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25 Jul 2010, 18:51
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If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy.
A. 40 and -30, respectively
B. 40 and -56, respectively
C. 42 and -30, respectively
D. 42 and -56, respectively
E. 49 and -56, respectively
A. 40 and -30, respectively
B. 40 and -56, respectively
C. 42 and -30, respectively
D. 42 and -56, respectively
E. 49 and -56, respectively
In equalities how to handle an expression which is squared
Does the above equation
(x+1)^2 <= 36 mean |x+1| < (+6 or -6)
I then get 4 equations.. and I am am not able to proceed. Can you someone please explain how such questions are to be handled.
Does the above equation
(x+1)^2 <= 36 mean |x+1| < (+6 or -6)
I then get 4 equations.. and I am am not able to proceed. Can you someone please explain how such questions are to be handled.
12 Aug 2010, 13:01
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If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy.
\((x+1)^2\leq{36}\) --> \({-\sqrt{36}}\leq{x+1}\leq{\sqrt{36}}\) --> \({-6}\leq{x+1}\leq{6}\) --> \({-7}\leq{x}\leq{5}\).
\((y-1)^2<{64}\) --> \({-\sqrt{64}}<{y-1}<{\sqrt{64}}\) --> \({-8}<{y-1}<{8}\) --> \({-7}<{y}<{9}\), as \(y\) is an integer we can rewrite this inequality as \({-6}\leq{y}\leq{8}\).
We should try extreme values of \(x\) and \(y\) to obtain min and max values of \(xy\):
Min possible value of \(xy\) is for \(x=-7\) and \(y=8\) --> \(xy=-56\);
Max possible value of \(xy\) is for \(x=-7\) and \(y=-6\) --> \(xy=42\).
Solving with absolute values gives the same results:
\((x+1)^2\leq{36}\) means \(|x+1|\leq{6}\) --> \({-7}\leq{x}\leq{5}\).
\((y-1)^2<{64}\) means \(|y-1|<{8}\) --> \({-7}<{y}<{9}\).
Hope it's clear.
General Discussion
One thing I notice is that you have to be careful with the exclusivity and inclusivity of ranges. In these questions, you will get it wrong if you thought y = 9 is included in the range.
