Application of Calculus

Imagine you are playing a football match and you kick the ball high into the air. At some point, the ball reaches its highest point before starting to fall back to the ground. How do we figure out exactly where that highest point is? This is where the first and second derivatives come into play. They help us uncover hidden details about a function, such as where it reaches its highest (maximum) or lowest (minimum) points. The second derivative helps us identify special points called saddle points, where the function flattens but does not have a peak or valley. These ideas are not just about solving equations on paper; they are the keys to understanding real-life problems in sports, economics, science and more!

In this section, you will learn how to use the second derivative to classify points as maximum, minimum or saddle points and then use that knowledge to sketch the curve of a function. You will also find the maximum and minimum values of functions to solve real-world problems. Let us dive in and unlock the secrets of curves and turning points!

KEY IDEAS

· Differentiation helps to determine how a quantity changes with respect to another quantity.

· Points of inflection are where the concavity of a function changes (from concave up to concave down or vice versa).

· Saddle points occur when the graph flattens at a critical point but does not result in a local maximum or minimum in all directions.

· When the rate of change is negative, the graph of the function falls as you move from left to right.

· When the rate of change is positive, the graph of the function rises as you move from left to right.

· When the rate of change is zero, the graph neither rises nor falls at that specific point.

In year one, we learnt about the importance of finding the derivative of a function.

It was discussed that the derivative of a function at a point represents the gradient of the tangent line that touches the curve of the function at that point. The gradient of a curve (other than straight lines) at different points on the curve differ, the derivatives are functions in themselves. We can therefore consider the nature of the gradient at different points on the curve.

1. Constant functions of the form f(x) = c , c ∈ R, representing the horizontal line, have a gradient of zero.

2. Functions of the form f(x) = mx + c,

a. These represent a straight line that slopes upwards from left to right if the value of m is positive, m > 0

b. These represent a straight line that slopes downwards from left to right if the values of m is negative,m < 0 .

Combining all these facts which state that given a function y = f(x), if y or f is a function of x , then the first derivative ᵈʸ_ _(dx) or fᴵ)x( is called the gradient function. The gradient at any point Q) x₁ , y₁( is obtained by substituting the value of x₁ and y₁ into the expression for ᵈʸ_ _(dx) .

Functions and derivative

1. For f(x), the gradient is represented by its derivative fᴵ)x(.

Behaviour: The derivative gives the rate of change of the function at any given point.

2. If fᴵ(a) > 0, the gradient is positive.

Behaviour: The function f(x) is increasing at x = a

3. If fᴵ(a) < 0, the gradient is negative.

Behaviour: The function f(x) is decreasing at x = a

4. If fᴵ(a) = 0, the function has a horizontal tangent at x = a .

Behaviour: This often indicate a local maximum, minimum or saddle point and is known as a stationary point.

Graphical Interpretation

1. Positive gradient means the graph of the function is rising as you move from left to right.

2. Negative gradient means the graph of the function is falling as you move from left to right.

3. Zero gradient means the graph is flat at the point, indicating no change in f (x).

Activity 12.1: Graphing functions Carry out the following activity in small groups.

Materials Needed:

1. Graph paper

2. Pen/pencil and paper

3. Ruler Steps:

1. Review basic concepts of:

a. Derivative, the derivative of a function f)x( represents the gradient (slope) of the function at any point x .

b. Positive gradient indicates the function is increasing at that point.

c. Negative gradient indicates the function is decreasing at that point.

d. Zero gradient indicates a horizontal tangent, possibly at a maximum, minimum or inflection point, but a stationary point.

2. Select a variety of functions to analyse, for example:

a. f(x) = x²b. f(x) = − x²+ 4

c. f(x) = x³− 3x²+ 2

3. Calculate derivatives of the functions:

a. Find the first derivative of each function

Example,f(x) = x², the derivative is f′(x) = 2x

b. Choose specific point to evaluate the gradients.

Example, for f(x) = x², evaluate f′(x) at x = − 1, x = 0 and x = 1

c. Calculate f′(− 1) = − 2, f′(0) = 0, f′′(1() = 2.

4. Interpret results:

a. Positive gradient means increasing function

b. Negative gradient means decreasing function

c. Zero gradient means horizontal tangent and a stationary point.

5. Graph the functions:

a. Use your graph paper and pencil and plot each function.

b. Mark the points where you evaluated the gradient.

c. Observe the behaviour of the function at these points.

6. Compare observations:

a. Compare your calculated gradients with visual slopes on the graph.

b. Confirm whether the function is increasing, decreasing or has a horizontal tangent at the chosen points.

Now let consider the following worked examples.

Example 12.1

If f(x) = x³− 3x²+ 2, find the gradient of f)x( at x = 0, x = 1, x = 3.

Solution

f(x) = x³− 3x²+ 2 fᴵ)x( = 3x²− 6x At x = 0 , fᴵ(0) = 0 At x = 1 , fᴵ(1) = 3(1)²− 6(1) = − 3 (Negative) At x = 3 , fᴵ(3) = 3(3)²− 6(3) = 27 − 18 = 9 (Positive) Interpretation:

The function has a horizontal tangent/stationary point at x = 0 The function is decreasing at x = 1 and increasing at x = 3 .

Example 12.2

If f(x) = 2x²+ 3x − 5 find the gradient of f)x( at x = − 1, x = 0, x = 2.

Solution

f(x) = 2x²+ 3x − 5 fᴵ(x) = 4x + 3 At x = − 1 , fᴵ(− 1) = 4(− 1) + 3 = − 1 (Negative) At x = 0 , fᴵ(0) = 4(0) + 3 = 3 (Positive) At x = 2 , fᴵ(2) = 4(2) + 3 = 11 (Positive) Interpretation:

The function is decreasing at x = − 1 and increasing at x = 0 and x = 2

Example 12.3

Find the gradient on the curve f(x) = ) x − 2(²− 4, at x = 1, x = 2 and x = 3

Solution

f(x) = ) x − 2(²− 4 = x²− 4x fᴵ(x) = 2)x − 2( At x = 1 , fᴵ(1) = 2(1 − 2) = 2(− 1) = − 2 (Negative) At x = 2 , fᴵ(2) = 2(2 − 2) = 2(0) = 0 (Zero) At x = 3 , fᴵ(3) = 2(3 − 2) = 2(1) = 2 (Positive) Interpretation:

The function is decreasing at x = 1, stationary at x = 2 and increasing at x = 3.

Example 12.4

Find the gradient of x²+ y²= 9 at the point where x = 1

Solution

Using implicit differentiation gives:

2x + 2y dy___ dx = 0 2y dy__ dx = − 2x , then divide through by 2y dy__ dx = − x__, this is the gradient function At x = 1 , we substitute the value of x = 1 into the original equation to find the value of y at this point:

( 1)²+ y²= 9 y²= 8 y = ± 2 √2 Substituting into the gradient function, at x = 1 and y = ± 2 √2 − 1/2√_ 2 = ± √__ 2/4

Worked Example 12.5

Find the coordinates of the point on the curve y = 2/x²at which its gradient is 1/2.

Solution

y = 2/x²= 2x⁻²dy_ dx = − 4x⁻³= −4_ x³When dy__ dx = 1/2 − 4_ x³= 1/2 x³= − 8 ∴ x = − 2 When x = − 2 y = 2/x²= 2_____ (− 2)²= 2/4 = 1/2 Therefore, the gradient is 1/2 at the point ) − 2, ¹_ ₂ (

A stationary point also known as a turning point is a point on the graph of a function where the gradient (slope) is zero or dy__ dx = 0. At this point, the function changes direction from increasing to decreasing, or vice versa, or is saddle point.

A saddle point is where the graph changes from convex to concave, or vice versa, but continues increasing or decreasing either side of the saddle. Therefore, a stationary point can be classified into three types: local maximum (increasing to decreasing), local minimum (decreasing to increasing) and saddle point (or point of inflection) where it continues increasing/decreasing either side of the saddle point.

Types of stationary point

1. Local maximum:

a. At a local maximum, the function changes direction from increasing to decreasing.

b. The derivative fᴵ(x) changes from positive to negative.

c. Graphically, it looks like a peak. Such as in Figure 1.

Figure 12.1: Mountain Odwoanoma in Kwahu-Ghana

2. Local minimum:

a. At a local minimum, the function changes direction from decreasing to increasing

b. The derivative fᴵ(x) changes from negative to positive.

c. Graphically, it looks like a valley.

3. Saddle point (Point of inflection)

a. At a saddle point, the function changes concavity, but not necessarily direction.

b. The derivative fᴵ(x) is zero, but it does not change sign

c. Graphically, it looks like a point where the curve flattens out momentarily.

Identifying Stationary Points

1. Find the first derivative: calculate the first derivative fᴵ(x) of the function.

2. Set the first derivative to zero and solve the equation fᴵ(x) = 0 to find the x

-coordinates of the stationary points.

3. To classify the stationary points, use the second derivative test.

a. Calculate the second derivative fᴵᴵ(x).

b. If fᴵᴵ(x) > 0 at a stationary point, it is a local minimum.

c. If fᴵᴵ(x) < 0 at a stationary point, it is a local maximum.

d. If fᴵᴵ(x) = 0 , the test is inconclusive and the point might be a saddle point.

Example 12.6

Find the coordinates of the stationary point(s) of f(x) = x³− 3x²+ 2.

Determine the nature of the stationary points.

Solution

f(x) = x³− 3x²+ 2 fᴵ(x) = 3x²− 6x Set the first derivative to zero:

3x²− 6x = 0 3x(x − 2) = 0 ∴ x = 0 or x = 2 When x = 0 f(0) = (0)³− 3(0)²+ 2 = 2 The stationary point is (0, 2( f(2) = (2)³− 3(2)²+ 2 = − 2 The stationary point is (2, − 2) Classify the nature of the stationary points:

fᴵᴵ(x) = 6x − 6 At x = 0 fᴵᴵ(0) = 6(0) − 6 = − 6 if fᴵᴵ(x) < 0 ⟹ Maximum Therefore, this is a local maximum.

At x = 2 if fᴵᴵ(x) > 0 ⟹ Minimum Therefore, this is a local minimum.

Example 12.7

Determine the coordinates and nature of the turning point of the function: y = −3x²+ 18x − 20

Solution

y = −3x²+ 18x − 20 dy_ dx = − 6x + 18 At the stationary point dy__ dx = 0 0 = − 6x + 18 x = 3 When x = 3 y = −3(3)²+ 18(3) − 20 = 7 The stationary point is (3, 7) fᴵᴵ(x) = − 6 < 0 ⟹ maximum Hence (3, 7) is maximum

Example 12.8

Determine the coordinates and the nature of the turning points of the function:

y = 2 x³− 3 x²− 12x + 18

Solution

y = 2 x³− 3 x²− 12x + 18 dy__ dx = 6 x²− 6x − 12 At the stationary points dy__ dx = 0 0 = 6 x²− 6x − 12 0 = x²− x − 2 0 = )x + 1()x − 2( x = − 1 or x = 2 When x = − 1 y = 2) − 1(³− 3) − 1(²− 12) − 1( + 18 = 25 When x = 2 y = 2) 2(³− 3) 2(²− 12)2( + 18 = − 2 Hence the stationary points are (−1, 25) and (2, − 2) To determine the nature of the turning points we find the second derivative:

fᴵᴵ(x) = d 2 y/d x²= 12x − 6 At x = − 1 d²y_ d x²= 12(− 1) − 6 = − 18 < 0 Hence ) − 1, 25( is a maximum point.

At x = 2 d²y/d x²= 12(2) − 6 = 18 > 0 Hence (2, − 2) is a minimum point

Example 12.9

Determine the coordinates and the nature of the turning points of the function:

y = 4 x³− 9 x²+ 6x − 2

Solution

y = 4 x³− 9 x²+ 6x − 2 dy_ dx = 12 x²− 18x + 6 At the stationary point dy__ dx = 0 0 = 12 x²− 18x + 6 0 = 2 x²− 3x + 1 0 = (2x − 1)(x − 1) x = 1/2 or x = 1 When x = 1/2 y = 4 (1/2)³− 9 (1_ 2) 2 + 6(1_

2) − 2 = − 3/4 When x = 1 y = 4 (1)³− 9 (1)²+ 6(1) − 2 = − 1 Hence the stationary points are (1/2, −3/4) and (1, − 1) To determine the nature of the turning points we find the second derivative:

d²y/d x²= 24x − 18 When x = 1/2 d²y_ d x²= 24(1_

2) − 18 = − 6 < 0 Hence (1/2, −3/4) is a maximum point When x = 1 d²y/d x²= 24(1) − 18 = 6 > 0 Hence (1 − 1) is a minimum point

Example 12.10

Determine the coordinates and the nature of the turning points of the function:

y = x³− 6 x²+ 12x − 5

Solution

y = x³− 6 x²+ 12x − 5 dy_ dx = 3x²− 12x + 12 At stationary point dy__ dx = 0 0 = 3x²− 12x + 12 0 = x²− 4x + 4 0 = )x − 2()x − 2( Which gives x = 2 When x = 2 y = ) 2(³− 6) 2(²+ 12(2) − 5 = 3 Hence the stationary point is (2, 3) To determine the nature of the turning point, we find the second derivative:

d²y/d x²= 6x − 12 When x = 2 d²y/d x²= 6(2) − 12 = 0 Since d²y/d x²= 0 , it follows that the stationary point (2, 3) is a point of inflection as the function is decreasing either side.

Example 12.11

The function f(x) = ax²+ bx + c has a gradient function 4x + 2 and is stationary when y = 1 . Find the values of a, b and c .

Solution

f(x) = ax²+ bx + c dy_ dx = 2ax + b But dy__ dx = 4x + 2 ∴ 2ax + b = 4x + 2 Equating the gradient to the differentiated function Comparing corresponding terms:

2a = 4 ⟹ a = 2 and b = 2 The stationary value of f)x( occurs when dy_ dx = 0 4x + 2 = 0 x = − 1/2 The stationary value f(x) = ax²+ bx + c fx = 2( − 1/2)²+ 2(− 1/2) + c f(x) = − 1/2 + c But the stationary value is when y = 1 .

− 1/2 + c = 1 c = 3/2 Therefore, a = 2, b = 2, c = 3/2

A polynomial function is a function that involves only positive integer exponents of a variable in an equation like the quadratic equation, cubic equation, etc. We will explore how to sketch such polynomials and the points to consider.

Follow these steps to sketch a curve.

1. Find the intercepts on the x and y axes.

For the intercept on the x− axis, put y = 0 and solve for x For the intercept on the y− axis, put x = 0 and solve for y

2. Find the turning point(s). At turning point put dy__ dx = 0 and solve for x . Substitute the value of x i nto the original equation y = f(x), to find the corresponding y− coordinate values. This establishes the coordinates of the stationary points.

3. Test for maximum and minimum, use the following conditions:

a. If d 2 y _ d x 2 > 0 )i.e., positive) the turning point is minimum)

b. If d 2 y _ d x 2 < 0 )i.e., negative) the turning point is maximum)

Example 12.12

Sketch the curve y = x³− 7x²+ 15x − 9 indicating clearly its points of intersection on the axes and its turning points.

Solution

The equation of the curve is y = x³− 7x²+ 15x − 9 For the intercept on the y− axis, put x = 0 y = − 9 The point (0, − 9) is the intercept on the y− axis For the intercept on the x− axis, put y = 0 x³− 7x²+ 15x − 9 = 0 By factorising the polynomial x³− 7x²+ 15x − 9 = 0 )x − 1()x − 3 (²= 0 x = 1 or 3 (repeated) The intercepts on the x− axis is (1, 0) and (3, 0).

Note, as (3, 0) is a repeated root this means the curve just touches the x-axis at this point, but does not cut through it.

y = x³− 7x²+ 15x − 9 dy_ dx = 3x²− 14x + 15 At stationary points dy__ dx = 0 3x²− 14x + 15 = 0 By solving x = 5/3 or 3 We investigate which of these points gives maximum or minimum points by finding the second derivative:

dy_ dx = 3x²− 14x + 15 d²y_ d x²= 6x − 14 At x = 3 d²y/d x²= 6(3) − 14 = 4 > 0 (Positive = local minimum) To find the corresponding y value substitute x = 3 into the expression for y.

When x = 3, y = 0 . Hence (3, 0) is the minimum turning point.

At x = 5/3 d²y/d x²= 6(5/3) − 14 = − 4 < 0 (Negative = local maximum) To find the corresponding y value substitute x = ⁵_ ₃ into the expression for y y = (5/3)³− 7(5_ 3) 2 + 15(5_

3) − 9 = 32/27 When x = 5/3, y = 32/27. Hence ) 5/3, 32/7 ( is the maximum turning point.

Sketch the graph with points (1, 0), (3, 0), (1.67, 1.19) and (0, -9)

Figure 12.2: Graphical illustration of y = x³− 7x²+ 15x − 9

Example 12.13

Sketch the curve y = x²− 4x + 3 indicating all its turning points and intercepts.

Solution

y = x²− 4x + 3 dy_ dx = 2x − 4 At turning point dy__ dx = 0 0 = 2x − 4 x = 2 Substituting x = 2 into the function to find y , we have y = (2)²− 4(2) + 3 = − 1 Hence, the turning point is (2, − 1) To determine the nature of the turning point we find the second derivative dy_ dx = 2x − 4 d²y/d x²= 2 > 0 , ∴ Positive minimum point and occurs at (2, − 1)

Figure. 12.3: Graphical illustration of y = x²− 4x + 3 For the y−intercept, put x = 0 y = (0)²− 4(0) + 3 = 3 For the x−intercept, put y = 0 0 = x²− 4x + 3 (x − 1)(x − 3) = 0 x = 1 and x = 3 Sketch the graph with points (1, 0), (3, 0), (0, 3) and (2, -1)

Example 12.14

Sketch the curve y = x ³− 9 x ²+ 15x − 7 indicating all its turning points and intercepts

Solution

y = x³− 9 x²+ 15x − 7 dy_ dx = 3x²− 18x + 15 At turning point dy__ dx = 0 0 = 3x²− 18x + 15 x²− 6x + 5 = 0 (x − 5)(x − 1) = 0 x = 5 and x = 1 Substituting x = 5 and x = 1 into the function to find y , we have y = ( 5)³− 9 (5)²+ 15(5) − 7 = − 32 y = (1)³− 9 (1)²+ 15(1) − 7 = 0 The turning points are (5, − 32) and (1,0) To determine the nature of the turning point we find the second derivative dy_ dx = 3x²− 18x + 15 d²y_ d x²= 6x − 18 At x = 5 d²y/d x²= 6(5) − 18 = 12 > 0 , Positive = minimum point and occurs at (5, − 32) At x = 1 d²y/d x²= 6(1) − 18 = − 12 < 0 , Positive = maximum point and occurs at (1,0) For the y−intercept, put x = 0 y = (0)³− 9 (0)²+ 15(x) − 7 = − 7 For the x−intercept, put y = 0 0 = x³− 9 x²+ 15x − 7 By the factor theorem:

f)x( = x³− 9 x²+ 15x − 7 f (1) = (1)³− 9(1)²+ 15(1) − 7 = 0 ⟹ (x − 1) is a factor of f(x) By the division of Polynomial x²− 8x + 7 x − 1 √______________ x³− 9 x²+ 15x − 7 x³− x²−8x²+ 15x −8x²+ 8x 7x − 7 7x − 7 0 0 f(x) = (x − 1)(x²− 8x + 7) = 0 Factorising the trinomial (x²− 8x + 7) = 0 :

(x − 1)(x − 1)(x − 7) = 0 Intercepts on the x− axis are x = 1 and x = 7 Sketch the graph with points (0, -7), (1, 0), (5, -32) and (7,0).

Figure 12.4: Graphical illustration of y = x³− 9 x²+ 15x − 7

Understanding the nature of gradients, investigating turning points and sketching polynomial functions is critical for solving real-life problems due to their ability to model and optimise real-world scenarios. Gradients help determine rates of change, essential for analysing phenomena such as speed in physics, profitability in economics and environmental changes in engineering. Turning points identify maxima or minima, which are vital in optimising resources, such as maximising agricultural yields or minimising production costs. Sketching polynomial functions provides a visual understanding of relationships and trends, aiding in decision-making, like predicting investment outcomes or modelling population growth. We will apply these concepts by solving in real-life problems.

Activity 12.2: Project on real life application of differentiation techniques

1. Working in pairs, or small groups, choose from one of the projects below:

a. Investigate how the time taken to walk from your dormitory to the dining hall varies depending on your walking speed. Use a stopwatch to record times for different walking speeds and calculate the rate of change of distance with respect to time.

b. Choose a fixed distance, use formula for speed = distance______ time

c. A shopkeeper in your school wants to determine the price of sachet water that will result in the highest sales revenue. Use basic mathematics to model and determine the price that maximises revenue using hypothetical data.

d. Note: Revenue = price × quantity

e. Observe the motion of a ball thrown vertically upward on the school field. Record the time it takes to reach its highest point and use this data to calculate the maximum height using basic physics and a quadratic model.

f. Note: Use motion under gravity formula

g. i.e., height(t) = Initial velocity - 1/2 (acceleration due to gravity) (time)²h. Investigate how the cost of printing exercise books changes as the number of books printed increases. Collect data to model this relationship and determine the number of books that minimises the cost while balancing production efficiency.

2. After choosing your project, define the variables involved in the problem.

3. Express the relationships between these variables using equations.

4. Apply the appropriate differentiation technique to the project of choice.

5. Interpret the results obtained.

6. Create a visual using graphs (manually or by software).

7. Write a clear, concise report summarising:

a. The real-life problem you chose.

b. The mathematical formulation of the problem.

c. The differentiation techniques used to solve the problem.

d. The interpretation of the results.

e. Graphs that visualise the solution.

f. Ensure that your report includes both the mathematical steps and the practical implications of your solution.

Let us go through some examples of how differentiation techniques are applied in real life.

Example 12.15

A rectangular cake dish is made by cutting out squares from the corners of a 12.5cm by 20cm rectangle of tin-plate and then folding the metal to form the container. What size squares must be cut out to produce the cake dish of maximum volume?

Solution

Step 1: Let m be the side lengths of the squares that are cut out.

Step 2: Determine the volume Volume = length × width × height = (20 − 2m)(12.5 − 2m)m = (250 − 40m − 25m + 4 m²)m = 250m − 65 m²+ 4 m³And 0 < m < 10

Step 3: Differentiate the volume dV_ dm = 12 m²− 130m + 250

Step 4: Equate the differentiated volume to zero and simplify 12 m²− 130m + 250 = 0 ∴ m = 25/3 , m = 5/2

Step 5: Find the second derivative to confirm which is the local maximum and which the local minimum. This is necessary as both values lie between 0 and 10.

d²V_ d m²= 24m − 130 When m = 25/3 , d 2 V/d m²= 70 ∴ a local minimum When m = ⁵_ ₂, ᵈ²ⱽ_ _(d) ₘ ₂ = − 70 ∴ a local maximum and the required value to maximise the volume of the cake tin.

Therefore, the maximum volume is obtained when 2.5cm squares are cut from the corners of the rectangle. (The minimum volume would be obtained when 8.33cm squares are cut from the corners of the rectangle.)

Example 12.16

A cocoa farmer in Ghana models the annual yield of cocoa (in tons) as a function of the amount of fertiliser applied, y = − 0.5 x²+ 10x + 20 where x is the amount of fertiliser (in kg).

Find the amount of fertiliser needed to maximise the yield and the maximum yield this gives.

Solution

Step 1: Differentiate y with respect to x dy_ dx = − x + 10

Step 2: To determine the value of x at turning point, equate dy__ dx to zero.

− x + 10 = 0 x = 10

Step 3: Substitute x = 10 into the original function.

y = − 0.5 (10)²+ 10(10) + 20 y = 70 tons The amount of fertiliser needed to maximise the yield is 10kg to give a maximum yield of 70 tons.

Figure 12.5: Graphical illustration of maximum cocoa yield

Example 12.17

A logistics company transports goods from Accra to Kumasi.

The cost function, C)x( = x ²− 8x + 1000, represents the cost in Ghana cedis per trip, where x is the number of trips made in a month.

Determine the number of trips to minimise the cost and find the minimum cost.

Solution

Step 1: Let C(x ) = y . Differentiate y with respect to x dy_ dx = 2x − 8

Step 2: To determine the value of x at turning point, equate dy__ dx to zero.

2x − 8 = 0 2x = 8 x = 4

Step 3: Substitute x = 4 into the original function.

C(4) = 4²− 8(4) + 1000 = 984 The minimum cost is GH¢ 984.00 and four trips are required.

Figure 12.5: Graphical illustration of cost of transporting logistics

Example 12.18

A rice farmer in South Tongu uses water to irrigate fields.

The productivity P)w( = − 2 w²+ 40w, where w is the water used in cubic meters per day, models the rice yield in kg. How much water should be used to maximise yield, and what is the maximum yield?

Solution

Step 1: Let P(w ) = v . Differentiate v with respect to w dv_ dw = − 4w + 40

Step 2: To determine the value of w at turning point, equate dv___ dw to zero.

− 4w + 40 = 0 − 4w = − 40 w = 10

Step 3: Substitute w = 10 into the original function.

P(10 ) = −2 (10)²+ 40(10) = 200 The farmer should use 10m³of water per day to irrigate the farm which will yield a maximum of 200kg of rice.

Figure 12.6: Graphical illustration of rice yield

Example 12.19

A market seller in Makola Market determines that the revenue from selling fruits is modelled using R(p ) = −3p²+ 30 p + 200 where p is the price per kg (in cedis).

a. What price maximises the revenue?

b. What is the maximum revenue?

Solution

Step 1: Let R(p ) = w . Differentiate w with respect to p dw_ dp = − 6p + 30

Step 2: To determine the value of p at turning point, equate ᵈʷ_ _(dp) to zero.

− 6p + 30 = 0 − 6p = − 30 p = 5

Step 3: Substitute p = 5 into the original function.

R(5) = − 3 (5)²+ 30(5)+ 200 = 275

a. A price of GH¢ 5.00 will maximise revenue.

b. The maximum revenue will be GH¢ 275.00.

Figure 12.7: Graphical illustration of Revenue

Example 12.20

A storage company designs cylindrical containers for shipping hazardous materials. Regulations require that the total height plus the circumference of the base must not exceed 96 inches.

Determine the dimensions of the cylindrical container that has the greatest possible volume while complying with these regulations.

Solution

Let h represent the height of the cylinder and r be the radius circular base

Step 1: Determine the constraints h + 2πr ≤ 96 , maximum will occur when h + 2πr = 96 ∴ h = 96 − 2πr

Step 2: State the volume formula for a cylinder V = π r²h

Step 3: Substitute the expression for h and simplify V = π r²( 96 − 2πr ) = 96 π r²− 2 π²r³Step 4: Differentiate V with respect to r dV_ dr = 192πr − 6 π²r²Step 5: Equate dV___ dr to zero and make r the subject.

192πr − 6 π²r²= 0 6πr (32 − πr) = 0 r = 32___ π

Step 6: Substitute r = ³²__(π) into V V = 96 π (³²_ π )²− 2 π²(³²_ π )³V = 32³___ π

Step 7: Calculate the value of the height h = 96 − 2π(³²__ π ) = 32 inches

Step 8: Determine the value of the circumference for the circular base 2πr = 2π(³²__ π ) = 64 inches The dimensions to yield a maximum volume are when the height is 32inches, circumference of 64inches and a radius of 32__ π inches.

Figure 12.8: Graphical illustration of packing volume

1. Find the coordinates of the point(s) on the given the curve at which its gradient has the given value.

a. y = )x + 3()x − 5( gradient = 0

b. y = 5 + 3x − 2x²gradient = − 3

c. y = x + 1/x gradient = 2

d. y = √x gradient = 2

2. Sketch the curve y = x³− 7 x²+ 15x − 9 , indicating clearly its points of intersection with the axes and its turning points.

3. The curve y = a x²+ bx + c passes through the point (1, 0) , y has a minimum value of −9/4 . when x = −1/2 ,find the values of a, b and c.

4. Find the gradient of the curve at the point (2, 1) on the curve x²y − 2x y²+ y²= 1 .

5. Given the implicit equation xy = 1 . Find the value of dy__ dx at the point (1, 1).

6. A curve, the gradient of which at any point is − 2x , passes through the point (0, 1) . Find the equation of the curve and sketch its graph, showing clearly the points where it cuts the x− axis.

7. The curve y = x²− ax + b has turning point at (1, 3) . Find the values of a and b .

8. Find the coordinate of the point to the curve y = x²+ x which has a gradient of − 1 .

9. When a manufacturer makes x items per day:

the cost function is C(x ) = 720 + 4 x + 0.02x²Ghana Cedis:

the price function is p(x ) = 15 – 0.002 x GH¢ per item.

Find the production level that will maximise profits.

10. A 25-foot ladder rests against a vertical wall. If the bottom of the ladder is sliding away from the base of the wall at the rate of 3 ft / s , how fast is the top of the ladder moving down the wall when the bottom of the ladder is 7 ft from the base?

11. Water is running out of a conical funnel at the rate of 1 i n³/ s. If the radius of the base of the funnel is 4 in and the height is 8 in , find the rate at which the water level is dropping when it is 2 in from the top.

(The formula for the volume, V , of a cone is ¹_ ₃ π r²h , where r is the radius of the base and h is the height.)

12. If a stone is thrown vertically upward with a velocity of 80ft / s , then its height after t seconds is s = 80t − 16 t².

a. What is the maximum height reached by the ball?

b. What is the velocity of the ball when it is 96ft above the ground on its way up and again on its way down?

13. Sketch the graph of y = x³− 3x indicating all the necessary points.