Consider the following for the next items that follow: Given that 4x2 + y2 = 9.
What is the maximum value of y?
3
Finding Maximum Value of y in an Ellipse Equation
The given equation is \(4x^2 + y^2 = 9\). We are asked to find the maximum value that the variable \(y\) can take.
This equation represents an ellipse centered at the origin. To understand the maximum possible values for \(x\) and \(y\), we can analyze the equation. The term \(4x^2\) is always greater than or equal to zero (\(4x^2 \ge 0\)), and similarly, the term \(y^2\) is always greater than or equal to zero (\(y^2 \ge 0\)).
The equation \(4x^2 + y^2 = 9\) tells us that the sum of \(4x^2\) and \(y^2\) is always equal to 9.
Calculating the Maximum Value of y
To find the maximum value of \(y\), we need to consider the condition under which \(y^2\) is maximized. Since \(4x^2\) is always non-negative, \(y^2\) will be maximized when \(4x^2\) is minimized. The minimum possible value for \(4x^2\) is 0, which occurs when \(x = 0\).
Let's substitute \(x = 0\) into the equation:
\(\qquad 4(0)^2 + y^2 = 9\)
\(\qquad 0 + y^2 = 9\)
\(\qquad y^2 = 9\)
To find the values of \(y\), we take the square root of both sides:
\(\qquad y = \pm\sqrt{9}\)
\(\qquad y = \pm 3\)
This gives us two possible values for \(y\) when \(x=0\): \(y = 3\) and \(y = -3\).
The values of \(y\) can range from -3 to 3. The maximum value among these is 3.
Alternatively, we can rearrange the equation to express \(y^2\) in terms of \(x^2\):
\(\qquad y^2 = 9 - 4x^2\)
For \(y\) to be a real number, \(y^2\) must be non-negative. Also, since \(4x^2 \ge 0\), the term \(9 - 4x^2\) will be largest when \(4x^2\) is smallest. The smallest value of \(4x^2\) is 0 (when \(x=0\)).
Maximum value of \(y^2 = 9 - 0 = 9\).
So, \(y^2 \le 9\). Taking the square root, we get \(|y| \le 3\). This means \(-3 \le y \le 3\).
The maximum value \(y\) can attain is 3.
Analyzing the Options for Maximum y
Let's look at the given options:
- Option 1: \(\frac{3}{2}\)
- Option 2: 3
- Option 3: 4
- Option 4: 6
Based on our calculation, the maximum value of \(y\) is 3. This matches Option 2.
| Condition | Equation | Result for y | Notes |
|---|---|---|---|
| To Maximize y | Set \(x=0\) | \(y^2 = 9 \implies y = \pm 3\) | Maximum y is 3 |
| To Maximize x | Set \(y=0\) | \(4x^2 = 9 \implies x^2 = \frac{9}{4} \implies x = \pm \frac{3}{2}\) | Maximum x is \(\frac{3}{2}\) |
The analysis confirms that the range of \(y\) is \([-3, 3]\), making the maximum value 3.
Revision Table: Understanding Maximum Values
| Concept | Explanation | Relation to \(4x^2 + y^2 = 9\) |
|---|---|---|
| Maximizing a Variable | To maximize one variable in an equation relating positive squared terms, minimize the other squared terms. | To maximize \(y\), minimize \(4x^2\) by setting \(x=0\). |
| Equation of Ellipse | An equation of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) represents an ellipse. Max \(x\) is \(\pm a\), max \(y\) is \(\pm b\). | \(4x^2 + y^2 = 9 \implies \frac{x^2}{9/4} + \frac{y^2}{9} = 1 \implies \frac{x^2}{(3/2)^2} + \frac{y^2}{3^2} = 1\). Here \(a=3/2\), \(b=3\). Max \(y\) is 3. |
Additional Information: Ellipse Geometry
The equation \(4x^2 + y^2 = 9\) describes an ellipse centered at the origin (0,0).
- The points where the ellipse crosses the x-axis occur when \(y=0\). \(4x^2 + 0^2 = 9 \implies 4x^2 = 9 \implies x^2 = \frac{9}{4} \implies x = \pm \frac{3}{2}\). These points are \((\frac{3}{2}, 0)\) and \((-\frac{3}{2}, 0)\). The semi-major or semi-minor axis along the x-axis has length \(\frac{3}{2}\).
- The points where the ellipse crosses the y-axis occur when \(x=0\). \(4(0)^2 + y^2 = 9 \implies y^2 = 9 \implies y = \pm 3\). These points are \((0, 3)\) and \((0, -3)\). The semi-major or semi-minor axis along the y-axis has length 3.
Since the length along the y-axis (3) is greater than the length along the x-axis (\(\frac{3}{2}\)), the major axis is along the y-axis. The vertices of the ellipse are at \((0, \pm 3)\) and the co-vertices are at \((\pm \frac{3}{2}, 0)\).
The maximum value of \(y\) is the positive y-intercept, which is 3. The minimum value of \(y\) is the negative y-intercept, which is -3.
Similarly, the maximum value of \(x\) is the positive x-intercept, which is \(\frac{3}{2}\). The minimum value of \(x\) is the negative x-intercept, which is \(-\frac{3}{2}\).
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