If 3 ≤ x ≤ 10 and 5 ≤ y ≤ 15 , then maximum value of \(\left(\frac{x}{y}\right)\) is-
To determine the maximum value of the expression \(\left(\frac{x}{y}\right)\) given the ranges for \(x\) and \(y\), we need to consider the properties of fractions. For a fraction to achieve its largest possible value, its numerator (the top part) should be as large as possible, and its denominator (the bottom part) should be as small as possible.
The problem provides specific ranges for the variables \(x\) and \(y\):
To maximize the fraction \(\left(\frac{x}{y}\right)\):
Now, we substitute these identified optimal values into the expression \(\left(\frac{x}{y}\right)\):
Maximum value of \(\left(\frac{x}{y}\right)\) = \(\frac{\text{Maximum value of } x}{\text{Minimum value of } y}\)
Substituting the specific values:
Maximum value = \(\frac{10}{5}\)
Maximum value = \(2\)
Therefore, the maximum value of the expression \(\left(\frac{x}{y}\right)\), based on the given ranges for \(x\) and \(y\), is 2.
| Variable | Given Range | Value Chosen for Maximum \(\left(\frac{x}{y}\right)\) |
|---|---|---|
| \(x\) | \(3 \le x \le 10\) | \(10\) (Maximum of its range) |
| \(y\) | \(5 \le y \le 15\) | \(5\) (Minimum of its range) |
This method ensures we obtain the largest possible ratio by maximizing the numerator and minimizing the denominator within their respective constraints.
Four small squares of side x are cut out of a square of side 12 cm to make a tray by folding the edges. What is the value of x so that the tray has the maximum volume?
If N is a four digit number formed by digits x 1, x 2, x 3and x 4, then maximum value of \(\frac{N}{x_{1}+x_{2}+x_{3}+x_{4}}\) is-
A wire of length 20 cm is to be bent into a rectangle. Which of the following statements is/are correct?
I. The rectangle of the largest area is the square.
II. It is possible to form a rectangle of an area of $27 \, \text{cm}^2$.
Select the answer using the code given below.
Consider the following statements :
Statement-I :
The function $f(x) = \frac{x^3 + 128}{x}$ has a minimum value 48 at $x = 4$.
Statement-II :
As $x$ increases through 4, $f'(x)$ changes sign from positive to negative.
Which one of the following is correct in respect of the above statements?