dy}{dx}

is positive for

x < 2

and negative for

x > 2

,

x = 2

corresponds to a local maximum. Substituting

x = 2

into

y = \frac{x}{x^2 + 4}

yields

y = \frac{2}{8} = \frac{1}{4}

.

Step-by-Step Solution

1

Differentiate

y = \frac{x}{x^2 + 4}

using the quotient rule

\frac{dy}{dx} = \frac{(x^2 + 4)(1) - x(2x)}{(x^2 + 4)^2} = \frac{4 - x^2}{(x^2 + 4)^2}

Stationary points occur where the first derivative

\frac{dy}{dx}

equals zero.

2

Find the stationary points by setting

\frac{dy}{dx} = 0

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

A fraction equals zero when its numerator is zero.

3

Determine the nature of the stationary point at

x = 2

\text{For } x < 2, \frac{dy}{dx} > 0; \text{ for } x > 2, \frac{dy}{dx} < 0 \implies x = 2 \text{ is a maximum point}

The derivative changes sign from positive to negative across a maximum point.

4

Evaluate

y

at

x = 2

y = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4}

Substituting the

x

-coordinate into the curve equation yields the maximum value of

y

.

Key Concept

Stationary Points, Maxima, and Minima

Question 2Question →

A curve is defined by the equation $y = x^3 - 3x^2 + k$ , where $k$ is a constant. If the local minimum value of $y$ on the curve is $2$ , what is the value of $k$ ?

Show answer & explanation

Answer: 6

Answer

The value of the constant

k

is

6

.

To find the constant

k

, differentiate the curve equation to obtain

\frac{dy}{dx} = 3x^2 - 6x

. Setting

\frac{dy}{dx} = 0

gives stationary points at

x = 0

and

x = 2

. Calculating the second derivative

\frac{d^2y}{dx^2} = 6x - 6

shows

\frac{d^2y}{dx^2} = 6 > 0

at

x = 2

, confirming that the local minimum occurs at

x = 2

. Substituting

x = 2

and the minimum value

y = 2

into

y = x^3 - 3x^2 + k

yields

2 = 8 - 12 + k

, which simplifies to

k = 6

.

Step-by-Step Solution

1

Differentiate

y = x^3 - 3x^2 + k

with respect to

x

.

\frac{dy}{dx} = 3x^2 - 6x

Stationary points occur where the first derivative equals zero.

2

Solve

\frac{dy}{dx} = 0

for

x

.

x = 0

or

x = 2

These

x

-values locate the turning points on the curve.

3

Evaluate the second derivative

\frac{d^2y}{dx^2} = 6x - 6

at each stationary point.

At

x = 0

,

\frac{d^2y}{dx^2} = -6

(local maximum); at

x = 2

,

\frac{d^2y}{dx^2} = 6

(local minimum).

A positive second derivative indicates a local minimum point.

4

Substitute

x = 2

and

y = 2

into the curve equation

y = x^3 - 3x^2 + k

.

2 = (2)^3 - 3(2)^2 + k \implies 2 = -4 + k \implies k = 6

The local minimum value of

y

is attained at

x = 2

.

Key Concept

Stationary Points, Maxima, and Minima

Question 3Question →

A curve is defined by the equation $y = 2x^3 - 9x^2 + 12x + 5$ . What is the value of $y$ at its maximum stationary point?

Show answer & explanation

Answer: 10

Answer

The value of

y

at the maximum stationary point is

10

.

To locate the maximum stationary point, set the first derivative

\frac{dy}{dx} = 6x^2 - 18x + 12

to zero, obtaining stationary values

x = 1

and

x = 2

. Testing in the second derivative

\frac{d^2y}{dx^2} = 12x - 18

gives

-6

at

x = 1

, confirming a local maximum. Substituting

x = 1

into the original cubic equation gives

y = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 10

.

Step-by-Step Solution

1

Differentiate

y

with respect to

x

to find the gradient function.

\frac{dy}{dx} = 6x^2 - 18x + 12

Stationary points occur where the gradient of the curve is zero.

2

Set the first derivative to zero and solve for

x

.

x = 1 \text{ or } x = 2

Factoring

6(x - 1)(x - 2) = 0

yields the

x

-coordinates of the turning points.

3

Determine the nature of the stationary points using the second derivative test.

\frac{d^2y}{dx^2} = 12x - 18; \quad \text{at } x = 1, \frac{d^2y}{dx^2} = -6 < 0

A negative second derivative indicates a local maximum stationary point.

4

Substitute

x = 1

into the original function to determine

y

.

y = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 10

Evaluating the curve function at the maximum

x

-coordinate provides the corresponding maximum

y

-value.

Key Concept

Stationary Points, Maxima, and Minima

Question 4Question →

A cubic curve is defined by the equation $y = ax^3 + bx^2 + cx + d$ . The curve has a point of inflexion at $(0, 1)$ and a stationary point at $(1, 5)$ . What is the local minimum value of the function?

-3

-7

5

-1

Show answer & explanation

Answer: $-3$

Answer

The local minimum value of the function is

-3

.

By applying the conditions for a point of inflexion (

y''=0

) and a stationary point (

y'=0

), the curve equation is identified as

y = -2x^3 + 6x + 1

. Solving

y' = 0

gives

x = -1

for the local minimum, resulting in a minimum value of

-3

.

Step-by-Step Solution

1

Use the point of inflexion and given point

(0, 1)

to find

b

and

d

.

d = 1

and

b = 0

.

Since

(0,1)

lies on the curve, substituting

x=0, y=1

gives

d = 1

. The second derivative is

y'' = 6ax + 2b

. At a point of inflexion

x = 0

,

y'' = 0

, giving

2b = 0 \implies b = 0

.

2

Use the stationary point condition at

(1, 5)

to determine

a

and

c

.

a = -2

and

c = 6

.

The equation reduces to

y = ax^3 + cx + 1

. Since

(1,5)

is on the curve,

a + c + 1 = 5 \implies a + c = 4

. Also,

y' = 3ax^2 + c = 0

at

x = 1

, so

3a + c = 0

. Solving

3a + c = 0

and

a + c = 4

simultaneously yields

a = -2

and

c = 6

.

3

Find all stationary points of

y = -2x^3 + 6x + 1

and test their nature using the second derivative.

Stationary points are at

x = 1

(maximum) and

x = -1

(minimum).

Setting

y' = -6x^2 + 6 = 0

yields

x^2 = 1 \implies x = \pm 1

. The second derivative is

y'' = -12x

. For

x = -1

,

y'' = 12 > 0

, confirming a local minimum.

4

Evaluate the function at

x = -1

to find the local minimum value.

y(-1) = -3

.

Substituting

x = -1

into

y = -2x^3 + 6x + 1

yields

y = -2(-1)^3 + 6(-1) + 1 = 2 - 6 + 1 = -3

.

Key Concept

Determining curve constants using stationary points and points of inflexion, followed by identifying local extrema.

Estimated Time:3m 0s

Question 5Question →

A curve is defined by the equation $y = x^3 + px^2 + qx + 5$ , where $p$ and $q$ are constants. If the curve has a stationary point with a local maximum at $x = -1$ and a local minimum at $x = 3$ , what is the value of $p + q$ ?

-12

12

-15

-6

Show answer & explanation

Answer: $-12$

Answer

The value of

p + q

is

-12

.

The derivative of

y = x^3 + px^2 + qx + 5

is

\frac{dy}{dx} = 3x^2 + 2px + q

. Setting

\frac{dy}{dx} = 0

at the stationary points

x = -1

and

x = 3

means

\frac{dy}{dx} = 3(x + 1)(x - 3) = 3x^2 - 6x - 9

. Comparing coefficients gives

2p = -6 \implies p = -3

and

q = -9

. Summing these constants gives

p + q = -12

. Evaluating the second derivative

\frac{d^2y}{dx^2} = 6x + 2p = 6x - 6

confirms a maximum at

x = -1

(

\frac{d^2y}{dx^2} = -12 < 0

) and a minimum at

x = 3

(

\frac{d^2y}{dx^2} = 12 > 0

).

Step-by-Step Solution

1

Find the derivative of the given function.

\frac{dy}{dx} = 3x^2 + 2px + q

Stationary points occur where the first derivative

\frac{dy}{dx} = 0

.

2

Use the stationary points

x = -1

and

x = 3

to form a quadratic equation for the derivative.

\frac{dy}{dx} = 3(x + 1)(x - 3) = 3(x^2 - 2x - 3) = 3x^2 - 6x - 9

Since

x = -1

and

x = 3

are roots of

\frac{dy}{dx} = 0

, the derivative must factor as

3(x - (-1))(x - 3)

.

3

Equate coefficients of the two derivative expressions to solve for

p

and

q

.

2p = -6 \implies p = -3

and

q = -9

Matching corresponding terms gives

2p = -6

and

q = -9

.

4

Calculate the required sum

p + q

.

p + q = -3 + (-9) = -12

Summing the calculated constants yields

-12

.

Key Concept

Determining parameters of a polynomial function from given stationary points using differentiation and coefficient matching.

Question 6Question →

A curve is defined by the equation $y = x^2 - 6x + 11$ . What is the minimum value of $y$ on this curve?

Show answer & explanation

Answer: 2

Answer

The minimum value of

y

on the curve is 2.

Differentiating

y = x^2 - 6x + 11

gives

\frac{dy}{dx} = 2x - 6

. Setting this derivative to zero yields

2x - 6 = 0

, so

x = 3

. Substituting

x = 3

into the original function gives

y = (3)^2 - 6(3) + 11 = 2

. Since

\frac{d^2y}{dx^2} = 2 > 0

, the point at

x = 3

is a local minimum, making 2 the minimum value of

y

.

Step-by-Step Solution

1

Find the first derivative of the curve function.

\frac{dy}{dx} = 2x - 6

Stationary points occur where the derivative is equal to zero.

2

Solve for the

x

-coordinate at the stationary point.

2x - 6 = 0 \implies x = 3

Setting the derivative to zero determines the input value where the slope is horizontal.

3

Calculate the corresponding

y

-value at

x = 3

.

y = (3)^2 - 6(3) + 11 = 2

Evaluating the original equation at

x = 3

yields the minimum value of

y

.

Key Concept

Finding the minimum value of a function using differentiation

Question 7Question →

An open rectangular box with a square base of side length $x\text{ cm}$ is to be constructed such that its total surface area is $108\text{ cm}^2$ . What is the maximum volume of the box in $\text{cm}^3$ ?

54\text{ cm}^3

72\text{ cm}^3

108\text{ cm}^3

432\text{ cm}^3

Show answer & explanation

Answer: $108\text{ cm}^3$

Answer

108\text{ cm}^3

For an open box with a square base of side

x

and height

h

, total surface area is

A = x^2 + 4xh = 108\text{ cm}^2

. Solving for

h

gives

h = \frac{108 - x^2}{4x}

. Substituting

h

into the volume formula gives

V(x) = x^2 h = 27x - \frac{1}{4}x^3

. Differentiating gives

\frac{dV}{dx} = 27 - \frac{3}{4}x^2

. Setting

\frac{dV}{dx} = 0

yields

x = 6

. The second derivative

\frac{d^2V}{dx^2} = -\frac{3}{2}x

evaluated at

x = 6

is

-9 < 0

, confirming a local maximum. Evaluating

V(6) = 27(6) - \frac{1}{4}(6^3) = 162 - 54 = 108\text{ cm}^3

yields the correct maximum volume.

Step-by-Step Solution

1

Set up the surface area equation for the open box and express height

h

in terms of

x

.

Surface area

A = x^2 + 4xh = 108 \implies h = \frac{108 - x^2}{4x}

.

An open box with a square base has 1 base face (

x^2

) and 4 vertical side faces (

xh

).

2

Formulate the volume function

V(x)

in terms of

x

.

V(x) = x^2 h = x^2 \left(\frac{108 - x^2}{4x}\right) = \frac{1}{4}(108x - x^3) = 27x - \frac{1}{4}x^3

.

Substitute the expression for

h

into the volume formula

V = x^2 h

.

3

Find the critical point by differentiating

V(x)

with respect to

x

and setting

\frac{dV}{dx} = 0

.

\frac{dV}{dx} = 27 - \frac{3}{4}x^2 = 0 \implies \frac{3}{4}x^2 = 27 \implies x^2 = 36 \implies x = 6\text{ cm}

.

Stationary points occur where the first derivative of the volume function equals zero.

4

Verify that

x = 6

gives a maximum volume and calculate

V(6)

.

\frac{d^2V}{dx^2} = -\frac{3}{2}x \implies \left.\frac{d^2V}{dx^2}\right|_{x=6} = -9 < 0

(maximum). Volume

V(6) = 27(6) - \frac{1}{4}(6^3) = 162 - 54 = 108\text{ cm}^3

.

The negative second derivative confirms a maximum turning point.

Key Concept

Optimization of physical quantities using the first and second derivative tests.

Estimated Time:2m 0s

Question 8Question →

The total profit $P(x)$ , in thousands of Naira, obtained from producing and selling $x$ hundred units of a commodity is modeled by the function $P(x) = -x^3 + 6x^2 + 15x - 8$ , where $x \ge 0$ . What is the maximum profit achievable?

Show answer & explanation

Answer: 92

Answer

The maximum profit achievable is 92 thousand Naira.

To find the maximum profit, we find the stationary points of

P(x) = -x^3 + 6x^2 + 15x - 8

by taking the derivative

P'(x) = -3x^2 + 12x + 15

and setting it to 0. Solving

-3(x-5)(x+1) = 0

with

x \ge 0

yields

x = 5

. Testing the second derivative gives

P''(5) = -18 < 0

, confirming

x = 5

is a maximum. Substituting

x = 5

into

P(x)

yields

P(5) = 92

.

Step-by-Step Solution

1

Differentiate the profit function

P(x)

with respect to

x

P'(x) = -3x^2 + 12x + 15

Stationary points occur where the rate of change of profit (the derivative) is equal to zero.

2

Set

P'(x) = 0

and solve for

x

-3(x^2 - 4x - 5) = 0 \implies (x - 5)(x + 1) = 0

, giving critical values

x = 5

and

x = -1

Factoring the quadratic equation yields the critical values of production level.

3

Filter critical values based on physical domain constraints

x = 5

(reject

x = -1

since production

x \ge 0

)

Production quantities cannot be negative in physical real-life contexts.

4

Perform the second derivative test to confirm the nature of the stationary point

P''(x) = -6x + 12 \implies P''(5) = -6(5) + 12 = -18

Since

P''(5) < 0

, the function achieves a local maximum at

x = 5

.

5

Substitute

x = 5

back into original function

P(x)

P(5) = -(5)^3 + 6(5)^2 + 15(5) - 8 = -125 + 150 + 75 - 8 = 92

Evaluating

P(5)

gives the maximum total profit value.

Key Concept

Optimization and Maximum Values using First and Second Derivatives

Question 9Question →

What is the x-coordinate of the maximum stationary point of the curve $y = \sin x + \cos x$ in the interval $0 \le x \le \pi$ ?

\frac{\pi}{4}

\frac{3\pi}{4}

\frac{\pi}{2}

\pi

Show answer & explanation

Answer: $\frac{\pi}{4}$

Answer

The x-coordinate of the maximum stationary point is

\frac{\pi}{4}

.

To find stationary points, we differentiate

y = \sin x + \cos x

to get

\frac{dy}{dx} = \cos x - \sin x

. Setting

\frac{dy}{dx} = 0

gives

\sin x = \cos x

, or

\tan x = 1

. Within

0 \le x \le \pi

, the solution is

x = \frac{\pi}{4}

. Evaluating the second derivative

\frac{d^2y}{dx^2} = -\sin x - \cos x

at

x = \frac{\pi}{4}

yields

-\sqrt{2} < 0

, confirming a local maximum.

Step-by-Step Solution

1

Differentiate the function

y = \sin x + \cos x

with respect to

x

.

\frac{dy}{dx} = \cos x - \sin x

Stationary points occur where the first derivative

\frac{dy}{dx} = 0

.

2

Set the first derivative to zero and solve for

x

in the interval

0 \le x \le \pi

.

\cos x - \sin x = 0 \implies \sin x = \cos x \implies \tan x = 1 \implies x = \frac{\pi}{4}

Dividing both sides by

\cos x

gives

\tan x = 1

, which has the solution

x = \frac{\pi}{4}

in the given domain.

3

Verify the nature of the stationary point using the second derivative test.

\frac{d^2y}{dx^2} = -\sin x - \cos x \implies \left.\frac{d^2y}{dx^2}\right|_{x=\frac{\pi}{4}} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2} < 0

A negative second derivative confirms that

x = \frac{\pi}{4}

is a local maximum point.

Key Concept

Finding stationary points and determining their nature using derivatives of trigonometric functions.

Question 10Question →

A cubic curve defined by $y = ax^3 + bx^2 + cx + d$ has a local maximum at $(-1, 10)$ and a point of inflexion at $(1, 2)$ . What is the value of $y$ at the local minimum of the curve?

Show answer & explanation

Answer: -6

Answer

The local minimum value of

y

on the curve is

-6

.

By setting up a system of equations using the conditions for the local maximum at

(-1, 10)

and the point of inflexion at

(1, 2)

, the cubic curve is uniquely determined as

y = 0.5x^3 - 1.5x^2 - 4.5x + 7.5

. Setting the derivative

\frac{dy}{dx} = 1.5(x-3)(x+1) = 0

gives

x = 3

as the

x

-coordinate of the local minimum. Evaluating

y(3)

gives

-6

.

Step-by-Step Solution

1

Differentiate the general cubic equation to obtain expressions for the first and second derivatives.

\frac{dy}{dx} = 3ax^2 + 2bx + c

and

\frac{d^2y}{dx^2} = 6ax + 2b

.

Stationary points require

\frac{dy}{dx} = 0

and points of inflexion require

\frac{d^2y}{dx^2} = 0

.

2

Apply the point of inflexion conditions at

(1, 2)

.

6a(1) + 2b = 0 \implies b = -3a

, and

a(1)^3 + b(1)^2 + c(1) + d = 2 \implies -2a + c + d = 2

.

At a point of inflexion, the second derivative is zero, and the point lies on the curve.

3

Apply the stationary point and coordinate conditions at the local maximum

(-1, 10)

.

3a(-1)^2 + 2b(-1) + c = 0 \implies 9a + c = 0 \implies c = -9a

, and

-a + b - c + d = 10 \implies 5a + d = 10

.

At a local maximum, the first derivative is zero, and the point lies on the curve.

4

Solve the system of linear equations for coefficients

a, b, c, d

.

a = 0.5

,

b = -1.5

,

c = -4.5

,

d = 7.5

.

Combining

-11a + d = 2

and

5a + d = 10

yields

16a = 8 \implies a = 0.5

.

5

Find the

x

-coordinate of the local minimum by solving

\frac{dy}{dx} = 0

.

1.5x^2 - 3x - 4.5 = 0 \implies 1.5(x - 3)(x + 1) = 0 \implies x = 3

(since

x = -1

is the local maximum).

Evaluating

\frac{d^2y}{dx^2}(3) = 6(0.5)(3) + 2(-1.5) = 6 > 0

confirms a local minimum at

x = 3

.

6

Calculate the value of

y

at

x = 3

.

y = 0.5(3)^3 - 1.5(3)^2 - 4.5(3) + 7.5 = -6

.

Substituting

x = 3

into the curve equation gives the value of

y

at the local minimum.

Key Concept

Determining polynomial coefficients from stationary and inflexion point conditions to find extreme values.

Question 11Question →

A curve has the equation $y = ax^3 + bx^2 + 12x + 1$ , where $a$ and $b$ are constants. If the curve has stationary points at $x = 1$ and $x = 2$ , what is the value of $a$ ?

2

6

-2

-6

Show answer & explanation

Answer: $2$

Answer

The value of

a

is

2

.

To find

a

, take the first derivative of the curve, yielding

\frac{dy}{dx} = 3ax^2 + 2bx + 12

. Setting this to zero gives a quadratic equation with roots

x = 1

and

x = 2

. The product of roots for a quadratic equation

Ax^2 + Bx + C = 0

is

\frac{C}{A}

. Therefore,

1 \times 2 = \frac{12}{3a}

, which simplifies to

2 = \frac{4}{a}

, giving

a = 2

.

Step-by-Step Solution

1

Find the derivative of the given curve with respect to

x

.

\frac{dy}{dx} = 3ax^2 + 2bx + 12

Stationary points occur where the first derivative

\frac{dy}{dx} = 0

.

2

Set the derivative to zero and substitute the stationary point locations

x = 1

and

x = 2

.

The roots of the quadratic equation

3ax^2 + 2bx + 12 = 0

are

x_1 = 1

and

x_2 = 2

.

Since stationary points are given at

x=1

and

x=2

, these values satisfy

\frac{dy}{dx} = 0

.

3

Apply the product of roots formula for a quadratic equation

Ax^2 + Bx + C = 0

, where

x_1 \cdot x_2 = \frac{C}{A}

.

1 \times 2 = \frac{12}{3a} \implies 2 = \frac{4}{a}

Equating the product of roots

1 \times 2 = 2

to

\frac{12}{3a}

isolates parameter

a

.

4

Solve for

a

.

a = 2

Multiplying both sides by

a

gives

2a = 4

, so

a = 2

.

Key Concept

Stationary points occur where

\frac{dy}{dx} = 0

. For a cubic curve, the derivative is a quadratic equation whose roots correspond to the

x

-coordinates of the stationary points.

Estimated Time:2m 0s

Question 12Question →

What is the $x$ -coordinate of the stationary point of the curve $y = 3x^2 - 12x + 7$ ?

2

-2

4

7

Show answer & explanation

Answer: $2$

Answer

The

x

-coordinate of the stationary point is

2

.

To find the stationary point, take the derivative of

y = 3x^2 - 12x + 7

with respect to

x

, obtaining

\frac{dy}{dx} = 6x - 12

. Setting the derivative equal to zero gives

6x - 12 = 0

, which solves to

x = 2

.

Step-by-Step Solution

1

Differentiate the equation of the curve with respect to

x

\frac{dy}{dx} = 6x - 12

Stationary points occur where the first derivative (gradient) of the function is equal to zero.

2

Set the derivative to zero and solve for

x

6x - 12 = 0 \implies 6x = 12 \implies x = 2

Solving this linear equation gives the exact value of

x

at which the tangent to the curve is horizontal.

Key Concept

Stationary Points of a Curve

Estimated Time:45s

Question 13Question →

A closed cylindrical metal container has a total surface area of $54\pi\text{ cm}^2$ . What radius, in centimeters, of the circular base will yield the maximum volume for the container?

Show answer & explanation

Answer: 3

Answer

The radius of the circular base that maximizes the volume is 3 cm.

To find the radius that yields maximum volume, we first express height

h

in terms of radius

r

using the total surface area formula

2\pi r^2 + 2\pi rh = 54\pi

, giving

h = \frac{27 - r^2}{r}

. Substituting this into the volume equation

V = \pi r^2 h

gives

V(r) = 27\pi r - \pi r^3

. Setting the first derivative

\frac{dV}{dr} = 27\pi - 3\pi r^2

to zero yields

3\pi r^2 = 27\pi

, so

r^2 = 9

and

r = 3\text{ cm}

. The second derivative

\frac{d^2V}{dr^2} = -6\pi r

evaluated at

r = 3

is

-18\pi

, which is strictly negative, confirming that

r = 3\text{ cm}

maximizes volume.

Step-by-Step Solution

1

Set up the surface area equation for a closed cylinder with the given value.

2\pi r^2 + 2\pi r h = 54\pi

A closed cylinder consists of two circular bases (

2\pi r^2

) and a curved lateral surface (

2\pi r h

).

2

Express

h

in terms of

r

.

h = \frac{27 - r^2}{r}

Dividing the surface area equation by

2\pi

yields

r^2 + rh = 27

, allowing

h

to be isolated.

3

Substitute

h

into the volume formula

V = \pi r^2 h

to write volume as a function of

r

only.

V(r) = 27\pi r - \pi r^3

To maximize volume using calculus, the volume equation must be expressed in terms of a single variable.

4

Differentiate

V(r)

with respect to

r

and set the derivative equal to zero to find stationary points.

\frac{dV}{dr} = 27\pi - 3\pi r^2 = 0 \implies r = 3

Maximum volume occurs at a stationary point where the first derivative is zero.

5

Verify that

r = 3

produces a maximum using the second derivative test.

\frac{d^2V}{dr^2} = -6\pi r

; at

r = 3

,

\frac{d^2V}{dr^2} = -18\pi < 0

A negative second derivative indicates a local maximum.

Key Concept

Optimization and Stationary Points in Mensuration

Estimated Time:2m 0s

Question 14Question →

A curve is defined by the equation $y = 4x + \frac{9}{x}$ for $x > 0$ . What is the $y$ -value at the minimum stationary point of the curve?

Show answer & explanation

Answer: 12

Answer

The

y

-value at the minimum stationary point is 12.

To find the minimum value of

y = 4x + \frac{9}{x}

for

x > 0

, set the first derivative

\frac{dy}{dx} = 4 - \frac{9}{x^2}

equal to

0

, yielding

x = 1.5

. The second derivative

\frac{d^2y}{dx^2} = \frac{18}{x^3}

is positive at

x = 1.5

, confirming a minimum stationary point. Evaluating the original equation at

x = 1.5

gives

y = 4(1.5) + \frac{9}{1.5} = 6 + 6 = 12

.

Step-by-Step Solution

1

Differentiate the given function

y = 4x + 9x^{-1}

with respect to

x

.

\frac{dy}{dx} = 4 - 9x^{-2} = 4 - \frac{9}{x^2}

Stationary points occur where the gradient (first derivative) equals zero.

2

Set the first derivative to zero and solve for

x

given the domain constraint

x > 0

.

4 - \frac{9}{x^2} = 0 \implies 4x^2 = 9 \implies x^2 = \frac{9}{4} \implies x = \frac{3}{2} = 1.5

Solving

\frac{dy}{dx} = 0

yields the

x

-coordinate of the turning point.

3

Evaluate the second derivative to confirm the turning point is a local minimum.

\frac{d^2y}{dx^2} = \frac{18}{x^3}. \text{ At } x = 1.5, \frac{d^2y}{dx^2} = \frac{18}{3.375} = 5.333 > 0

A positive second derivative indicates that the stationary point is a local minimum.

4

Substitute

x = 1.5

back into the original equation

y = 4x + \frac{9}{x}

to find the corresponding

y

-value.

y = 4(1.5) + \frac{9}{1.5} = 6 + 6 = 12

The question asks for the

y

-value of the curve at the minimum stationary point.

Key Concept

Stationary Points, Maxima, and Minima

Estimated Time:2m 0s

Question 15Question →

What is the value of $y$ at the local minimum stationary point of the curve $y = x^3 - 3x^2 - 9x + 15$ for $x > 0$ ?

Show answer & explanation

Answer: -12

Answer

The value of

y

at the local minimum stationary point is

-12

.

To find the local minimum point of

y = x^3 - 3x^2 - 9x + 15

, set the derivative

\frac{dy}{dx} = 3x^2 - 6x - 9

equal to

0

. Solving

3(x - 3)(x + 1) = 0

yields

x = 3

for

x > 0

. The second derivative

\frac{d^2y}{dx^2} = 6x - 6

equals

12 > 0

at

x = 3

, confirming a local minimum. Substituting

x = 3

back into the original curve equation yields

y = (3)^3 - 3(3)^2 - 9(3) + 15 = -12

.

Step-by-Step Solution

1

Differentiate

y = x^3 - 3x^2 - 9x + 15

with respect to

x

\frac{dy}{dx} = 3x^2 - 6x - 9

Stationary points occur where the slope (first derivative) is zero.

2

Set

\frac{dy}{dx} = 0

and solve for

x

x = 3 \text{ or } x = -1

Factoring

3(x^2 - 2x - 3) = 0

gives

(x - 3)(x + 1) = 0

.

3

Apply the second derivative test at

x = 3

(since

x > 0

)

\frac{d^2y}{dx^2} = 6(3) - 6 = 12 > 0

A positive second derivative confirms that

x = 3

is a local minimum.

4

Evaluate

y

at

x = 3

in the original curve equation

y = (3)^3 - 3(3)^2 - 9(3) + 15 = -12

Substituting

x = 3

into

y(x)

gives the

y

-coordinate of the minimum point.

Key Concept

Stationary Points and Local Minima of Polynomial Curves

Question 16Question →

What is the $x$ -coordinate of the maximum stationary point of the curve $y = \sin(2x) - x$ in the interval $0 \le x \le \pi$ ?

\frac{\pi}{6}

\frac{5\pi}{6}

\frac{\pi}{3}

\pi

Show answer & explanation

Answer: $\frac{\pi}{6}$

Answer

The

x

-coordinate of the maximum stationary point is

\frac{\pi}{6}

.

Differentiating

y = \sin(2x) - x

gives

\frac{dy}{dx} = 2\cos(2x) - 1

. Setting this derivative to zero yields

\cos(2x) = \frac{1}{2}

, giving

x = \frac{\pi}{6}

and

x = \frac{5\pi}{6}

in the given interval. Checking the second derivative

\frac{d^2y}{dx^2} = -4\sin(2x)

at

x = \frac{\pi}{6}

gives

-2\sqrt{3} < 0

, which confirms that the maximum stationary point occurs at

\frac{\pi}{6}

.

Step-by-Step Solution

1

Find the first derivative of

y = \sin(2x) - x

using the chain rule.

\frac{dy}{dx} = 2\cos(2x) - 1

Stationary points occur where the rate of change

\frac{dy}{dx} = 0

.

2

Set

\frac{dy}{dx} = 0

and solve for

x

within

0 \le x \le \pi

.

2\cos(2x) - 1 = 0 \implies \cos(2x) = \frac{1}{2} \implies 2x = \frac{\pi}{3}

or

2x = \frac{5\pi}{3} \implies x = \frac{\pi}{6}

or

x = \frac{5\pi}{6}

The trigonometric equation

\cos(\theta) = \frac{1}{2}

has solutions

\frac{\pi}{3}

and

\frac{5\pi}{3}

in

[0, 2\pi]

.

3

Evaluate the second derivative

\frac{d^2y}{dx^2}

to determine the nature of the stationary points.

\frac{d^2y}{dx^2} = -4\sin(2x)

. At

x = \frac{\pi}{6}

,

\frac{d^2y}{dx^2} = -4\sin\left(\frac{\pi}{3}\right) = -2\sqrt{3} < 0

.

A negative second derivative (

\frac{d^2y}{dx^2} < 0

) indicates a local maximum.

Key Concept

Determining maximum stationary points of trigonometric functions using the first and second derivative tests.

Question 17Question →

What is the maximum value of the curve $y = \frac{x}{x^2 + 4}$ ?

2

\frac{1}{4}

-\frac{1}{4}

\frac{1}{2}

Show answer & explanation

Answer: $\frac{1}{4}$

Answer

The maximum value of the curve is

\frac{1}{4}

.

Differentiating

y = \frac{x}{x^2 + 4}

using the quotient rule gives

\frac{dy}{dx} = \frac{4 - x^2}{(x^2 + 4)^2}

. Setting

\frac{dy}{dx} = 0

yields critical points at

x = \pm 2

. Substituting

x = 2

into the original function gives

y = \frac{2}{8} = \frac{1}{4}

, which is the maximum value of the function.

Step-by-Step Solution

1

Differentiate

y = \frac{x}{x^2 + 4}

with respect to

x

using the quotient rule.

\frac{dy}{dx} = \frac{(x^2 + 4)(1) - x(2x)}{(x^2 + 4)^2} = \frac{4 - x^2}{(x^2 + 4)^2}

Stationary points occur where the first derivative equals zero.

2

Set

\frac{dy}{dx} = 0

to solve for the stationary points.

4 - x^2 = 0 \implies x^2 = 4 \implies x = 2 \text{ or } x = -2

A rational expression equals zero when its numerator is zero.

3

Evaluate

y

at each critical point to determine the function values.

For

x = 2

:

y = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4}

. For

x = -2

:

y = \frac{-2}{(-2)^2 + 4} = -\frac{1}{4}

.

The question asks for the maximum value of

y

on the curve.

4

Compare the stationary values to select the maximum.

The maximum value is

\frac{1}{4}

at

x = 2

.

Since

\frac{1}{4} > -\frac{1}{4}

,

x = 2

corresponds to the maximum point.

Key Concept

Stationary Points and Maxima/Minima of Rational Functions

Estimated Time:1m 30s

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