A new game show on TV has 100 boxes numbered 1, 2, . . . , 100 in a row, each containing a mystery prize. The prizes are items of different types, a, b, c, . . . , in decreasing order of value. The most expensive item is of type a, a diamond ring, and there is exactly one of these. You are told that the number of items at least doubles as you move to the next type. For example, there would be at least twice as many items of type b as of type a, at least twice as many items of type c as of type b and so on. There is no particular order in which the prizes are placed in the boxes.
Q1: What is the minimum possible number of different types of prizes?
Solution:
Ans: 2 It is given that the most expensive item is a diamond ring of type a and there is exactly one of these. Since the item b should be at least twice. The minimum number of items will be obtained when a = 1 and b = 99, which means there are only two different types of items.
Q2: What is the maximum possible number of different types of prizes?
Solution:
Ans: 6 It is given that the most expensive item is a diamond ring of type a and there is exactly one of these. Since the number of items of type b should be at least twice of that of a and the number of items of type c should be at least twice of that of b and so on. So the maximum number of different types of items of a, b, and c will be obtained when a = 1, b = 2, c = 4, d = 8, e = 16, f = 69. Hence the maximum number of different types of items will be 6. If the number of items is 7, then the minimum number of prizes should be 1 + 2 + 4 + 8 + 16 + 32 + 64 = 127 which is more than 100. Hence 6 is the answer.
Q3: Which of the following is not possible? (a) There are exactly 75 items of type e. (b) There are exactly 30 items of type b. (c) There are exactly 45 items of type c. (d) There are exactly 60 items of type d.
Solution:
Ans: (c) Option (a): There are exactly 75 items of type e. a = 1, b = 2, c = 4, d = 8, e = 85. Here the maximum value of e = 85. Hence it can take the value 75. An example of such case is a = 1, b = 2, c = 4, d = 18, e = 75 Option (b): There are exactly 30 items of type b. a = 1 b = 30 and c = 69. Hence this case is also possible. Option (c): There are exactly 45 items of type c. Since the value of d should be at least 90, it means that d is not present because 45 + 90 will be more than 100 (maximum number of items). Only a, b and c are present. The maximum value of b = 22 and a =1, but 45 + 22 + 1 = 68, which is less than 100. So this case is not possible. Option (d): There are exactly 60 items of type d. d = 60, c = 30, b = 9 and a = 1. a + b + c + d = 100. Hence this case is possible. C is the answer.
Q4: You ask for the type of item in box 45. Instead of being given a direct answer, you are told that there are 31 items of the same type as box 45 in boxes 1 to 44 and 43 items of the same type as box 45 in boxes 46 to 100. What is the maximum possible number of different types of items? (a) 5 (b) 6 (c) 4 (d) 3
Solution:
Ans: (a) The total number of items from 1 to 100, which are of same type as in box 45 = 31 + 1 + 43 = 75 Now to maximize the number of items, a = 1, b = 2, c = 4, d = 18 and e = 75(given) There can be maximum 5 types of items. If we consider number of items to be 6, then minimum number of items of 5th type will be 16, 1 + 2 + 4 + 8 + 16 + 75 = 106 which is more than 100.
Ans. Maxima and minima refer to the highest and lowest points on a function, respectively. A maximum point is a point where the function value is greater than the values at nearby points, while a minimum point is where the function value is less than those nearby points. These concepts are essential in optimization problems, where one seeks to find the best solution under given constraints.
2. How do you determine if a critical point is a maximum or minimum?
Ans. To determine whether a critical point is a maximum or minimum, you can use the second derivative test. If the second derivative of the function at that point is positive, the function is concave up, indicating a local minimum. If the second derivative is negative, the function is concave down, indicating a local maximum. If the second derivative is zero, the test is inconclusive, and you may need to use other methods.
3. What is the importance of the first derivative in finding maxima and minima?
Ans. The first derivative of a function provides information about the slope of the function. By finding where the first derivative is equal to zero, you can identify critical points, which are potential locations for maxima and minima. Analyzing the sign of the first derivative before and after these critical points helps determine whether the function is increasing or decreasing, thus identifying whether the critical point is a maximum or minimum.
4. Can you explain the concept of global maxima and minima?
Ans. Global maxima and minima refer to the absolute highest and lowest points of a function over its entire domain. A function may have multiple local maxima and minima, but the global maximum is the highest value of the function across all points, while the global minimum is the lowest. Finding global extrema can be more complex, especially for functions defined over a closed interval or those involving constraints.
5. What role do constraints play in optimization problems involving maxima and minima?
Ans. Constraints limit the possible solutions in optimization problems. When a function is maximized or minimized subject to constraints, methods such as Lagrange multipliers can be used. This technique allows you to find the extrema of a function while satisfying the given constraints, ensuring that the solution is feasible within the defined boundaries of the problem.
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