Applications of maxima and minima

Applications of maxima and minima

• 1.
  Nature of Points Applications Maxima, Minima, Point of Inflection
• 2.
  Recap We saw howto find the coordinates of a turning point: Differentiate Set f’(x) = 0 Solve to find x Substitute x into the original equation to find y. f’(x) or dy = 0 f’(x) or dy = 0 dx dx f’’(x) or d2y Positive f’’(x) or d2y Negative dx2 dx2
• 3.
  Sketching curves Find thestationary point(s):  Find an expression for and put it equal to 0, then solve the resulting equation to find the x co-ordinate(s) of the stationary point(s).  Find and substitute each value of x to find the kind of stationary point(s). (+ suggests a minimum, – a maximum, 0 could be either or a point of inflection)  Use the curve’s equation to find the y co-ordinate(s) of the stationary point(s). Find the point(s) where the curve meets the axes:  Substitute x = 0 in the curve’s equation to find the y co-ordinate of the point where the curve meets the y axis.  Substitute y = 0 in the curve’s equation. If possible, solve the equation to find the x co-ordinate(s) of the point(s) where the curve meets the x axis. Sketch the curve, then use a graphic calculator to check.
• 4.
  We can usethe calculator to check We can find the y values using graph mode In run mode F2 d/dx enter function , x value EXE will give original gradient OPTN F3 d/dx 2 enter function , x value F4 → calc d2y EXE will give dx 2
• 5.
  We can usethe calculator to check In graph mode F1 → Ycal Enter function F6 → Draw Enter x value EXE Shift F5 → G solv F6 →> Repeat for second x value
• 6.
  y = 2x 3 − 3 x 2 − 12 x − 15 y 10 – 10 10 x – 10 – 20 – 30 – 40 – 50
• 7.
  Rules of Howto Approach Applications of Maxima and Minima 1.If possible draw a diagram •Use given information to form an equation about the situation. You may have to use other variables. One equation is about the given situation you are asked about. One relates y and x. Watch out for Pythagoras statements about perimeter, area, volume. 3.Express y in terms of x (or x in terms of y whichever is simpler) -I.e. Express one variable in terms of the other 4.Differentiate and put equal to zero (you are finding the turning point of the graph either max or min) 5.Solve to find x 6.Re read the question 7. Put x back in original equations to find other variables as required by the question 8.Check that the value that you have found is in fact a 2 max or min as required. I.e. find d y 2 or f ′′( x) dx If it is <0, it is a max, If it is >0, it is a min
• 8.
  APPLICATION EXAMPLES WITHSOLUTIONS A box with an open top can be made from a square sheet of cardboard with sides 12cm, by cutting a square of side x from each corner, and folding along the dotted lines. x x 12 cm What is the maximum volume? 1
• 9.
  Sides are 12cm: What is the length / width of the box in terms of x ? L = 12 – 2x Volume of the box , V=L*L*x = (12 – 2x)2x = (144 – 24x – 24x + 4x2)x = (144 – 48x + 4x2)x =144x – 48x2 + 4x3 2
• 10.
  Volume =144x –48x2 + 4x3 Maximum volume - differentiate : dV = 144 – 96x + 12x2 dx At maximum : dV = 0 dx 144 – 96x + 12x2 = 0 12(12 – 8x + x2) = 0 12(x – 6)(x – 2) = 0 So x = 6 or x = 2 3
• 11.
  Find the natureof the turning Differentiate again : points dV = 144 – 96x + 12x2 dx d2V = -96 +24x dx2 Substitute in x = 6 d2V = -96 +24*6 = 48 dx2 >0 Minimum Substitute in x = 2 d2V = -96 +24*2 = -48 dx2 <0 Maximum 4
• 12.
  Maximum at x= 2 Substitute in x = 2 to find maximum volume: V =144x – 48x2 + 4x3 V = 144*2 – 48*22 + 4*23 V = 128 cm3 NB don’t forget the units 5
• 13.
  Typical Example Afarmer wishes to fence a paddock using an existing wall as one side of the paddock. He has 100m of fencing and wants to know the dimensions of the paddock enclosing the maximum area. P = 100 A = xy y P = 2x + y A = x(100 − 2 x) x x 2 x + y = 100 = 100 x − 2 x 2 y = 100 − 2 x If area is to be a maximum ∴ we must differentiate and put it = 0 y = 100 − 2 x dA y = 100 − 2(25) = 100 − 4 x dx y = 50 100 − 4 x = 0 Dimensions are 25m by 50m 4 x = 100 x = 25 ∴ maximum area is 25 × 50 = 1250m 2 7
• 14.
  Example: Ex 18.3# 6 #6) Find the maximum area of a rectangle y with a perimeter of 120cm x Perimeter = 120cm A = xy x 2 x + 2 y = 120 A = x(60 − x) y x + y = 60 A = 60 x − x 2 y = 60 − x dA If A is to be a maximum ∴ =0 dx dA Question asked for the maximum = 60 − 2 x dx area ∴ A = x × y ∴ 60 − 2 x = 0 A = 30 × 30 2 x = 60 x = 30 = 900cm 2 ∴ y = 60 − x If we wish to prove this is a maximum, then we will y = 30 d2y differentiate again 2 = −2 as this is <0 ∴ this a maximum dx 6
• 15.
  Drawing the graphfor Ex 18.3 Question 6 1000 y A = 60 x − x 2 900 Area 800 Of 700 paddock 600 500 400 300 200 100 – 10 10 20 30 40 50 60 70 x X=side of paddock 11