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FROM Veritas Prep Blog: MaxMin Strategies: Focus on Extremes 
In the last two weeks, we discussed some max min strategies. Today, let’s look at another maxmin question in which we apply the strategy of focusing on the extremes. The largest or the smallest values are often found at the extremes of a given range. Question: If x and y are integers such that (x+1)^2 is less than or equal to 36 and (y1)^2 is less than 64, what is the sum of the maximum possible value of xy and the minimum possible value of xy? (A) 16 (B) 14 (C) 0 (D) 14 (E) 16 Solution: To get the sum of the maximum and minimum possible values of xy, we need to know the maximum and minimum values of xy. For those, we need to find the values that x and y can take. So first, we should review the information given: (x + 1)^2 <= 36 (y – 1)^2 < 64 We need to find the values that x and y can take. There are many ways of doing that. We can solve the inequality using the wave method discussed in this post or using the concept of absolute values. Let’s discuss both the methods. Wave method to solve inequalities: Solve for x: (x + 1)^2 <= 36 (x + 1)^2 – 6^2 <= 0 (x + 1 + 6)(x + 1 – 6) <= 0 (x + 7)(x – 5) <= 0 7 <= x <= 5 (Using the wave method) Solve for y: (y – 1)^2 < 64 (y – 1)^2 – 8^2 < 0 (y – 1 + 8)(y – 1 – 8) < 0 (y + 7)(y – 9) < 0 7 < y < 9 (Using the wave method) Or you can solve taking the square root on both sides Solve for x: (x + 1)^2 <= 36 x + 1 <= 6 6 <= x + 1 <= 6 (discussed in your Veritas Algebra book) 7 <= x <= 5 So x can take values: 7, 6, 5, 4, … 3, 4, 5 Solve for y: (y – 1)^2 < 64 y – 1 < 8 8 < y – 1 < 8 (discussed in your Veritas Algebra book) 7 < y < 9 So y can take values: 6, 5, 4, 3, … 6, 7, 8. Now that we have the values of x and y, we should try to find the minimum and maximum values of xy. Note that the values of xy can be positive as well as negative. The minimum value will be the negative value with largest absolute value (largest negative) and the maximum value will be the positive value with the largest absolute value. Minimum value – For the value to be negative, one and only one of x and y should be negative. Focus on the extreme values: if x is 7 and y is 8, we get xy = 56. This is the negative value with largest absolute value. Maximum value – For the value to be positive, both x and y should have the same signs. If x = 7 and y = 6, we get xy = 42. This is the largest positive value. The sum of the maximum value of xy and minimum value of xy is 56 + 42 = 14 Answer (B) Try to think of it in terms of a number line. x lies in the range 7 to 5 and y lies in the range 6 to 8. The range is linear so the end points give us the maximum/minimum values. Think of what happens when you plot a quadratic – the minimum/maximum could lie anywhere. Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! 

