Practical Problem-Solving Using Differentiation: Recognizing Maxima and Minima
Recognize optimization problems that can be solved with differentiation for finding maxima and minima. Learn to represent shapes algebraically, create formulas for volume and surface area, and apply differentiation to solve practical problems. Practice simplifying expressions and deriving equations for maximum values.
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Optimisation Problems
Optimisation problems BAT apply differentiation to solve optimization problems BAT manipulate algebra to form equations that can be differentiated KUS objectives Starter: Simplify these expressions (to polynomials) 1 2?2?3/2 4? 1 2?3 1 3?5 2?3 6?2 3? 1 2?3 2?3 1000 ? ?2 300 200? 1 60000 1 ? 1 ?3 200 ? ? 1000? 2 1000?2 20 ??2 ? ? ?? 20? 3 5 ?2 300 1 (2?)3 36 ? 8? 2400?2 9? 1
You need to be able to recognise practical problems that can be solved by using the idea of maxima and minima Whenever you see a question asking about the maximum value or minimum value of a quantity, you will most likely need to use differentiation at some point. Most questions will involve creating a formula, for example for Volume or Area, and then calculating the maximum value of it. WB45 : Represent the volume and surface area of the following shapes algebraically. Then, find the surface area in terms of x (red) and r (yellow). r h h V = 8 mm3 V = 20 m3 x x
WB46a 1) Try to make formulae using the information you have A large tank (shown) is to be made from 54m2 of sheet metal. It has no top. Show that the Volume of the tank will be given by: x x y = 2 V x y Formula for the Volume Formula for the Surface Area (no top) = + 2 2 3 SA x xy 2 3 = 3 18 V x x = + 2 54 2 3 x xy 2) Find a way to remove a constant, in this case y . We can rewrite the Surface Area formula in terms of y. 54 2 3 x xy = + 3) Substitute the SA formula into the Volume formula, to replace y. 2 x y = 54 2 = 2 2 3 x xy V 2 4 54 3 2 3 x x x x 2 54 2 3 x = V = y x 2 54 2 2 x = V x 3 x 2 3 = 3 18 V x x 2 4 54 2 x x = V 3 x
WB46b V V = 18x 2/3x3 A large tank (shown) is to be made from 54m2 of sheet metal. It has no top. Show that the Volume of the tank will be given by: x x y 2 3 = 3 18 V x x x = + 2 54 2 3 x xy b) Calculate the values of x that will give the largest volume possible, and what this Volume is. The graph above shows the formula for the volume V, in terms of x (that we just worked out!) So we need to calculate the value of x where the gradient is 0 differentiate! Before attempting this think about what we are doing: What does the graph look like?
WB46c A large tank (shown) is to be made from 54m2 of sheet metal. It has no top. Show that the Volume of the tank will be given by: x x y b) 2 3 2 3 = 3 18 V x x = 3 18 V x x Differentiate = + 2 dV dx 54 2 3 x xy = 18 2 2 x b) Calculate the values of x that will give the largest volume possible, and what this Volume is. Set equal to 0 18 2 = 2 0 x Rearrange = 2 18 x = 2x Solve 3 2 3 = 3 18 We want the volume V x x Sub the x value in 2 3 V = 3 18(3) (3) = 3 36 V m
WB47a A wire of length 2m is bent into the shape shown, made up of a Rectangle and a Semi-circle. y a) Find an expression for y in terms of x. c) Find the maximum possible Area x 2 x b) Show that the Area is: x A = (8 4 ) x x y 8 a) Find the length of the semi-circle, as this makes up part of the length. + x 2 = 2yx + Rearrange to get y alone 2 x = 2 2 x y 2 Divide by 2 x x = 1 y 2 4
y WB47b x 2 x b) Show that the Area is: x A = (8 4 ) x x y 8 b) Work out the areas of the Rectangle and Semi-circle separately. 2 x A= xy + Rectangle Semi Circle 8 Replace y xy 2 2 r x x 2 x A= + 2 1 x x 2 4 x 2 8 2 2 Expand 2 2 x x A= + x 2 x 2 4 8 2 4 2 2 x x A= x 2 x 2 8 Factorise 8 x = (8 4 ) A x x 8
y WB47c x 2 x b) Show that the Area is: x A = (8 4 ) x x y 8 c) Use the formula we have for the Area x = = + 4 + 4 4 4x x x = (8 4 ) A x x Factorise 8 ( = ) Expand 2 2 x x Divide by (4 + ) = A x 0.56 x 2 8 x Differentiate = 2 0.28 A m dA dx = 1 x 4 Set equal to 0 2 x = 1 0 x 8 2 Multiply by 8 8 8 4 4 x = = 0 0 x x x Divide by 2
WB48 A cuboid has a rectangular cross section where the length of the rectangle is equal to twice its width, x cm. The volume of the cuboid is 192 cubic cm a) Show that the total length, L cm, of the twelve edges of the cuboid is given by ? = 12? +384 ?2 b) Use calculus to justify that the value of L that you have found is a minimum ?????? = ? 2? ? = 192 Rearranges to ? =96 ?2 ? ? Length of all sides L = 4(? + 2? + ?) 2? L = 4 3? +96 ?2 384 ?2 L = 12? + QED
WB49The depth of water, y metres, in a tank after time t hours is given by ? =1 ?? ?? b) c) verify that y has stationary value when ? = 2 and show if it is a max or min d) find the rate of change of the depth of the water, in metres per hour, when ? = 1 hence, determine, with a reason, whether the depth of water is increasing or decreasing when ? = 1 8?4 6?2+ 20?,0 ? 4. Find: a) ?2? ??2 ?2? ??2=3 ?? ??= 1 2?3 12? + 20 2?2 12 ?) ?) ?? ??=1 ??2=3 ?) when ? = 2 2(2)3 12(2) + 20 = 0 so a stationary point 2(2)2 12 = 6 < 0 so a Maximum and ?2? ?) rate of change is the gradient when t = 1 ?? the rate of change is positive so the depth of water is increasing ??=17 2
Practice 1 1. Where x is the length of the container. Find the value of x which makes the surface area a maximum and hence find this maximum area The surface area A, of a container is given by the equation ? = 50? 4?2 2. A right circular cylinder of base radius r metres and height h metres is to be made so that the sum of its radius and height is 12m. Show that the volume of the cylinder is given by ? = 12??2 ??3 Find the maximum volume of the cylinder 3. An open topped metal tank of square base has volume 108?3 Give that the square base has sides of length ? ?, show that a) The height of the tank is given by 108 b) The external surface of the tank is ?2+432 c) Given also that the surface area of the tank is a minimum, find the value of ? and minimum surface area ?2 ? m2 4. A closed box in the shape of a cuboid has a rectangular base with dimensions ? ?? by 2? ?? The total surface area of the box is 300 cm2 Show that the volume of the box is given by ? = 100? 4?3 Find the dimensions of the box which make the volume a maximum 3
Practice 1 5. A closed right circular cylinder of base radius r cm and height h cm has a volume of 54 cm3. Show that S, the total surface area of the cylinder, is given by ? =108? + 2??2 ? Hence find the radius and height which makes the surface area a minimum 6. An open topped cylindrical cake tin is to have a volume of 1000? cm3 a) Find an expression for the height h, of the tin in terms of the radius of the base of the tin, r cm Show that the area of metal, A cm2, used in making the tin is given by ? = ??2+2000? b) ? c) Find the minimum value of A in terms of ? and justify your answer
Practice 1 SOLUTIONS 1. Where x is the length of the container. Find the value of x which makes the surface area a maximum and hence find this maximum area ?? ??= 0 ? ?? ? =50 The surface area A, of a container is given by the equation ? = 50? 4?2 ? = 50? 4?2 8= 6.25 ?? ??= 50 8? ? = 50 6.25 4 6.252= 156.25 2. A right circular cylinder of base radius r metres and height h metres is to be made so that the sum of its radius and height is 12m. Show that the volume of the cylinder is given by ? = 12??2 ??3 Find the maximum volume of the cylinder = 12 ? ?? ??= 0 ? ?? ? = 0 ,8 ? = ??212 ? = 12??2 ??3 ?? ??= 24?? 3??2 ? = 12?(8)2 ? 83= 256?
Practice 1 SOLUTIONS 3. An open topped metal tank of square base has volume 108?3 Give that the square base has sides of length ? ?, show that a) The height of the tank is given by 108 b) The external surface of the tank is ?2+432 c) Given also that the surface area of the tank is a minimum, find the value of ? and minimum surface area a) ? = ?2 = 108 ?2 So =108 ? m2 ?2 c) ??? ??= 2? 432? 2 b) Surface area = bottom +4 sides S? = ?2+ (4(?) ?? = ?2+432 108 ?2 ??? ??= 0 ? ?? ? = 3 432 2 = 6 ?? = 108 ? 4. A closed box in the shape of a cuboid has a rectangular base with dimensions ? ?? by 2? ?? The total surface area of the box is 300 cm2 Show that the volume of the box is given by ? = 100? 4?3 Find the dimensions of the box which make the volume a maximum 3 3? = 100? 4?3 50? 1 2 S? = 4?2+ 6? = 300 ? = ? 2? 3 So =300 4?2 = 50? 1 2 ?? ??= 100 4?2 =0 when ? = 5 3? 6? So dimensions ? = 5, 2? = 10, =20 3= 6.67
Practice 1 SOLUTIONS 5. A closed right circular cylinder of base radius r cm and height h cm has a volume of 54 cm3. Show that S, the total surface area of the cylinder, is given by ? =108? ? Hence find the radius and height which makes the surface area a minimum + 2??2 ??? ?? = 4?? 108? 2 54 ??2 ? = ??2 = 54 ?? = 2??2+ 2?? ??? ?? 3108 = 0 ? ?? ? = = 3 4 54 ??2 = ?? = 2??2+108 54 ??2= 1.91 ? = 6. An open topped cylindrical cake tin is to have a volume of 1000? cm3 a) Find an expression for the height h, of the tin in terms of the radius of the base of the tin, r cm Show that the area of metal, A cm2, used in making the tin is given by ? = ??2+2000? b) ? c) Find the minimum value of A in terms of ? and justify your answer ?? ??= 2?? 2000?? 2 1000 ?2 ? = ??2+ 2?? a) ? = ??2 = 1000? ??? ?? 32000? 2? = 0 ? ?? ? = = 10 =1000 ?2 ? = ??2+2000? ? = ?(10)2+2000? = 300? ? 10
KUS objectives BAT apply differentiation to solve optimization problems BAT manipulate algebra to form equations that can be differentiated self-assess One thing learned is One thing to improve is
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