Solutions to Tutorial Sheet 10 Topics: Applications: Optimization + Modelling

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1 The University of Sydney School of Mathematics and Statistics Solutions to Tutorial Sheet 10 Topics: Applications: Optimization + Modelling MATH1111: Introduction to Calculus Semester 1, 2012 Web Page: Lecturer: Clio Cresswell 1. Let F() = (a) Find all critical points of F(). (b) Classify the critical points of F() as local maima, local minima or neither. (c) What are the maimum and minimum values of F() on the interval ? (a) F () = , and F () = 0 when = 1 or = 2, hence the critical points of F() occur at = 1 and = 2. (b) F () = 12 18, and F (1) = 6 < 0, F (2) = 6 > 0. Hence the critical point at = 1 is a local maimum, and the critical point at = 2 is a local minimum. (c) We calculate the function values of the endpoints, as well as the critical points lying in the given interval. We have F( 1.5) = 41, F(1.5) = 8.5 and F(1) = 9. Hence the maimum value the function obtains on this interval is 9 at = 1, and the minimum value is 41 at = For each of the following functions, find the global maimum and global minimum on the given interval. (a) y = for 4 0. (b) f(θ) = θ+sinθ for 1 θ 4. (c) g(t) = tlnt for 2 t 3. (d) H() = e for 0 5. (a) y = 2 3, and y = 0 when = 3 2. Now y( 4) = 1, y( 3 2 ) = 21 4, and y(0) = 3, so the global minimum is 1 occuring at = 4 and the global maimum is 21 4 occuring at = 3 2. (b) f (θ) = 1 + cosθ, and f (θ) = 0 when θ = π in this interval. Now f (θ) = sinθ, and f (π) = 0, so the critical point at θ = π is neither a local maimum nor a local minimum, but f(1) = 1.84, f(π) = π and f(4) = 3.24, so the global minimum is 1.84 occuring at θ = 1 and the global maimum is 3.24 occuring at θ = 4. (c) g (t) = 1+lnt, and g (t) = 0 when lnt = 1, i.e. when t = e which is outside out interval. Now g(2) = 1.39 and g(3) = 3.30, so the global minimum is 1.39 occuring at t = 2 and the global maimum is 3.30 occuring at t = 3. (d) H () = (1 )e, and H () = 0 when = 1. Now H(0) = 0, H(1) = e and H(5) = 5e , hence the global maimum is occuring at = 1 and the global minimum is 0 occuring at = 0. Copyright c 2012 The University of Sydney 1

2 3. The population of termites in a termite mound is modelled by the equation where t is measured in days. f(t) = t, (a) What is the initial population of the termites? (b) How many days will it take for the population to reach 9000? (c) Find f (t) and f (t). (d) What happens to the growth rate of the population as t? What happens to the population? (a) (b) This occurs for t so that 9000 = , i.e. when = 1000, which occurs when 1+t 1+t t = 8.9 days. (c) f(t) = (1+t) 1, f (t) = 9900(1+t) 2, f (t) = 19800(1+t) 3. (d) As t, f (t) 0, and f(t) A principal at a primary school wants to fence off an area of a large playground for a vegetable garden, using an eisting wall as one side: y She has 20m of fencing available. (a) What are the values of and y that maimize the area of the garden? (b) What is the maimum area of the garden? (a) We have 2 +y = 20 and A = y = (20 2) = Now da d da = 0 when = 5. Hence = 5 and y = 10. d (b) 50 m 2. = 200 4, and 5. An open rectangular bo with square base is to be made from 48 cm 2 of material. What dimensions will result in a bo with the largest possible volume? Letbethelengthofanedgeofthebaseandy betheheightofthebo. Considering the surface area we have 48 = 2 +4y and so y = 12 (1/4). Let V be the volume. Then V = 2 y = 12 (1/4) 3 and V = 12 (3/4) 2 where V = 0 = ±4. V = (3/2) and so = 4 gives V 0 and is therefore where V is maimised. So the dimensions will be cm. 2

3 6. A television manufacturing firm needs to design an open-topped bo with a square base. The bo must hold 32 cm 3. Find the dimensions of the bo that can be built with the minimum amount of materials. The volume V is given as V = base height. With square base of length and height y we have V = 2 y = 32 or y = 32/ 2 with the amount of materials M given by M = 2 +4y = /. M = = 0 = 4 where M 0 and y = 2. So the dimensions of the bo that will minimise materials are cm. 7. A sheet of metal 3m by 4m will be made into a bo by cutting equal-sized squares from each corner and folding up the four edges. What will be the dimensions of the bo with the largest volume? Let be the length of the side of the square cut from each corner. The volume, V, of the bo is given by V = (4 2)(3 2) where V = and V = 0 = 7± ,1.77. Note that since cutting 2 from both sides of the sheet of metal But in any case V = and at = 0.57, V 0 (while for = 1.77 V 0). So the bo with the largest volume will have dimensions m. 8. A fence must be built in a large field to enclose a rectangular area of 15,625m 2. One side of the area is bounded by an eisting fence; no fence is needed there. Material for the fence costs $2 per metre for the two ends, and $4 per metre for the side opposite the eisting fence. Find the cost, C, of the least epensive fence. C = y and y = So y = 15625/ and C = C = where C = 0 = ±125. C = and so = 125 gives C 0 and local minimum where C = A container in theshapeof a circular cylinder with no top has surface area 3πcm 2. What height, h, and base radius, r, will maimise the volume, V, of the cylinder? Let S be the surface area. Then S = πr 2 3π πr2 + (2πr)h = 3π or h =. 2πr Now V = πr 2 h = (3/2)πr (1/2)πr 3 and V = (3/2)π (3/2)πr 2 where V = 0 r = ±1. V = 3πr and so r = 1 V 0 and so therefore r = 1 maimises V where h = A young man is at a point on a river bank. He wants to get to his cabin that is on the other side of the river and is 8km west and 3km north of where he currently stands. The river runs east to west. He can travel at 4km/hr on the river but jog at 5km/hr on the land. How far up the river bank should he go before crossing in order to reach his cabin in minimum time? Let 8 be the distance along the river bank to be travelled and T be the total time taken to get to his cabin. Then T = distance/speed = and 5 4 T = where T = = 5 2 = 16 or = ±4. Note that 0 8 and that T(0) = 2.35, T(4) = 2.05 and T(8) = So the young man should travel 4km up the river bank before crossing in order to minimise his travel time. 11. A small furniture business signs a contract with a customer to deliver up to 400 chairs on a regular basis, with the eact number varying from week to week. The price is set at $90 per chair for the first 300 chairs, and every chair purchased above 300 reduces the price of every chair in the order by $

4 (a) Suppose the order is between 0 and 300 chairs. What is the minimum and maimum amount the customer must pay for their order? (b) Now suppose the order is between 300 and 400 chairs. Write an equation epressing the cost, C, of an order of n chairs as a function of n. (c) What is the minimum and maimum amount the customer must pay for their order if they order between 300 and 400 chairs? (d) Has the business signed a bad contract? Why or why not? (a) The customer pays $0 for an order of 0 chairs, and $27000 for an order of 300 chairs. (b) C = n( (n 300)) = 165n 0.25n 2. dc dc (c) = n, and = 0 when n = 330. Now C(300) = 27000, C(330) = 27225, and dn dn C(400) = 26000, hence the customer pays the most for 330 chairs and the least for 400 chairs in this interval. (d) Any order of chairs over 330 is going to become progressively cheaper, so the customer is getting more chairs for less money. This is almost certainly a bad contract for the business to have signed. 12. (*) The efficiency of a screw, E, is given by E = θ µθ2 µ+θ, where θ > 0 is the angle of pitch of the thread, and µ > 0 is the coefficient of friction of the material. What value of θ maimises E? de dθ = (µ+θ)(1 2µθ) (θ µθ2 )(1) (µ+θ) 2 = µ(θ2 +2µθ 1) (µ+θ) 2, and de dθ = 0 when θ2 +2µθ 1 = 0. Using the quadratic formula, θ = 2µ± 4µ and since θ > 0 we have θ = µ+ µ = µ± µ 2 +1, 13. (*) For some positive constant C, a patient s tempterature change, T, after being given a D milligram dose of a particular drug, is given by ( C T = 2 D ) D 2. 3 (a) What dosage maimises this temperature change? (b) The sensitivity of the patient to the drug is defined as dt. What dosage maimises dd sensitivity? (a) dt dd = DC D2, and dt dd = 0 when D = C. Now d2 T dd 2 = C 2D, and d2 T dd2(c) = C < 0, so the critical point is a maimum. 4

5 (b) To maimise dt dd, we must find the critical points of d2 T dd2, which occur when C 2D = 0, i.e. D = C 2. To check this is a maimum, we have d3 T = 2 < 0. dd3 14. (*) A boy pulls a billy cart which, together with its load, weighs m kg. The force he eerts in pulling this cart varies with the angle, θ, that his arm makes with his body. The least force he must eert to move the cart is modelled by the equation F = mgµ sinθ+µcosθ, where µ is the coefficient of friction, and g is the force due to gravity. If µ = 0.2 and g = 9.8, find the maimum and minimum values of F for 0 θ π 2 (give your answer in terms of m). We have F = 1.96m(sinθ+0.2cosθ) 1, and df dθ = 1.96m(sinθ +0.2cosθ) 2 (cosθ 0.2sinθ). Now df dθ = 0 when cosθ 0.2sinθ = 0, i.e. when tanθ = 5. In the interval we are considering, the one solution to this equation is θ = Now F(0) = 9.8m, F(1.37) = 1.92m and F( π 2 ) = 1.96m, hence the maimum value is 9.8m occuring at θ = 0 and the minimum value is 1.92m occuring at θ = The questions labelled with an asterisk are drawn from Calculus: Single and Variable, by Deborah Hughes-Hallett, Andrew M. Gleason, William G. McCallum et al., John Wiley & Sons, 5th ed. 5