Maxima And Minima Problems
Mathematics is like checkers in being suitable for the young, not too difficult, amusing, and without peril to the state. (Plato)
Key concepts from curve sketching
The gradient = The derivative of a curve of the related function at a point 𝑚𝑚 𝑑𝑑𝑑𝑑 evaluated at the point 𝑑𝑑𝑑𝑑 Gradient
> 0 > 0 < 0 < 0 leans′ right leans′ right leans′ left leans′ left increasing𝑦𝑦 increasing𝑦𝑦 decreasing𝑦𝑦 decreasing𝑦𝑦 Concavity < 0 > 0 > 0 < 0 concave′′ down concave′′ up concave′′ up concave′′ down 𝑦𝑦 𝑦𝑦 Stationary point 𝑦𝑦= 𝑦𝑦 A point on the curve where the tangent′ is parallel to the -axis the gradient is𝒚𝒚 zero𝟎𝟎 Nature of stationary points 𝑥𝑥 (Relative) (Relative)↔ Maximum turning Minimum turning Horizontal point of inflexion point point
= 0 = 0 = 0 = 0 ′ What tells us′ the nature of the stationary′ points? ′ 𝑦𝑦 These are𝑦𝑦 the tests to justify 𝑦𝑦your answers: 𝑦𝑦 < 0 > 0 = 0 = 0 ′′ ′′ concavity′′ changes about the′′ point 𝑦𝑦 𝑦𝑦 𝑦𝑦 𝑦𝑦 Before Before Ist derivative test < 0 > 0 After′′ After′′ 𝑦𝑦 > 0 𝑦𝑦 < 0 Note:′′ the gradient hasn’t ′′changed Before Before 𝑦𝑦 𝑦𝑦 > 0 < 0 Before Before After′ After′ > 0 < 0 𝑦𝑦 < 0 𝑦𝑦 > 0 After′ After′ ′ ′ 𝑦𝑦 > 0 𝑦𝑦 < 0 𝑦𝑦 𝑦𝑦 ′ ′ 𝑦𝑦 𝑦𝑦 Prepared by Sue Millet for HSC Revision Day UOW Maxima and Minima Problems o Read the question, and annotate. (drawing any given diagrams as you read the question can help to improve understanding of the situation.) o Identify the variables (and any constants that are pretending to be variables.) o Focus on two variables –the quantity ( ) you want to maximise/minimise and one other variable ( ) -the question will help you to choose these variables. 𝑄𝑄 𝑥𝑥 o Use the given information to reduce the variables -to eliminate a variable make it the subject of an equation then substitute for it. o Write an equation for the quantity you want to maximise/minimise in terms of your second variable ( = ( )). The question will usually tell you this equation. 𝑄𝑄 𝑓𝑓 𝑥𝑥 o Simplify the equation, if this is possible and sensible. o Find the derivative ( ) 𝑑𝑑𝑑𝑑 Find the stationary point value ( = 0 ) o 𝑑𝑑𝑑𝑑 Test the stationary point value to establish the maximum/minimum𝑑𝑑𝑑𝑑 using o 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓𝑓𝑓 𝑥𝑥 , if reasonable 2 𝑑𝑑 𝑄𝑄 2 sign𝑑𝑑𝑥𝑥 of on each side of your solution, if the first derivative is too complex𝑑𝑑𝑑𝑑 eg you already used the quotient rule 𝑑𝑑𝑑𝑑 o Read the question. o Answer the question. o Throughout the process use any given answers to work towards all or part of the solution.
Curve Sketching
Read the question, annotate and follow the directions in each part. These will generally require you to: o Find the derivative ( ) 𝑑𝑑𝑑𝑑 o Find the stationary points ( = 0 , 𝑑𝑑𝑑𝑑 ( )) 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑦𝑦′ 𝑎𝑎𝑎𝑎𝑎𝑎 o Test the stationary points to establish their nature (maximum turning point/minimum𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓𝑓𝑓 𝑥𝑥 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 turning𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇 𝒚𝒚 point/𝑜𝑜𝑜𝑜 𝑖𝑖𝑖𝑖 𝑚𝑚 horizontal𝑚𝑚𝑚𝑚 𝑏𝑏𝑏𝑏 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 point𝑓𝑓 of𝑥𝑥 inflexion using , if is reasonable sign of on each side of your solution, if the first derivative is too 𝑦𝑦complex′′ eg you already used the quotient rule 𝑦𝑦′ o Find intercepts -definitely = 0, if examiners want = 0 they’ll usually ask. 𝑦𝑦 𝑤𝑤ℎ𝑒𝑒𝑒𝑒 𝑥𝑥 o Find points of inflexion = 0 if required. You must𝑥𝑥 𝑤𝑤testℎ𝑒𝑒𝑒𝑒 ′′𝑦𝑦 either side of the point to show concavity changes 𝑦𝑦 ′′ 𝑦𝑦 o Determine endpoints if given a domain. o Read the question. o Answer the question. o Remember the maximum/minimum value may be an endpoint value
Related skills
o algebraic manipulation including confidence with expanding, factorising quadratics, indices, fractions, fractional indices,… o (Use of reference sheet for) differentiation, including application of the product, quotient and function of a function or chain rules
uv’+vu’
vu’-uv’ v2
derivative outer ×derivative inner
o these questions are often at the end of the exam targeting Band 6 but can also occur earlier and with good exam technique you should always be able to access some if not all of these marks
Have we learnt anything else about maximum and minimum values?
The minimum value of a perfect square number/ expression is zero.
Example: What is the range of = + 7 or = ( 4) + 7? 2 2 Since the square term𝑦𝑦 is always𝑥𝑥 non𝑦𝑦- 𝑥𝑥 − 7 for both functions. In an exam I’d say negative (≥0) 𝑦𝑦 ≥ 0 for all = + 7 0 + 7 = 7 ( 2 4) 0 for all 2 = ( 4) + 7 0 + 7 = 7 𝑥𝑥 ≥ 2 𝑥𝑥 ∴ 𝑦𝑦 𝑥𝑥 ≥ 2 This𝑥𝑥 − is called≥ justify𝑥𝑥ing∴ 𝑦𝑦your 𝑥𝑥answer.− ≥
Have we learnt anything else about maximum and minimum values?
The square root sign indicates only the positive square root
Example: What is the range of = + 7
Since the square root𝑦𝑦 term√𝑥𝑥 is non- 7. In an exam I’d say negative (≥0) 𝑦𝑦 ≥ 0 for all = + 7 0 + 7 = 7
Another√𝑥𝑥 ≥ justify𝑥𝑥 your∴ 𝑦𝑦 answer.√𝑥𝑥 ≥
So what about ? What is the range of = 7
0 for all 𝑦𝑦= 7 − √𝑥𝑥 7 + 0 = 7
Have−√𝑥𝑥 ≤ we learnt𝑥𝑥 anything∴ 𝑦𝑦 − else√𝑥𝑥 about≤ maximum and minimum values?
The sine and cosine functions have maximum and minimum values
1 1 and 1 1
− ≤ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ≤ − ≤ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ≤ Example: What is the maximum value of 4 5 ? 1 1 5 5 5 notice the sign reversal −when𝑠𝑠𝑠𝑠𝑠𝑠 you𝑠𝑠 multiply by a negative. 9− ≤4 𝑠𝑠𝑠𝑠𝑠𝑠5𝑠𝑠 ≤ 1 1≥ −4𝑠𝑠𝑠𝑠𝑠𝑠5𝑠𝑠 ≥ − 9 The≥ maximum− 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ≥value− is 9 − ≤ − 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ≤
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