UNIT 1:

ANALYTICAL METHODS FOR ENGINEERS

Unit code:

A/601/1401

QCF Level:

4

Credit value:

15 OUTCOME 3 - CALCULUS TUTORIAL 2 – MAXIMA AND MINIMA

The calculus: the concept of the limit and continuity; definition of the derivative; derivatives of standard functions; notion of the derivative and rates of change; differentiation of functions using the product, quotient and function of a function rules; integral calculus as the calculation of area and the inverse of differentiation; the indefinite integral and the constant of integration; standard integrals and the application of algebraic and trigonometric functions for their solution; the definite integral and area under curves Further differentiation: second order and higher derivatives; logarithmic differentiation; differentiation of inverse trigonometric functions; differential coefficients of inverse hyperbolic functions Further integration: integration by parts; integration by substitution; integration using partial fractions Applications of the calculus: e.g. maxima and minima; points of inflexion; rates of change of temperature; distance and time; electrical capacitance; rms values; electrical circuit analysis; ac theory; electromagnetic fields; velocity and acceleration problems; complex stress and strain; engineering structures; simple harmonic motion; centroids; volumes of solids of revolution; second moments of area; moments of inertia; rules of Pappus; radius of gyration; thermodynamic work and heat energy Engineering problems: e.g. stress and strain; torsion; motion; dynamic systems; oscillating systems; force systems; heat energy and thermodynamic systems; fluid flow; ac theory; electrical signals; information systems; transmission systems; electrical machines; electronics

This tutorial devoted to maxima and minima should be studied if you have not covered this work previously. You should judge your progress by completing the self assessment exercises. On completion of this tutorial you should be able to do the following.      

Define a maximum and minimum point of a function. Define a turning point. Define a point of Inflection. Use differential calculus to determine these points. Use differential calculus to determine if a point is a maximum or minimum. Solve practical problems by applying the theory.

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MAXIMA AND MINIMA 1.

INTRODUCTION

There are many engineering problems where a value peaks or dips. For example the power of an engine under constant throttle conditions peaks at a certain speed. Another example is a cylindrical canister to contain a certain volume. The amount of material needed can be minimised by choosing the correct diameter and length. This tutorial is about finding the critical values that make a function reach a maximum or minimum value and differential calculus is the way to do it. 2. MAXIMUM POINT Consider the function y = 10x – 2x2 Plotting y against x from x = 0 to x = 5 gives the following graph.

Figure 1 From the graph we can clearly see that there is a peak at point A and that this is the maximum value of y. The question is, what is the value of x and y at this maximum point? At point A the gradient is horizontal and hence zero so if we equate dy/dx to zero we can find the point. Here is the method. y = 10x – 2x2 dy/dx = 10 – 4x

Equate to zero 10 – 4x = 0

4x = 10

x = 10/4 = 2.5 and hence y = 10 x 2.5 – 2 x 2.52 = 25 – 12.5 = 12.5 This result is confirmed by the graph.

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WORKED EXAMPLE No.1 What is the largest area of a rectangular field that can be enclosed by a fence with a perimeter of 200 m? SOLUTION Let the rectangle be x m long and y m wide. The area enclosed is A = x y………….. (1) The perimeter is 2x + 2y = 200 m ……. (2) We must eliminate one variable, say y, from equation (1) by substitution from equation 2. 2y = 200 - 2x

y = 100 – x ………... (3)

Substitute (3) into (1)

A = x (100 – x) = 100x – x2

Plotting A against x reveals that the maximum area is 2500 m 2 when x = 50 m.

Figure 2 Prove this by differentiation. A  100x  x 2 dA  100  2x dx For a max or min dA/dx = 0

hence 100 – 2x = 0

100 = 2x

x = 100/2 = 50 m A=100x – x2 A = 100.50 – 502 = 5000 – 2500 = 2500 mm2

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3. Minimum point Consider the function y = x2 - 4x Plotting y against x from x = 0 to x = 4 gives the following graph.

Figure 3 From the graph we can clearly see that there is a minimum value of y at B. The question is, what is the value of x and y at this minimum point? At point B the gradient is horizontal and hence zero. If we equate dy/dx to zero we can find the point. dy/dx = 2x – 4 Equate to zero 2x – 4 = 0

4 = 2x

x = 4/2 =2

Substitute x = 2 into the function to find the value of y. y = x2 - 4x = 22 – 8 = -4 This result is confirmed by the graph.

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SELF ASSESSMENT EXERCISE No.1

1. Determine the maximum area and the lengths of the sides of a rectangle that can be enclosed by a rectangular perimeter of 8 m length. (Answer 2 m x 2 m and 4 m2) 2. The velocity of a missile is related to time by the equation v = t2 – 4t. Calculate the time at which the velocity is a minimum and determine this minimum. (Answers 2 and – 4) 3. The current in a device is related to time by the equation i = 4t 2 – 8t. Find the time at which the current the minimum and determine this minimum. (Answers 1 and – 4) 4. Find the value of u that makes q a maximum when they are related by the equation q = 10u – 5u2 + 2. Find this maximum value of q. (Answers u = 1 q = 7)

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4. TURNING POINTS and POINTS of INFLECTION Consider the function y = x3 – 5x2 +5x + 2. The graph for x = 0 to x = 4 is shown below.

Figure 4 At point A the graph changes from up to down and at B it changes from down to up. They look like a maximum and minimum point but strictly speaking they are not as the value of y exceeds this value later on (and is smaller for negative values of y). Such points are called turning points and they may be found in the same way as a max or min point. Examining the graph we see the turning points occur at about x = 0.6 and 2.7 but we need to use calculus to find them precisely. The important thing to note is that at A and B the gradient of the graph is horizontal so the value of dy/dx must be zero. This enables us to find the value of x and y at these points. Here is how to do it. dy y  x 3  5x 2  5x  2  3x 2  10x  5 dx At the turning points dy/dx is zero so equate to zero as follows. dy  3x 2  10x  5 0  3x 2  10x  5 dx This is a quadratic equation and we must solve it to find the two values of x. QUADRATIC EQUATION

ax 2  bx  c  0  b  b 2  4ac x 2a In this case a = 3, b = -10 and c = 5 so solving we get

x

 b  b 2  4ac 2a

 (10)  (10) 2  4.3.5 10  100  60 10  40 x   2.3 6 6

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There are two possible solutions because all positive numbers have a positive and a negative square root. 40 = 6.324 10  40 10  6.324 x  6 6 10  6.324 16.324 10  6.324 3.676 x   2.721 or x    0.613 6 6 6 6 Hence the turning points occur at x =2.721 and 0.613. The graph tells us which is A and B. Note that without the graph you could not be sure which is the maximum and which a minimum. We need further studies to find out how to do this. POINT OF INFLECTION A point of inflection is where the tangent at a point crosses the graph and it can be shown that at these points the second derivative is zero. For the equation y = x3 – 5x2 +5x + 2 plotted on figure 4, there is such a point at C. This may be determined from the second derivative. d2y  6x  10  0 x  10/6  1.667 dx 2 An important point about this is that you can have a point of inflection where the gradient is zero and so they could be confused with a maximum or minimum point.

SELF ASSESSMENT EXERCISE No.2 1. Find the values of x where the maximum and minimum occur for the following function. Also find the point(s) of inflection. y = x3 - 6x2 + 9x + 10 Sketch the graph and determine which is the maximum and which the minimum points. (Answers x = 1 gives a max and x = 3 is a min and x = 2 is a point of inflection) 2. Find the values of x where y is a maximum or minimum for the following function. Also find the point(s) of inflection. y = x3 - 5x2 - 8x (Answers x = 4 gives a min and x = -0.667 is a max x = 1.6667 is a point of inflection)

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WORKED EXAMPLE No.2 A cylindrical vessel is to be made from thin metal plate to contain 15 litres (0.015 m 3) of liquid. Find the dimensions of the cylinder that make the surface area a minimum including the two ends. SOLUTION We need to set up two equations, one for the volume and one for the surface area. Volume = A L = D2L/4 0.015 = D2L/4 ………….. (1) Surface Area = A = wall + 2 ends (see diagram) A = DL +2 D2/4 = DL +D2/2…………. (2) Now we decide whether to first find L or D by max or min theory. Let us decide to find D. From (1) find L in terms of D 0.015 x 4 0.06 L  ...............(3) πD 2 πD 2 0.06 πD 2 Substitute this into (2) A  πD x  2 πD 2

πD 2 ................(4) 2 For a minimum area the differential coefficient dA/dD = 0. Differentiate with respect to D A  0.06D -1 

Figure 5

dA  0.06D  2  πD dD 0.06D  2  πD 0.01909  D

3

Equate to zero 0.06 D2

0.06  D3 π

 πD

1 D  (0.01909) 3

0  0.06D  2  πD

 0.2673m

We are not sure that this value gives a minimum value of Area without plotting the graph but assuming it is we can now find the value of L from (3). L = 0.267 m (i.e. equal to the diameter. Now we can find the area from (2) or (4). A = 0.3367 m2

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SELF ASSESSMENT EXERCISE No.3 1. A rectangular trough is to be made from a sheet of metal 5 m long and 4 m wide by cutting a square of side x m from each corner and turning up the ends and sides. Show that the volume is given by V = 4x3 - 18x2 + 20x. Find the value of x that makes the volume a maximum. (Answer 0.736 m)

2. The power transmitted by a pulley belt system is given by the following equation.

P  (Fv  ρAv2 )(1  e μθ ) P is the power v is the velocity of the belt.  is the density of the belt material A is the cross sectional area of the belt F is the maximum force in the belt  is the coefficient of friction  is the angle of contact

 = 1200 kg/m3 A = 800 x 10-6 m2 F = 500 N  = 0.3  = 1.2 radian

Determine the velocity that makes the power a maximum and calculate this power. (Answers v = 260.4 m/s and P = 7787 W)

3. The power developed by a Pelton Wheel water turbine is given by the following equation.

P  mu(v  u)(1  kcos θ) P is the power m is the mass flow rate u is the velocity of the buckets v is the velocity of the water jet k is the blade friction coefficient  is the angle of deflection

m = 40 kg/s v = 20 m/s k = 0.98  = 165o

Determine the velocity of the bucket that will make the power a maximum and calculate the maximum power. (Answers 10 m/s and 31.1 kW)

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5. DETERMINING IF A POINT IS A MAXIMA OR MINIMA Consider the equation V = 4x3 - 18x2 + 20x from question 1 previous. The equation is plotted below (in red) and as predicted the maximum point occurs at x = 0.736. Differentiating the equation we

dV  12x 2  36x  20 dx

This equation represents the gradient of the graph at any point x. If we plot this as well (in blue), we dV can see that as expected, the gradient is zero at the max and min points. dx

Figure 6 If we differentiate a second time we shall get an equation representing the gradient of the new graph, i.e. the gradient of the gradient. When we do this we write it as follows. d 2V  24x  36 dx 2 Plotting this graph (in green) produces a straight line and reveals that where the original function d 2V was a maximum, the value of is negative and where it is a minimum, the value is positive. dx 2 This is always the case and so we can determine if a point is a maximum or minimum by evaluating d 2V at that point. dx 2

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WORKED EXAMPLE No.3 Find the turning points of the following function and determine which is the maximum and which the minimum. y = 2x3 – 20x2 + 4x SOLUTION

Differentiate once dy  6x 2  40x  4 dx Equate to zero 6x 2  40x  4  0 Solve using the quadratic equation a  6 b  -40 c  4 2 - b  b 2  4ac 40  (-40)  (4)(6)(4) 40  1504   2a (2)(6) 12 40  38.781 x  6.565 or 0.102 12 Differentiate again.

x

d2y dx 2

 12x - 40

Evaluate when x  6.565 and

d2y

 38.78 dx 2 Since this is positive the point x  6.565 must be a minimum. Evaluate with x  0.102 and

d2y

 38.78 dx 2 Since this is negative, the point must be a maximum. The graphs of the functions show these answers are correct.

Figure 7

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SELF ASSESSMENT EXERCISE No.4 Find the turning points of the following functions and determine which is the maximum and which is the minimum point. 1.

y = 2x3 – 12x2 + 10x

(Answers 3.528 Min and 0.472 max) 2.

p = 4q3 – 20q2 + q +10

(Answers 3.308 Min and 0.025 max)

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