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Re: What is the difference between themaximumand theminimumvalue of
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16 Feb 2024, 15:22
\((x-2)^2 = 9\), so \(x-2 = ±3\) ---> x = -1 and 5
\((y-3)^2 = 25\), so \(y-3 = ±5\) ---> y = -2 and 8
You could calculate the values for every variation of \(\frac{x}{y}\), which isn't a bad strategy since there's only 4:
x = -1, y = -2: \(\frac{-1}{-2} = \frac{1}{2} = \frac{4}{8}\)
x = -1, y = 8: \(\frac{-1}{8}\)
x = 5, y = -2: \(\frac{5}{-2} = \frac{-20}{8}\)
x = 5, y = 8: \(\frac{5}{8}\)
... but a way to save time is to consider what would result in the minimums and maximums. Since there is a negative value for both x and y, the minimum value of \(\frac{x}{y}\) is the one that is MOST negative, i.e., using the value of x with the greatest absolute value to maximize the numerator. Vice versa for maximum value of x/y.
Either way, we find that the minimum value of x/y is \(\frac{-20}{8}\) and the maximum value of x/y is \(\frac{5}{8}\). The difference between them is 25/8.