Comprehensive Guide to Maxima and Minima in Calculus

Example: Locating Stationary Points and Classifying Them

Find the stationary points of the function \( y = 5x^3 + 15x^2 + 18x + 7 \) and classify each point.

Solution:

First, compute the first derivative:

\[ \frac{dy}{dx} = 15x^2 + 30x + 18 \]

Set the derivative equal to zero to find stationary points:

\[ 15x^2 + 30x + 18 = 0 \implies 5x^2 + 10x + 6 = 0 \]

Using the quadratic formula:

\[ x = \frac{-10 \pm \sqrt{10^2 - 4 \times 5 \times 6}}{2 \times 5} = \frac{-10 \pm \sqrt{100 - 120}}{10} \]

Since the discriminant is negative, there are no real stationary points. Hence, the function has no local maxima or minima.

Example: Identifying Turning Points on a Curve

Consider a function with stationary points at \( x = 2, 4, 6 \). At \( x=2 \) and \( x=4 \), the function changes from increasing to decreasing and decreasing to increasing respectively, but at \( x=6 \), the function remains decreasing before and after. Classify these points.

Solution:

• Points at \( x=2 \) and \( x=4 \) are turning points because the function changes direction.

• Point at \( x=6 \) is a stationary point but not a turning point since the function does not change its increasing/decreasing behavior.

Example: Using Derivative Tests to Find Local Extrema

Find the local maxima and minima of the function \( f(x) = 2x^4 - 8x^3 + 6x^2 + 1 \).

Solution:

First derivative:

\[ f'(x) = 8x^3 - 24x^2 + 12x \]

Set \( f'(x) = 0 \):

\[ 8x^3 - 24x^2 + 12x = 0 \implies 4x(2x^2 - 6x + 3) = 0 \]

So, \( x = 0 \) or solve \( 2x^2 - 6x + 3 = 0 \):

\[ x = \frac{6 \pm \sqrt{36 - 24}}{4} = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt{3}}{4} \]

Second derivative:

\[ f''(x) = 24x^2 - 48x + 12 \]

Evaluate at critical points:

• At \( x=0 \): \( f''(0) = 12 > 0 \) → local minimum.

• At \( x = \frac{6 + 2\sqrt{3}}{4} \approx 2.37 \): \( f''(2.37) > 0 \) → local minimum.

• At \( x = \frac{6 - 2\sqrt{3}}{4} \approx 0.63 \): \( f''(0.63) < 0 \) → local maximum.

Example: Verifying Maxima and Minima Properties

Consider the function \( g(x) = x^3 - 3x^2 + 4 \). Show that maxima and minima alternate and identify them.

Solution:

First derivative:

\[ g'(x) = 3x^2 - 6x \]

Set \( g'(x) = 0 \):

\[ 3x(x - 2) = 0 \implies x = 0, 2 \]

Second derivative:

\[ g''(x) = 6x - 6 \]

Evaluate at critical points:

• At \( x=0 \): \( g''(0) = -6 < 0 \) → local maximum.

• At \( x=2 \): \( g''(2) = 6 > 0 \) → local minimum.

This confirms the alternating pattern of maxima and minima.

Example 1: Finding Turning Points of a Cubic Function

Determine the turning points of \( y = 3x^3 + 9x^2 + 12x + 5 \).

Solution:

Calculate the first derivative:

\[ \frac{dy}{dx} = 9x^2 + 18x + 12 \]

Set derivative to zero:

\[ 9x^2 + 18x + 12 = 0 \implies 3x^2 + 6x + 4 = 0 \]

Discriminant:

\[ \Delta = 6^2 - 4 \times 3 \times 4 = 36 - 48 = -12 < 0 \]

No real roots, so no turning points exist.

Example 2: Identifying Local Extrema of a Quartic Function

Find the local maxima and minima of \( f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1 \).

Solution:

First derivative:

\[ f'(x) = 4x^3 - 12x^2 + 12x - 4 \]

Set \( f'(x) = 0 \):

\[ 4x^3 - 12x^2 + 12x - 4 = 0 \implies x^3 - 3x^2 + 3x - 1 = 0 \]

This factors as:

\[ (x - 1)^3 = 0 \]

So, \( x = 1 \) is the only critical point.

Second derivative:

\[ f''(x) = 12x^2 - 24x + 12 \]

Evaluate at \( x=1 \):

\[ f''(1) = 12 - 24 + 12 = 0 \]

Since the second derivative is zero, the test is inconclusive. Using the first derivative test or analyzing the function shows \( x=1 \) is a point of inflection, not a maximum or minimum.

Example 3: Maximizing Curved Surface Area of a Cylinder in a Cone

Prove that the radius of the right circular cylinder with the greatest curved surface area inscribed in a cone is half the radius of the cone.

Solution:

Let the cone have radius \( R \) and height \( H \), and the cylinder have radius \( r \) and height \( h \).

The curved surface area of the cylinder is:

\[ S = 2\pi r h \]

From similar triangles, height of cylinder:

\[ h = H \frac{R - r}{R} \]

Substitute \( h \) into \( S \):

\[ S = 2\pi r \times H \frac{R - r}{R} = \frac{2\pi H}{R} (rR - r^2) \]

Differentiate \( S \) with respect to \( r \):

\[ \frac{dS}{dr} = \frac{2\pi H}{R} (R - 2r) \]

Set derivative to zero for maxima/minima:

\[ R - 2r = 0 \implies r = \frac{R}{2} \]

Second derivative:

\[ \frac{d^2S}{dr^2} = \frac{2\pi H}{R} (-2) = -\frac{4\pi H}{R} < 0 \]

Since second derivative is negative, \( r = \frac{R}{2} \) gives maximum curved surface area.

Example 4: Calculating Maximum Height of a Thrown Object

A ball is thrown vertically with height given by \( h = -6t^2 + 12t + 3 \). Find its maximum height.

Solution:

Differentiate height with respect to time:

\[ \frac{dh}{dt} = -12t + 12 \]

Set derivative to zero to find time at maximum height:

\[ -12t + 12 = 0 \implies t = 1 \text{ second} \]

Calculate height at \( t=1 \):

\[ h = -6(1)^2 + 12(1) + 3 = -6 + 12 + 3 = 9 \text{ meters} \]

The maximum height reached is 9 meters.

Example 5: Determining Local Maxima and Minima of a Cubic Polynomial

Find the local maxima and minima of \( f(x) = 3x^3 - 24x^2 + 45x - 10 \).

Solution:

First derivative:

\[ f'(x) = 9x^2 - 48x + 45 \]

Set \( f'(x) = 0 \):

\[ 9x^2 - 48x + 45 = 0 \implies x^2 - \frac{16}{3}x + 5 = 0 \]

Using quadratic formula:

\[ x = \frac{\frac{16}{3} \pm \sqrt{\left(\frac{16}{3}\right)^2 - 4 \times 1 \times 5}}{2} \]

\[ = \frac{\frac{16}{3} \pm \sqrt{\frac{256}{9} - 20}}{2} = \frac{\frac{16}{3} \pm \sqrt{\frac{256 - 180}{9}}}{2} = \frac{\frac{16}{3} \pm \frac{\sqrt{76}}{3}}{2} \]

Approximate roots:

\[ x \approx \frac{5.33 \pm 2.91}{2} \]

So, \( x_1 \approx 4.12 \), \( x_2 \approx 1.21 \).

Second derivative:

\[ f''(x) = 18x - 48 \]

Evaluate at critical points:

• At \( x=1.21 \): \( f''(1.21) = 18(1.21) - 48 = 21.78 - 48 = -26.22 < 0 \) → local maximum.

• At \( x=4.12 \): \( f''(4.12) = 18(4.12) - 48 = 74.16 - 48 = 26.16 > 0 \) → local minimum.