# G3 Using Graphs to Solve Activities Equations

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G3 Using Graphs to Solve Activities Equations

Mathematics SKE, Strand G UNIT G3 Using Graphs to Solve Equations: Activities G3 Using Graphs to Solve Equations Activities Activities G3.1 Maximum Volume of an Open-Topped Box G3.2 Dipstick Problem Notes and Solutions (2 pages) © CIMT, Plymouth University UNIT G3 Using Graphs to Solve Equations: Activities Mathematics SKE, Strand G ACTIVITY G3.1 Maximum Volume of an Open-Topped Box We want to make cardboard trays for displaying home-made goods at a local bazaar. The sheets of card available are of size 10 cm × 12 cm 10 cm If we cut out a one cm 2 square from each corner and fold the sides up, we have a tray of volume 8 × 10 × 1 = 80 cm 3 1 cm Can you form a tray with a larger volume? 12 cm 1. Construct trays by cutting out squares of side 2 cm, 3 cm, etc. For each tray, determine the volume. Which tray has maximum volume? x We can find an even larger volume by considering cutting a square of side x cm from the card, as shown opposite. 2. x What is the length and breadth of the base of the tray? Show that the volume is given by V = 4 x 3 − 44 x 2 + 120 x 3. Draw a graph of V against x, plotting V for x = 0. 0.5, 1, 1.5, ..., 5 x x Estimate the value of x which gives the maximum value. We can generalise these results further by considering the same problem of finding the tray of maximum volume that can be cut from a square of side a cm. 4. a Show that the volume of the tray is given by V = 4 x 3 − 4a x 2 + a 2 x x We can non-dimensionalise the problem by taking a = 1 . a x x x 5. With a = 1 , plot a graph of V against x for x = 0, 0.1, 0.2, ..., 0.5 What value of x gives a maximum value for V ? © CIMT, Plymouth University UNIT G3 Using Graphs to Solve Equations: Activities Mathematics SKE, Strand G ACTIVITY G3.2 Dipstick Problem Service stations very rarely run out of fuel. This is due partly to efficient deliveries but also to precise stock control. Each type of fuel (petrol and diesel) is stored in an underground tank and the amount in each tank is carefully monitored using some form of dipstick. It is easy to measure the height, say h, of fuel in the tank. However, the volume will be proportional to the cross-sectional area – not the height. Suppose the cross-section is a circle (it is, in fact, usually elliptical, but a circle is a good approximation). You need to find the relationship between area, A, and height, h, and so provide a ready reckoner to convert height to area. For simplicity, take r = 1 m . For values of h from 0 to 1, you need to find the angle θ and the area of fuel. O θ r Problem 1 Show that cos θ = 1 − h . Problem 2 Show that the area of the sector OAB is given by πθ 180 Problem 3 Show that the area of the triangle OAB is r−h B A h (1 − h) sin θ Problem 4 Deduce the area of the cross-section of fuel and express this as a fraction, A′ , of the complete cross-sectional area of the tank. Problem 5 (a) (b) Problem 6 Problem 7 Using the equation in Problem 1, find the value of θ for each value of h in the table opposite. Use the formula deduced in Problem 4 to find the area fractions. h θ° 0 0 Area fraction 0 0.1 25.84 0.019 0.2 ... ... Plot a graph of A′ (vertical axis) against height h (horizontal axis). ... ... ... ... ... ... Use your graph to estimate the height that corresponds to an area fraction of 1.0 90.00 0.500 (a) 0.05 (b) 0.10 Extension Construct a dipstick for this problem, from which you can read off the area fraction against every height value. © CIMT, Plymouth University UNIT G3 Using Graphs to Solve Equations: Activities Mathematics SKE, Strand G ACTIVITIES G3.1 and G3.2 Notes and Solutions Notes and solutions given only where appropriate. G3.1 One of the best ways to approach this problem is to set it as a practical exercise for group work. 1. x 1 2 V 80 2. 3 96 72 4 5 32 0 x = 2 gives maximum volume. l = 12 − 2 x, b = 10 − 2 x and V = x (12 − 2 x ) (10 − 2 x ) ( = x 120 − 44 x + 4 x 2 ) = 120 x − 44 x 2 + 4 x 3 3. 4. The graph should give a value of x about 1.8. V = x (a − 2 x ) (a − 2 x ) ( = x a 2 − 4a x + 4 x 2 ) = a 2 x − 4a x 2 + 4 x 3 5. With a = 1, V = x − 4 x 2 + 4 x 3 and x ≈ 0.17 gives the maximum value of V. G3.2 1. From the triangle, cos θ = 1− h =1− h 1 2θ ⎞ θπ 2. Cross-sectional area = ⎛ × π12 = ⎝ 360 ⎠ 180 1 3. Area of triangle = 2 × ⎛ × 1 × (1 − h) × sin θ ⎞ = (1 − h) sin θ ⎝2 ⎠ 4. Area of fuel cross-section = Area fraction = © CIMT, Plymouth University θπ − (1 − h) sin θ 360 θ (1 − h) sin θ − 360 π Mathematics SKE, Strand G ACTIVITIES 5. UNIT G3 Using Graphs to Solve Equations: Activities G3.1 and G3.2 h θ° Area fraction 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0 25.84 36.87 45.57 53.13 60.00 66.42 72.54 78.46 84.26 90.00 0 0.019 0.052 0.094 0.142 0.196 0.252 0.312 0.374 0.436 0.500 Notes and Solutions 6. 0.5 0.4 A′ 0.3 0.2 0.1 0 0.2 0.4 0.6 0.8 1.0 h 7. (a) Approximately 0.2 (b) Approximately 0.32 Extension This is the lower half of the dipstick Height © CIMT, Plymouth University Area