# IA_2019s1_Tute8_sol.pdf - Mathematics IA Tutorial 8...

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Mathematics IA Tutorial 8 Semester 1, 2019 Algebra 1. Minimise the function f ( x, y ) = 3 x - y subject to the following condi- tions: x + y 1 x + y 4 - 2 x + 3 y 3 4 x + y 10 x, y 0 Solution: - 1 . 1 . 2 . 3 . - 1 . 1 . 2 . 3 . 0 The vertices are (1 , 0) , (0 , 1) , ( 5 2 , 0) , ( 9 5 , 11 5 ) and (2 , 2). The values of f on each of these are: f (1 , 0) = 3 , f (0 , 1) = - 1 , f ( 5 2 , 0) = 15 2 = 7 . 5 , f ( 9 5 , 11 5 ) = 16 5 = 3 . 2 , f (2 , 2) = 4 Therefore the minimum is - 1 at (0 , 1).

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2. Use the definition of the integral to evaluate Z 2 0 (2 - x ) 2 dx . Solution: We must partition [0 , 2] into n subintervals, so Δ x = 2 /n . It is useful to consider the graph of this function: - 4 - 3 - 2 - 1 1 2 3 4 5 6 7 - 3 - 2 - 1 1 2 3 4 5 6 0 Note in particular that the function is decreasing on the interval [0 , 2], so the minimum of each subinterval is at the right endpoint, and the maximum at the left endpoint. If we denote the i th subinterval by [ x i - 1 , x i ], then x i = 2 i/n . The minimum value on the i th subinterval is given by m i = (2 - x i ) 2 = (2 - 2 i/n ) 2 = 4 - 8 i/n + 4 i 2 /n 2 .
Then L n = n X i =1 m i Δ x = n X i =1 (4 - 8 i/n + 4 i 2 /n 2 )(2 /n ) = (8 /n ) n X i =1 1 - (16 /n 2 ) n X i =1 i + (8 /n 3 ) n X i =1 i 2 = 8 - (16 /n 2 ) n 2 + n 2 + (8 /n 3 ) n (2 n 2 + 3 n + 1) 6 = 8 - 8 - 8 n + 16 6 + 4 n + 4 3 n 2 = 8 3 - 4 n + 4 3 n 2 The maximum value on the i th subinterval is given by M i = (2 - x i - 1 ) 2 = (2 - 2( i - 1) /n ) 2 . Rather than expand this, we use the fact that n i =1 g ( i - 1) = g (0) + g (1) + . . . + g ( n - 1) = g (0) + n - 1 i =1 g ( i ), for any function g . Then U n = n X i =1 M i Δ x = n X i =1 (2 - 2( i - 1) /n ) 2 (2 /n ) = (2 /n ) n - 1 X i =0 (2 - 2 i/n ) 2 = (2 /n ) . 4 + (2 /n ) n - 1 X i =1 (4 - 8 i/n + 4 i 2 /n 2 ) = 8 /n + (8 /n ) n - 1 X i =1 1 - (16 /n 2 ) n - 1 X i =1 i + (8 /n 3 ) n - 1 X i =1 i 2 = 8 n + 8 - 8 n - 16 n 2 n 2 - n 2 + 8 n 3 n (2 n 2 - 3 n + 1) 6 = 8 - 8 + 8 n + 16 6 - 4 n + 4 3 n 2 = 8 3 + 4 n + 4 3 n 2 We then see that as n → ∞ , both U n and L n approach 16 6 , so R 2 0 (2 - x ) 2 dx = 8 3 .

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Practice Questions 1. (a) Write down the definition of a convex set. (b) Define a vertex of a convex set. Solution: (a) A convex set C is a set of points (or region) in R n such that for any two points in C , the line segment joining the two points lies completely in C . (b) A point P with position vector -→ OP = v is a vertex of the convex set C if v can not be written in the form (1 - t ) u + t w 0 t 1 for any (distinct) points u , w C except when v = u ( t = 0) or v = w ( t = 1). 2. Sketch the following sets in R 2 and determine their vertices. (a) { ( x 1 , x 2 ) | x 1 + 3 x 2 6 , 2 x 1 - x 2 4 , x 1 0 , x 2 0 } (b) { ( x 1 , x 2 ) | x 1 + x 2 2 , x 1 - x 2 4 , x 1 1 , x 2 0 } (c) { ( x 1 , x 2 ) | - x 1 + 4 x 2 5 , x 1 - 4 x 2 2 , x 1 + 2 x 2 2 } Solution: Graphs: (a) (b) (c)
Vertices: (a): (2,0), (6,0), (18/7,8/7) (b) (1,1), (2,0), (4,0) (c) (2,0), (-1/3,7/6) 3. Sketch the following sets in R 2 . Determine whether they are bounded or unbounded, and whether they are convex or not. In the case where the set is not convex, give an example of a line segment that violates the definition of convexity. (a) { ( x, y ) | x + y 1 , x - y 2 , x 0 } (b) { ( x, y ) | 1 ≤ | x | ≤ 2 , | y - 3 | ≤ 2 } (c) { ( x 1 , x 2 ) | x 2 1 + x 2 2 4 and x 1 + x 2 1 } (d) { ( x 1 , x 2 ) | x 2 1 + x 2 2 1 and x 1 + x 2 4 } Solution: (a) Unbounded, convex b) bounded, not convex.

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